Ellipse Questions in English

(D) The equation of the line is $x + \sqrt{3}y = 2\sqrt{3}$,which can be written as $y = -\frac{1}{\sqrt{3}}x + 2$. Here,the slope $m = -\frac{1}{\sqrt{3}}$.
We check the point $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ for each curve.
For option $D$: $x^{2} + 9y^{2} = 9$.
Substitute the point: $\left(\frac{3\sqrt{3}}{2}\right)^{2} + 9\left(\frac{1}{2}\right)^{2} = \frac{27}{4} + \frac{9}{4} = \frac{36}{4} = 9$. The point lies on the curve.
The equation of the tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at $(x_{1}, y_{1})$ is $\frac{xx_{1}}{a^{2}} + \frac{yy_{1}}{b^{2}} = 1$.
For $x^{2} + 9y^{2} = 9$,we have $\frac{x^{2}}{9} + y^{2} = 1$.
Tangent at $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ is $\frac{x \cdot \frac{3\sqrt{3}}{2}}{9} + y \cdot \frac{1}{2} = 1$.
$\frac{\sqrt{3}x}{6} + \frac{y}{2} = 1 \implies \sqrt{3}x + 3y = 6 \implies x + \sqrt{3}y = 2\sqrt{3}$.
This matches the given line.

(A) The given curves are $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ (an ellipse) and $x^{2} + y^{2} = \frac{31}{4}$ (a circle).
Let the slope of the common tangent be $m$.
The equation of a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $y = mx \pm \sqrt{a^{2}m^{2} + b^{2}}$.
Here $a^{2} = 9$ and $b^{2} = 4$,so the tangent is $y = mx \pm \sqrt{9m^{2} + 4}$.
The equation of a tangent to the circle $x^{2} + y^{2} = r^{2}$ is $y = mx \pm r\sqrt{1 + m^{2}}$.
Here $r^{2} = \frac{31}{4}$,so the tangent is $y = mx \pm \frac{\sqrt{31}}{2}\sqrt{1 + m^{2}}$.
For the lines to be identical,the constant terms must be equal:
$9m^{2} + 4 = \frac{31}{4}(1 + m^{2})$.
Multiplying by $4$,we get $36m^{2} + 16 = 31 + 31m^{2}$.
$5m^{2} = 15$.
$m^{2} = 3$.

(B) The equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$. Here $a^{2}=8$ and $b^{2}=4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{4}{8}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
The foci are $S(-ae, 0)$ and $S'(ae, 0)$,where $ae = \sqrt{8} \cdot \frac{1}{\sqrt{2}} = 2$. So,$S(-2, 0)$ and $S'(2, 0)$.
The line perpendicular to $x+2y=0$ has the form $2x-y+k=0$,or $y=2x+k$.
The condition for tangency $y=mx+c$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}+b^{2}$.
Here $m=2$,so $c^{2} = 8(2^{2}) + 4 = 32+4 = 36$,which gives $c = \pm 6$.
Since $P$ is in the second quadrant,the tangent must have a positive $y$-intercept,so $y=2x+6$.
The point of tangency $P(x_{1}, y_{1})$ is given by $(-\frac{a^{2}m}{c}, \frac{b^{2}}{c}) = (-\frac{8 \cdot 2}{6}, \frac{4}{6}) = (-\frac{8}{3}, \frac{2}{3})$.
The area $A$ of $\Delta SPS'$ with base $SS'=4$ and height $|y_{1}| = \frac{2}{3}$ is $A = \frac{1}{2} \cdot 4 \cdot \frac{2}{3} = \frac{4}{3}$.
Finally,$(5-e^{2}) \cdot A = (5 - \frac{1}{2}) \cdot \frac{4}{3} = \frac{9}{2} \cdot \frac{4}{3} = 6$.

(D) Given equation: $x^{2}+9 y^{2}-4 x+3=0$
Rearranging the terms: $(x^{2}-4 x)+(9 y^{2})+3=0$
Completing the square for $x$: $(x^{2}-4 x+4)+9 y^{2}+3-4=0$
$(x-2)^{2}+(3 y)^{2}=1$
This is the equation of an ellipse: $\frac{(x-2)^{2}}{1^{2}}+\frac{y^{2}}{(1/3)^{2}}=1$
For the ellipse $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$,the range of $x$ is $[h-a, h+a]$ and the range of $y$ is $[k-b, k+b]$.
Here,$h=2, a=1, k=0, b=1/3$.
Therefore,$x \in [2-1, 2+1] = [1, 3]$ and $y \in [0-1/3, 0+1/3] = [-\frac{1}{3}, \frac{1}{3}]$.
Thus,$x$ and $y$ lie in $[1, 3]$ and $[-\frac{1}{3}, \frac{1}{3}]$ respectively.

(C) The equation of the tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{(2a)^{2}}=1$ at point $(b \cos \theta, 2a \sin \theta)$ is given by $\frac{x \cos \theta}{b} + \frac{y \sin \theta}{2a} = 1$.
The $x$-intercept is $A = (\frac{b}{\cos \theta}, 0)$ and the $y$-intercept is $B = (0, \frac{2a}{\sin \theta})$.
The area of the triangle $\Delta OAB$ is given by $\text{Area} = \frac{1}{2} \times |x_{\text{intercept}}| \times |y_{\text{intercept}}| = \frac{1}{2} \times \frac{b}{\cos \theta} \times \frac{2a}{\sin \theta} = \frac{ab}{\sin \theta \cos \theta} = \frac{2ab}{\sin 2\theta}$.
Since the minimum value of $\sin 2\theta$ is $1$ (for $\theta = \frac{\pi}{4}$),the minimum area is $2ab$.
Given that the minimum area is $kab$,we have $kab = 2ab$,which implies $k = 2$.

(B) The given line is $12 x \cos \theta + 5 y \sin \theta = 60$.
Dividing by $60$,we get $\frac{12 x \cos \theta}{60} + \frac{5 y \sin \theta}{60} = 1$,which simplifies to $\frac{x \cos \theta}{5} + \frac{y \sin \theta}{12} = 1$.
Comparing this with the tangent equation of an ellipse $\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$,we identify $a = 5$ and $b = 12$.
The equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,which is $\frac{x^{2}}{25} + \frac{y^{2}}{144} = 1$.
Multiplying by $3600$,we get $144 x^{2} + 25 y^{2} = 3600$.

(C) Let the point on the ellipse be $P(2 \cos \theta, 3 \sin \theta)$.
Let the fixed point be $Q(-3, -5)$.
Let $(h, k)$ be the mid-point of the segment $PQ$. Then:
$h = \frac{2 \cos \theta - 3}{2} \implies 2 \cos \theta = 2h + 3 \implies \cos \theta = \frac{2h + 3}{2}$
$k = \frac{3 \sin \theta - 5}{2} \implies 3 \sin \theta = 2k + 5 \implies \sin \theta = \frac{2k + 5}{3}$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$\left(\frac{2h + 3}{2}\right)^2 + \left(\frac{2k + 5}{3}\right)^2 = 1$
$\frac{4h^2 + 12h + 9}{4} + \frac{4k^2 + 20k + 25}{9} = 1$
Multiplying by $36$:
$9(4h^2 + 12h + 9) + 4(4k^2 + 20k + 25) = 36$
$36h^2 + 108h + 81 + 16k^2 + 80k + 100 = 36$
$36h^2 + 16k^2 + 108h + 80k + 145 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $36x^2 + 16y^2 + 108x + 80y + 145 = 0$.

(A) Given curves are $C_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $C_2: x^2 + y^2 = ab$.
Let the point of intersection be $(x_1, y_1)$.
For $C_1$,differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \implies y'_1 = -\frac{b^2 x_1}{a^2 y_1}$.
For $C_2$,differentiating with respect to $x$: $2x + 2yy' = 0 \implies y'_2 = -\frac{x_1}{y_1}$.
Solving the equations for intersection: $x_1^2 = \frac{a^2 b}{a+b}$ and $y_1^2 = \frac{a b^2}{a+b}$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{y'_1 - y'_2}{1 + y'_1 y'_2} \right|$.
Substituting the slopes: $\tan \theta = \left| \frac{-\frac{b^2 x_1}{a^2 y_1} + \frac{x_1}{y_1}}{1 + \frac{b^2 x_1^2}{a^2 y_1^2}} \right| = \left| \frac{x_1 y_1 (a^2 - b^2)}{a^2 y_1^2 + b^2 x_1^2} \right|$.
Substituting $x_1^2$ and $y_1^2$: $\tan \theta = \left| \frac{a-b}{\sqrt{ab}} \right|$.
Since $a > b$,$\theta = \tan^{-1} \left( \frac{a-b}{\sqrt{ab}} \right)$.

(C) For ellipse $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,the eccentricity is $e^{2} = 1 - \frac{b^{2}}{a^{2}}$.
For ellipse $E_{2}$,the foci are $(0, b)$ and $(0, -b)$,so it is a vertical ellipse with $2c = 2b$,hence $c = b$. The vertices are at $(\pm a, 0)$,so the semi-major axis is $A = a$. The equation of $E_{2}$ is $\frac{x^{2}}{B^{2}} + \frac{y^{2}}{A^{2}} = 1$,where $A = a$. Since $c^{2} = A^{2} - B^{2}$,we have $b^{2} = a^{2} - B^{2}$,so $B^{2} = a^{2} - b^{2}$.
The eccentricity of $E_{2}$ is $e^{2} = 1 - \frac{B^{2}}{A^{2}} = 1 - \frac{a^{2} - b^{2}}{a^{2}} = \frac{b^{2}}{a^{2}}$.
Since both ellipses have the same eccentricity $e$,we have $e^{2} = \frac{b^{2}}{a^{2}}$.
From the first equation,$e^{2} = 1 - e^{2}$,which implies $2e^{2} = 1$,so $e = \frac{1}{\sqrt{2}}$.
Wait,re-evaluating the condition: The foci of $E_{2}$ are $(0, b)$ and $(0, -b)$,so $c_{2} = b$. The vertices of $E_{2}$ are $(\pm a, 0)$,so the semi-minor axis is $a$. The major axis is along the $y$-axis,so $A_{2} = c_{2}/e = b/e$. The relation $c_{2}^{2} = A_{2}^{2} - B_{2}^{2}$ gives $b^{2} = (b/e)^{2} - a^{2}$.
Thus $a^{2} = \frac{b^{2}}{e^{2}} - b^{2} = b^{2}(\frac{1-e^{2}}{e^{2}})$.
Since $1-e^{2} = b^{2}/a^{2}$,we have $a^{2} = b^{2}(\frac{b^{2}/a^{2}}{e^{2}}) = \frac{b^{4}}{a^{2}e^{2}}$,so $a^{4}e^{2} = b^{4}$,which means $a^{2}e = b^{2}$.
Substituting $b^{2} = a^{2}e$ into $e^{2} = 1 - b^{2}/a^{2}$,we get $e^{2} = 1 - e$,or $e^{2} + e - 1 = 0$.
Solving for $e > 0$,we get $e = \frac{-1 + \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 + \sqrt{5}}{2}$.

(D) Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through $\left(\sqrt{\frac{3}{2}}, 1\right)$,we have $\frac{3}{2a^{2}} + \frac{1}{b^{2}} = 1$.
Given eccentricity $e = \frac{1}{\sqrt{3}}$,so $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{1}{3}$,which implies $\frac{b^{2}}{a^{2}} = \frac{2}{3}$ or $a^{2} = \frac{3}{2}b^{2}$.
Substituting $a^{2}$ in the first equation: $\frac{3}{2(\frac{3}{2}b^{2})} + \frac{1}{b^{2}} = 1$ $\Rightarrow \frac{1}{b^{2}} + \frac{1}{b^{2}} = 1$ $\Rightarrow b^{2} = 2$.
Then $a^{2} = \frac{3}{2}(2) = 3$.
The ellipse equation is $\frac{x^{2}}{3} + \frac{y^{2}}{2} = 1$.
The focus $F(\alpha, 0)$ is given by $\alpha = ae = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1$. So $F = (1, 0)$.
The circle equation with center $(1, 0)$ and radius $\frac{2}{\sqrt{3}}$ is $(x-1)^{2} + y^{2} = \frac{4}{3}$.
From the ellipse equation,$y^{2} = 2(1 - \frac{x^{2}}{3})$.
Substitute $y^{2}$ into the circle equation: $(x-1)^{2} + 2 - \frac{2x^{2}}{3} = \frac{4}{3}$ $\Rightarrow x^{2} - 2x + 1 + 2 - \frac{2x^{2}}{3} = \frac{4}{3}$ $\Rightarrow \frac{x^{2}}{3} - 2x + \frac{5}{3} = 0$ $\Rightarrow x^{2} - 6x + 5 = 0$.
Solving $(x-5)(x-1) = 0$,we get $x=1$ or $x=5$. Since the circle radius is $\frac{2}{\sqrt{3}} \approx 1.15$,$x=5$ is outside the ellipse.
For $x=1$,$y^{2} = 2(1 - \frac{1}{3}) = 2(\frac{2}{3}) = \frac{4}{3}$,so $y = \pm \frac{2}{\sqrt{3}}$.
The points are $P(1, \frac{2}{\sqrt{3}})$ and $Q(1, -\frac{2}{\sqrt{3}})$.
$PQ^{2} = (1-1)^{2} + (\frac{2}{\sqrt{3}} - (-\frac{2}{\sqrt{3}}))^{2} = (\frac{4}{\sqrt{3}})^{2} = \frac{16}{3}$.

(C) The equation of the ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$,where $a=2$ and $b=1$.
Let the point of tangency be $P(2 \cos \theta, \sin \theta)$.
The equation of the tangent at $P$ is $\frac{x \cos \theta}{2} + y \sin \theta = 1$,or $x \cos \theta + 2y \sin \theta = 2$.
The tangents at the extremities of the major axis are $x = -2$ and $x = 2$.
For $B$,substitute $x = -2$: $-2 \cos \theta + 2y \sin \theta = 2 \Rightarrow y = \frac{1 + \cos \theta}{\sin \theta} = \cot \frac{\theta}{2}$. So,$B = (-2, \cot \frac{\theta}{2})$.
For $C$,substitute $x = 2$: $2 \cos \theta + 2y \sin \theta = 2 \Rightarrow y = \frac{1 - \cos \theta}{\sin \theta} = \tan \frac{\theta}{2}$. So,$C = (2, \tan \frac{\theta}{2})$.
The equation of the circle with $BC$ as diameter is $(x - x_B)(x - x_C) + (y - y_B)(y - y_C) = 0$.
$(x + 2)(x - 2) + (y - \cot \frac{\theta}{2})(y - \tan \frac{\theta}{2}) = 0$
$x^{2} - 4 + y^{2} - y(\tan \frac{\theta}{2} + \cot \frac{\theta}{2}) + \tan \frac{\theta}{2} \cot \frac{\theta}{2} = 0$
$x^{2} + y^{2} - y(\tan \frac{\theta}{2} + \cot \frac{\theta}{2}) - 3 = 0$.
Checking the point $(\sqrt{3}, 0)$: $(\sqrt{3})^{2} + 0^{2} - 0 - 3 = 3 - 3 = 0$. Thus,the circle passes through $(\sqrt{3}, 0)$.

(C) The image of the point $(2, 1)$ with respect to the $y$-axis is $(-2, 1)$.
The reflected ray passes through $(-2, 1)$ and $(5, 3)$.
The slope of the reflected ray is $m = \frac{3 - 1}{5 - (-2)} = \frac{2}{7}$.
The equation of the reflected ray is $y - 3 = \frac{2}{7}(x - 5)$,which simplifies to $2x - 7y + 11 = 0$.
Let the equation of the other directrix be $2x - 7y + \lambda = 0$.
The distance between the two directrices of an ellipse is $\frac{2a}{e}$.
The distance of the focus from the directrix is $\frac{a}{e} - ae = \frac{a(1 - e^2)}{e} = \frac{8}{\sqrt{53}}$.
Given $e = \frac{1}{3}$,we have $\frac{a(1 - 1/9)}{1/3} = 3a \times \frac{8}{9} = \frac{8a}{3} = \frac{8}{\sqrt{53}}$,so $a = \frac{3}{\sqrt{53}}$.
The distance between the two directrices is $\frac{2a}{e} = 2 \times \frac{3}{\sqrt{53}} \times 3 = \frac{18}{\sqrt{53}}$.
The distance between the parallel lines $2x - 7y + 11 = 0$ and $2x - 7y + \lambda = 0$ is $\frac{|\lambda - 11|}{\sqrt{2^2 + (-7)^2}} = \frac{|\lambda - 11|}{\sqrt{53}}$.
Equating the distances: $\frac{|\lambda - 11|}{\sqrt{53}} = \frac{18}{\sqrt{53}}$,so $|\lambda - 11| = 18$.
This gives $\lambda - 11 = 18$ or $\lambda - 11 = -18$,so $\lambda = 29$ or $\lambda = -7$.
The equations are $2x - 7y + 29 = 0$ or $2x - 7y - 7 = 0$.

(C) The center of the ellipse is $C(3, -4)$.
The focus is $S(4, -4)$ and the vertex is $A(5, -4)$.
Since the $y$-coordinates are the same,the major axis is horizontal.
The distance from the center to the vertex is $a = |5 - 3| = 2$.
The distance from the center to the focus is $ae = |4 - 3| = 1$.
Thus,$e = \frac{1}{2}$.
Using $b^{2} = a^{2}(1 - e^{2})$,we get $b^{2} = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
The equation of the ellipse is $\frac{(x - 3)^{2}}{4} + \frac{(y + 4)^{2}}{3} = 1$.
The condition for the line $y = mx - 4$ to be a tangent to the ellipse $\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1$ is $(y - k) = m(x - h) \pm \sqrt{a^{2}m^{2} + b^{2}}$.
Here,$h = 3, k = -4, a^{2} = 4, b^{2} = 3$.
The line is $y + 4 = mx - 3m$,so $y - (-4) = m(x - 3) - 3m$.
Comparing with the tangent form,the constant term is $-3m = \pm \sqrt{4m^{2} + 3}$.
Squaring both sides,$9m^{2} = 4m^{2} + 3$.
$5m^{2} = 3$.

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b=2$. The vertices are $(\pm a, 0)$. Let one vertex be $(a, 0)$.
Let the other two vertices be $(-a \cos \theta, 2 \sin \theta)$ and $(-a \cos \theta, -2 \sin \theta)$.
The base of the triangle is $4 \sin \theta$ and the height is $a + a \cos \theta = a(1 + \cos \theta)$.
The area $A = \frac{1}{2} \times (4 \sin \theta) \times a(1 + \cos \theta) = 2a \sin \theta (1 + \cos \theta)$.
To maximize $A$,let $f(\theta) = \sin \theta (1 + \cos \theta)$.
$f'(\theta) = \cos \theta (1 + \cos \theta) - \sin^2 \theta = \cos \theta + \cos^2 \theta - (1 - \cos^2 \theta) = 2 \cos^2 \theta + \cos \theta - 1 = 0$.
$(2 \cos \theta - 1)(\cos \theta + 1) = 0$.
Since $\theta \neq \pi$,we have $\cos \theta = \frac{1}{2}$,so $\sin \theta = \frac{\sqrt{3}}{2}$.
$A_{\max} = 2a \left(\frac{\sqrt{3}}{2}\right) \left(1 + \frac{1}{2}\right) = 2a \left(\frac{\sqrt{3}}{2}\right) \left(\frac{3}{2}\right) = \frac{3\sqrt{3}}{2} a$.
Given $A_{\max} = 6\sqrt{3}$,so $\frac{3\sqrt{3}}{2} a = 6\sqrt{3} \Rightarrow a = 4$.
Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(A) The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$,which simplifies to $x^{2}+2y^{2}=4$.
Substituting $y=x+1$ into the ellipse equation:
$x^{2}+2(x+1)^{2}=4$
$x^{2}+2(x^{2}+2x+1)=4$
$3x^{2}+4x-2=0$.
Let the roots be $x_{1}$ and $x_{2}$. Then $|x_{1}-x_{2}| = \frac{\sqrt{D}}{|a|} = \frac{\sqrt{16 - 4(3)(-2)}}{3} = \frac{\sqrt{40}}{3}$.
The length of the chord $PQ$ is given by $PQ = |x_{1}-x_{2}| \sqrt{1+m^{2}}$,where $m$ is the slope of the line $y=x+1$,so $m=1$.
$PQ = \frac{\sqrt{40}}{3} \sqrt{1+1^{2}} = \frac{\sqrt{40}}{3} \sqrt{2} = \frac{\sqrt{80}}{3}$.
Since $PQ$ is the diameter of the circle,$2r = PQ = \frac{\sqrt{80}}{3}$,so $r = \frac{\sqrt{80}}{6}$.
Therefore,$(3r)^{2} = 9r^{2} = 9 \times \frac{80}{36} = \frac{80}{4} = 20$.

(C) Let $P = (4, 3)$ and $Q = (2\cos\theta, \sqrt{2}\sin\theta)$ be a point on the ellipse $x^{2} + 2y^{2} = 4$.
Let $D(h, k)$ be the midpoint of $PQ$.
Then,$h = \frac{4 + 2\cos\theta}{2} = 2 + \cos\theta \implies \cos\theta = h - 2$.
And $k = \frac{3 + \sqrt{2}\sin\theta}{2} \implies \sqrt{2}\sin\theta = 2k - 3 \implies \sin\theta = \frac{2k - 3}{\sqrt{2}}$.
Using the identity $\cos^{2}\theta + \sin^{2}\theta = 1$,we get:
$(h - 2)^{2} + \left(\frac{2k - 3}{\sqrt{2}}\right)^{2} = 1$
$(h - 2)^{2} + \frac{4(k - 1.5)^{2}}{2} = 1$
$(h - 2)^{2} + 2(k - 1.5)^{2} = 1$
Dividing by $1$,we get $\frac{(h - 2)^{2}}{1} + \frac{(k - 1.5)^{2}}{1/2} = 1$.
This is an ellipse with $a^{2} = 1$ and $b^{2} = 1/2$.
Since $a^{2} > b^{2}$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1/2}{1}} = \sqrt{1 - 1/2} = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

(B) The circle is $x^{2} + y^{2} = \frac{9}{4}$ and the parabola is $y^{2} = 4x$.
The tangent to the parabola $y^{2} = 4x$ is $y = mx + \frac{1}{m}$.
The distance from the origin $(0,0)$ to this tangent must equal the radius of the circle,which is $\frac{3}{2}$.
$\frac{|m(0) - 0 + 1/m|}{\sqrt{m^{2} + 1}} = \frac{3}{2} \Rightarrow \frac{1}{|m|\sqrt{m^{2} + 1}} = \frac{3}{2}$.
Squaring both sides: $\frac{1}{m^{2}(m^{2} + 1)} = \frac{9}{4} \Rightarrow 9m^{4} + 9m^{2} - 4 = 0$.
Solving for $m^{2}$: $(3m^{2} - 1)(3m^{2} + 4) = 0$. Since $m^{2} > 0$,we have $m^{2} = \frac{1}{3}$,so $m = \pm \frac{1}{\sqrt{3}}$.
The common tangents are $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$ and $y = -\frac{1}{\sqrt{3}}x - \sqrt{3}$.
These tangents intersect at the point $Q$ on the $x$-axis where $y=0$. Setting $y=0$ in $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$ gives $x = -3$. Thus,$Q = (-3, 0)$.
The length $OQ = |-3| = 3$. Given the ellipse has semi-minor axis $b = 3$ and semi-major axis $a = 6$.
Eccentricity $e$ is given by $b^{2} = a^{2}(1 - e^{2})$ $\Rightarrow 9 = 36(1 - e^{2})$ $\Rightarrow 1 - e^{2} = \frac{1}{4}$ $\Rightarrow e^{2} = \frac{3}{4}$.
Length of latus rectum $l = \frac{2b^{2}}{a} = \frac{2 \times 9}{6} = 3$.
Therefore,$\frac{l}{e^{2}} = \frac{3}{3/4} = 4$.

(A) Given the ellipse equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $a>b$ and eccentricity $e = \frac{1}{4}$.
We know $e^{2} = 1 - \frac{b^{2}}{a^{2}}$,so $\frac{1}{16} = 1 - \frac{b^{2}}{a^{2}}$,which gives $\frac{b^{2}}{a^{2}} = \frac{15}{16}$ or $b^{2} = \frac{15}{16}a^{2}$.
The ellipse passes through $\left(-4 \sqrt{\frac{2}{5}}, 3\right)$,so $\frac{(-4 \sqrt{2/5})^{2}}{a^{2}} + \frac{3^{2}}{b^{2}} = 1$.
Substituting the values: $\frac{16 \times (2/5)}{a^{2}} + \frac{9}{b^{2}} = 1 \Rightarrow \frac{32}{5a^{2}} + \frac{9}{b^{2}} = 1$.
Substitute $b^{2} = \frac{15}{16}a^{2}$: $\frac{32}{5a^{2}} + \frac{9 \times 16}{15a^{2}} = 1$.
$\frac{32}{5a^{2}} + \frac{144}{15a^{2}} = 1$ $\Rightarrow \frac{96 + 144}{15a^{2}} = 1$ $\Rightarrow \frac{240}{15a^{2}} = 1$.
$16 = a^{2}$.
Then $b^{2} = \frac{15}{16} \times 16 = 15$.
Thus,$a^{2} + b^{2} = 16 + 15 = 31$.

(C) The first region is the area between the circle $x^{2} + y^{2} = a^{2}$ and the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$. The area is $\pi a^{2} - \pi ab = 30\pi$,which simplifies to $a^{2} - ab = 30$.
The second region is the area between the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and the circle $x^{2} + y^{2} = b^{2}$. The area is $\pi ab - \pi b^{2} = 18\pi$,which simplifies to $ab - b^{2} = 18$.
Adding these two equations: $(a^{2} - ab) + (ab - b^{2}) = 30 + 18$,so $a^{2} - b^{2} = 48$.
Subtracting the second from the first: $(a^{2} - ab) - (ab - b^{2}) = 30 - 18$,which gives $a^{2} - 2ab + b^{2} = 12$.
Thus,$(a - b)^{2} = 12$.

(A) The ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ meets the line $\frac{x}{7}+\frac{y}{2\sqrt{6}}=1$ on the $x$-axis. Setting $y=0$ in the line equation,we get $x=7$. Since the ellipse passes through $(7, 0)$,we have $a=7$.
The ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ meets the line $\frac{x}{7}-\frac{y}{2\sqrt{6}}=1$ on the $y$-axis. Setting $x=0$ in the line equation,we get $y=-2\sqrt{6}$. Since the ellipse passes through $(0, -2\sqrt{6})$,we have $b^{2}=(-2\sqrt{6})^{2}=24$,so $b=2\sqrt{6}$.
The eccentricity $e$ is given by $e^{2}=1-\frac{b^{2}}{a^{2}}$.
Substituting the values,$e^{2}=1-\frac{24}{49} = \frac{25}{49}$.
Therefore,$e=\frac{5}{7}$.

(B) The equation of the ellipse is $2x^{2} + 3y^{2} = 5$,which can be written as $\frac{x^{2}}{5/2} + \frac{y^{2}}{5/3} = 1$.
Here $a^{2} = \frac{5}{2}$ and $b^{2} = \frac{5}{3}$.
The equation of any tangent to the ellipse is $y = mx \pm \sqrt{a^{2}m^{2} + b^{2}} = mx \pm \sqrt{\frac{5}{2}m^{2} + \frac{5}{3}}$.
Since the tangent passes through $(1, 3)$,we have $3 - m = \pm \sqrt{\frac{5}{2}m^{2} + \frac{5}{3}}$.
Squaring both sides: $(3 - m)^{2} = \frac{5}{2}m^{2} + \frac{5}{3}$.
$9 - 6m + m^{2} = \frac{5}{2}m^{2} + \frac{5}{3}$.
Multiplying by $6$: $54 - 36m + 6m^{2} = 15m^{2} + 10$.
$9m^{2} + 36m - 44 = 0$.
Let $m_{1}$ and $m_{2}$ be the roots of this quadratic equation.
Then $m_{1} + m_{2} = -\frac{36}{9} = -4$ and $m_{1}m_{2} = -\frac{44}{9}$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left|\frac{m_{1} - m_{2}}{1 + m_{1}m_{2}}\right|$.
$|m_{1} - m_{2}| = \sqrt{(m_{1} + m_{2})^{2} - 4m_{1}m_{2}} = \sqrt{(-4)^{2} - 4(-\frac{44}{9})} = \sqrt{16 + \frac{176}{9}} = \sqrt{\frac{144 + 176}{9}} = \sqrt{\frac{320}{9}} = \frac{8\sqrt{5}}{3}$.
$1 + m_{1}m_{2} = 1 - \frac{44}{9} = -\frac{35}{9}$.
$\tan \theta = \left|\frac{8\sqrt{5}/3}{-35/9}\right| = \left|\frac{8\sqrt{5}}{3} \times \frac{-9}{35}\right| = \frac{24\sqrt{5}}{35} = \frac{24\sqrt{5}}{7 \times 5} = \frac{24}{7\sqrt{5}}$.
Thus,$\theta = \tan^{-1}\left(\frac{24}{7\sqrt{5}}\right)$.

(D) The given equation is $x^{2} + 4y^{2} + 2x + 8y - \lambda = 0$.
Completing the square,we get $(x^{2} + 2x + 1) + 4(y^{2} + 2y + 1) = \lambda + 1 + 4$.
$(x + 1)^{2} + 4(y + 1)^{2} = \lambda + 5$.
Dividing by $\lambda + 5$,we get $\frac{(x + 1)^{2}}{\lambda + 5} + \frac{(y + 1)^{2}}{(\lambda + 5)/4} = 1$.
Here,$a^{2} = \lambda + 5$ and $b^{2} = \frac{\lambda + 5}{4}$,so $a = \sqrt{\lambda + 5}$ and $b = \frac{\sqrt{\lambda + 5}}{2}$.
The length of the latus rectum is $\frac{2b^{2}}{a} = 4$.
Substituting the values,$\frac{2(\lambda + 5)/4}{\sqrt{\lambda + 5}} = 4$.
$\frac{\lambda + 5}{2\sqrt{\lambda + 5}} = 4 \implies \frac{\sqrt{\lambda + 5}}{2} = 4 \implies \sqrt{\lambda + 5} = 8$.
Squaring both sides,$\lambda + 5 = 64$,so $\lambda = 59$.
The length of the major axis $l = 2a = 2\sqrt{\lambda + 5} = 2(8) = 16$.
Therefore,$\lambda + l = 59 + 16 = 75$.

(A) The given ellipse is $\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1$. Here,$a^{2} = 2$ and $b^{2} = 4$. Since $b^{2} > a^{2}$,the major axis is along the $y$-axis.
Eccentricity $e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{2}{4}} = \frac{1}{\sqrt{2}}$.
The foci are $(0, \pm be) = (0, \pm 2 \times \frac{1}{\sqrt{2}}) = (0, \pm \sqrt{2})$.
Given $S$ is the focus on the negative major axis,$S = (0, -\sqrt{2})$.
The chord of contact of tangents from $R(\sqrt{2}, 2\sqrt{2}-2)$ to the ellipse is given by $\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1$.
Substituting $x_{1} = \sqrt{2}, y_{1} = 2\sqrt{2}-2, a^{2} = 2, b^{2} = 4$,we get $\frac{x \sqrt{2}}{2} + \frac{y(2\sqrt{2}-2)}{4} = 1$,which simplifies to $\frac{x}{\sqrt{2}} + \frac{y(\sqrt{2}-1)}{2} = 1$.
Solving this with the ellipse equation,we find the points of contact $P$ and $Q$ as $(1, \sqrt{2})$ and $(\sqrt{2}, 0)$.
Now,$SP^{2} = (1-0)^{2} + (\sqrt{2} - (-\sqrt{2}))^{2} = 1^{2} + (2\sqrt{2})^{2} = 1 + 8 = 9$.
$SQ^{2} = (\sqrt{2}-0)^{2} + (0 - (-\sqrt{2}))^{2} = 2 + 2 = 4$.
Therefore,$SP^{2} + SQ^{2} = 9 + 4 = 13$.

(B) The set $S$ represents points $(x, y)$ with $x, y \in N$ (natural numbers) inside or on the ellipse $\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9} \leq 1$.
Since $x, y \geq 1$,we test integer values for $x$ and $y$:
For $x=1: 9(-2)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 36 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 108$ $\Rightarrow (y-4)^2 \leq 6.75$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.
For $x=2: 9(-1)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 9 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 135$ $\Rightarrow (y-4)^2 \leq 8.43$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.
For $x=3: 9(0)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 9$. Possible $y \in \{1, 2, 3, 4, 5, 6, 7\}$.
For $x=4: 9(1)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 9 + 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 8.43$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.
For $x=5: 9(2)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 36 + 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 6.75$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.
Now check which of these points satisfy $T: (x-7)^2 + (y-4)^2 \leq 36$.
Points $(x, y)$ in $S$ are: $(1, 1..6), (2, 1..6), (3, 1..7), (4, 1..6), (5, 1..6)$.
Checking $(x-7)^2 + (y-4)^2 \leq 36$ for these points:
For $x=1: (-6)^2 + (y-4)^2 \leq 36$ $\Rightarrow 36 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 0$ $\Rightarrow y=4$. Point $(1, 4)$.
For $x=2: (-5)^2 + (y-4)^2 \leq 36$ $\Rightarrow 25 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 11$ $\Rightarrow y \in \{1, 2, 3, 4, 5, 6, 7\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)
For $x=3: (-4)^2 + (y-4)^2 \leq 36$ $\Rightarrow 16 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 20$ $\Rightarrow y \in \{0..8\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6, 7\}$. ($7$ points)
For $x=4: (-3)^2 + (y-4)^2 \leq 36$ $\Rightarrow 9 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 27$ $\Rightarrow y \in \{-1..9\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)
For $x=5: (-2)^2 + (y-4)^2 \leq 36$ $\Rightarrow 4 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 32$ $\Rightarrow y \in \{-1..9\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)
Total points $= 1 + 6 + 7 + 6 + 6 = 26$.

(B) The family of lines passing through the intersection of $bx + 10y - 8 = 0$ and $2x - 3y = 0$ is given by $(bx + 10y - 8) + \lambda(2x - 3y) = 0$.
Since the line passes through $(1, 1)$,we have $(b + 10 - 8) + \lambda(2 - 3) = 0$,which gives $b + 2 - \lambda = 0$,so $\lambda = b + 2$.
Substituting $\lambda$ back,the line equation is $(b + 2(2))x + (10 - 3(b + 2))y - 8 = 0$,which simplifies to $(b + 4)x + (4 - 3b)y - 8 = 0$.
The line is tangent to the circle $x^2 + y^2 = \frac{16}{17}$,so the perpendicular distance from the origin $(0, 0)$ to the line equals the radius $r = \frac{4}{\sqrt{17}}$.
Thus,$\frac{|-8|}{\sqrt{(b + 4)^2 + (4 - 3b)^2}} = \frac{4}{\sqrt{17}}$.
Squaring both sides,$\frac{64}{b^2 + 8b + 16 + 16 - 24b + 9b^2} = \frac{16}{17} \implies \frac{4}{10b^2 - 16b + 32} = \frac{1}{17}$.
$10b^2 - 16b + 32 = 68 \implies 10b^2 - 16b - 36 = 0 \implies 5b^2 - 8b - 18 = 0$.
Wait,re-evaluating the intersection: $(b+2\lambda)x + (10-3\lambda)y - 8 = 0$. For $(1,1)$,$b+2\lambda + 10-3\lambda - 8 = 0 \implies b+2-\lambda = 0 \implies \lambda = b+2$.
Line: $(b+2(b+2))x + (10-3(b+2))y - 8 = 0 \implies (3b+4)x + (4-3b)y - 8 = 0$.
Distance: $\frac{8}{\sqrt{(3b+4)^2 + (4-3b)^2}} = \frac{4}{\sqrt{17}} \implies \frac{2}{\sqrt{9b^2+24b+16+16-24b+9b^2}} = \frac{1}{\sqrt{17}} \implies \frac{4}{18b^2+32} = \frac{1}{17} \implies 18b^2+32 = 68 \implies 18b^2 = 36 \implies b^2 = 2$.
For the ellipse $\frac{x^2}{5} + \frac{y^2}{2} = 1$,$a^2 = 5$ and $b^2 = 2$. Since $a^2 > b^2$,$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2}{5}} = \sqrt{\frac{3}{5}}$.

(C) Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b < a$. The coordinates are $C = (0, b)$,$F_1 = (-ae, 0)$,and $B = (a, 0)$.
Since $\angle F_1CB = 90^{\circ}$,the product of the slopes of $CF_1$ and $CB$ is $-1$.
Slope of $CF_1 = \frac{0 - b}{-ae - 0} = \frac{-b}{-ae} = \frac{b}{ae}$.
Slope of $CB = \frac{0 - b}{a - 0} = \frac{-b}{a}$.
Therefore,$(\frac{b}{ae}) \times (\frac{-b}{a}) = -1$.
$\frac{b^2}{a^2e} = 1 \Rightarrow b^2 = a^2e$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2e = a^2(1 - e^2)$.
$e = 1 - e^2 \Rightarrow e^2 + e - 1 = 0$.
Solving for $e$ using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.
Since eccentricity $e > 0$,we have $e = \frac{\sqrt{5} - 1}{2}$.

(B) The equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$.
The equation of the tangent at the point $(3 \cos \theta, 2 \sin \theta)$ is $\frac{x}{3} \cos \theta + \frac{y}{2} \sin \theta = 1$.
The intercepts of this tangent on the coordinate axes are $(3 \sec \theta, 0)$ and $(0, 2 \operatorname{cosec} \theta)$.
The four tangents form a rhombus with vertices at $(\pm 3 \sec \theta, 0)$ and $(0, \pm 2 \operatorname{cosec} \theta)$.
The area of this quadrilateral (rhombus) is given by $A(\theta) = 4 \times \left( \frac{1}{2} \times |3 \sec \theta| \times |2 \operatorname{cosec} \theta| \right)$.
$A(\theta) = 12 \sec \theta \operatorname{cosec} \theta = \frac{12}{\sin \theta \cos \theta} = \frac{24}{2 \sin \theta \cos \theta} = \frac{24}{\sin 2 \theta}$.
Since $0 < \theta < \frac{\pi}{2}$,the maximum value of $\sin 2 \theta$ is $1$ (at $\theta = \frac{\pi}{4}$).
Therefore,the minimum value of $A(\theta) = \frac{24}{1} = 24$.

(D) Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$.
Since $\triangle ABC$ is equilateral,the angle $\angle BAC = 60^{\circ}$.
Due to the symmetry of the ellipse,the line $AO$ (the $y$-axis) bisects $\angle BAC$,so $\angle OAF_2 = 30^{\circ}$.
In the right-angled $\triangle AOF_2$,we have $\tan(30^{\circ}) = \frac{OF_2}{OA}$.
Here,$OF_2 = ae$ and $OA = b$.
So,$\frac{1}{\sqrt{3}} = \frac{ae}{b}$,which implies $b = \sqrt{3}ae$.
Squaring both sides,$b^2 = 3a^2e^2$.
Using the relation $b^2 = a^2(1-e^2)$,we get $a^2(1-e^2) = 3a^2e^2$.
Dividing by $a^2$,we have $1-e^2 = 3e^2$,which simplifies to $4e^2 = 1$.
Thus,$e^2 = \frac{1}{4}$,and since $e>0$,$e = \frac{1}{2}$.

(C) Let the equation of the ellipse be $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. Since it passes through $(0,0)$,$(1,0)$,and $(0,2)$,we substitute these points:
$1$) $\frac{h^2}{a^2} + \frac{k^2}{b^2} = 1$
$2$) $\frac{(1-h)^2}{a^2} + \frac{k^2}{b^2} = 1 \implies \frac{1-2h+h^2}{a^2} + \frac{k^2}{b^2} = 1$. Subtracting $(1)$ from $(2)$ gives $\frac{1-2h}{a^2} = 0 \implies h = \frac{1}{2}$.
$3$) $\frac{h^2}{a^2} + \frac{(2-k)^2}{b^2} = 1 \implies \frac{h^2}{a^2} + \frac{4-4k+k^2}{b^2} = 1$. Subtracting $(1)$ from $(3)$ gives $\frac{4-4k}{b^2} = 0 \implies k = 1$.
Thus,the center is $(\frac{1}{2}, 1)$. The equation is $\frac{(x-1/2)^2}{a^2} + \frac{(y-1)^2}{b^2} = 1$. Substituting $(0,0)$ gives $\frac{1/4}{a^2} + \frac{1}{b^2} = 1$.
Since a focus lies on the $Y$-axis $(x=0)$,the distance from the center $(\frac{1}{2}, 1)$ to the focus $(0, 1)$ is $ae = \frac{1}{2}$.
Thus $a^2e^2 = \frac{1}{4}$. Using $b^2 = a^2(1-e^2)$,we have $\frac{1}{4a^2} + \frac{1}{a^2(1-e^2)} = 1 \implies \frac{1-e^2+4}{4a^2(1-e^2)} = 1$.
Using $a^2 = \frac{1}{4e^2}$,we get $\frac{5-e^2}{4(\frac{1}{4e^2})(1-e^2)} = 1 \implies \frac{e^2(5-e^2)}{1-e^2} = 1 \implies 5e^2 - e^4 = 1 - e^2 \implies e^4 - 6e^2 + 1 = 0$.
Solving for $e^2$: $e^2 = \frac{6 \pm \sqrt{36-4}}{2} = 3 \pm \sqrt{8} = 3 \pm 2\sqrt{2}$.
Since $e < 1$,$e^2 = 3 - 2\sqrt{2} = (\sqrt{2}-1)^2$. Thus $e = \sqrt{2}-1$.

(A) Let the floor be the $x$-axis and the vertex of the angle be the origin $(0,0)$. The wall is a line making an angle of $135^{\circ}$ with the floor,so its equation is $y = -x$ (or $x+y=0$).
Let the ladder have endpoints $P(x_1, 0)$ on the floor and $Q(x_2, y_2)$ on the wall. Since $Q$ is on $y = -x$,$Q$ is $(x_2, -x_2)$.
The length of the ladder is $l$,so $(x_1 - x_2)^2 + (0 - (-x_2))^2 = l^2$,which simplifies to $(x_1 - x_2)^2 + x_2^2 = l^2$.
The mid-point $(h, k)$ of the ladder is $\left(\frac{x_1+x_2}{2}, \frac{0-x_2}{2}\right) = \left(\frac{x_1+x_2}{2}, -\frac{x_2}{2}\right)$.
From this,$x_2 = -2k$ and $x_1+x_2 = 2h$,so $x_1 = 2h - x_2 = 2h + 2k$.
Substituting into the length equation: $(2h+2k - (-2k))^2 + (-2k)^2 = l^2$,which is $(2h+4k)^2 + 4k^2 = l^2$.
Expanding gives $4h^2 + 16hk + 16k^2 + 4k^2 = l^2$,or $4h^2 + 16hk + 20k^2 = l^2$.
The general equation of an ellipse is $Ax^2 + Bxy + Cy^2 = F$. The area of such an ellipse is given by $\frac{2\pi F}{\sqrt{4AC - B^2}}$.
Here $A=4, B=16, C=20, F=l^2$. The denominator is $\sqrt{4(4)(20) - (16)^2} = \sqrt{320 - 256} = \sqrt{64} = 8$.
Area $= \frac{2\pi l^2}{8} = \frac{\pi l^2}{4}$.

(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.
The ends of the major axis are $A'(-a, 0)$ and $A(a, 0)$.
The foci are $S'(-ae, 0)$ and $S(ae, 0)$.
Given that the foci and the ends of the major axis are equally spaced,we have the segments $A'S'$,$S'S$,and $SA$ equal in length.
$A'S' = S'S = SA = k$ (say).
The total length of the major axis is $A'A = 2a$.
Thus,$k + k + k = 2a \implies 3k = 2a \implies k = \frac{2a}{3}$.
The distance between the foci is $S'S = 2ae = k = \frac{2a}{3}$.
Therefore,$e = \frac{1}{3}$.
We know the relation $b^2 = a^2(1 - e^2)$.
Given $b = 2\sqrt{2}$,so $b^2 = 8$.
$8 = a^2(1 - (\frac{1}{3})^2) = a^2(1 - \frac{1}{9}) = a^2(\frac{8}{9})$.
$8 = \frac{8a^2}{9} \implies a^2 = 9 \implies a = 3$.
The length of the semi-major axis is $3$.

(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where the center is at $(0, b)$ relative to the diameter.
Since it touches the diameter $y=0$,the ellipse is $\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} = 1$.
The semi-circle is $x^2 + y^2 = r^2$ for $y \ge 0$.
Substituting $x^2 = a^2(1 - \frac{(y-b)^2}{b^2})$ into the circle equation:
$a^2(1 - \frac{(y-b)^2}{b^2}) + y^2 = r^2$
$a^2 - \frac{a^2}{b^2}(y^2 - 2by + b^2) + y^2 = r^2$
$(1 - \frac{a^2}{b^2})y^2 + \frac{2a^2}{b}y + (a^2 - a^2 - r^2) = 0$
$(1 - \frac{a^2}{b^2})y^2 + \frac{2a^2}{b}y - r^2 = 0$.
For tangency,the discriminant $D = 0$:
$(\frac{2a^2}{b})^2 - 4(1 - \frac{a^2}{b^2})(-r^2) = 0$
$\frac{4a^4}{b^2} + 4r^2 - \frac{4a^2r^2}{b^2} = 0$
$a^4 + b^2r^2 - a^2r^2 = 0 \Rightarrow b^2 = \frac{a^2r^2 - a^4}{r^2} = a^2(1 - \frac{a^2}{r^2})$.
Area $A = \pi ab = \pi a^2 \sqrt{1 - \frac{a^2}{r^2}}$.
Let $u = a^2$,then $A^2 = \pi^2 u^2(1 - \frac{u}{r^2}) = \pi^2(u^2 - \frac{u^3}{r^2})$.
For maximum area,$\frac{d(A^2)}{du} = 2u - \frac{3u^2}{r^2} = 0 \Rightarrow u = \frac{2r^2}{3}$.
So $a^2 = \frac{2r^2}{3}$ and $b^2 = \frac{2r^2}{3}(1 - \frac{2}{3}) = \frac{2r^2}{9}$.
Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2r^2/9}{2r^2/3}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}$.

(C) The given equation is $e x^2 + \pi y^2 - 2 e^2 x - 2 \pi^2 y + e^3 + \pi^3 = \pi e$.
Rearranging the terms,we get $e(x^2 - 2ex) + \pi(y^2 - 2\pi y) = \pi e - e^3 - \pi^3$.
Completing the square,$e(x^2 - 2ex + e^2) + \pi(y^2 - 2\pi y + \pi^2) = \pi e - e^3 - \pi^3 + e^3 + \pi^3$.
This simplifies to $e(x - e)^2 + \pi(y - \pi)^2 = \pi e$.
Dividing by $\pi e$,we get $\frac{(x - e)^2}{\pi} + \frac{(y - \pi)^2}{e} = 1$.
This is the equation of an ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ where $a^2 = \pi$ and $b^2 = e$.
Since $\pi > e$,the major axis is along the $x$-direction,so $a = \sqrt{\pi}$.
For any point $P$ on an ellipse,the sum of the distances from the foci $P S_1 + P S_2$ is constant and equal to the length of the major axis,which is $2a$.
Therefore,$P S_1 + P S_2 = 2 \sqrt{\pi}$.

(A) Given the equation of the ellipse: $4x^2 + 9y^2 - 8x - 36y + 15 = 0$.
Completing the square,we get: $4(x^2 - 2x + 1) + 9(y^2 - 4y + 4) = -15 + 4 + 36$.
$4(x - 1)^2 + 9(y - 2)^2 = 25$.
Dividing by $25$,we have: $\frac{(x - 1)^2}{25/4} + \frac{(y - 2)^2}{25/9} = 1$.
Let $X = x - 1$ and $Y = y - 2$. The expression becomes $X^2 + Y^2$.
This represents the square of the distance from the center $(1, 2)$ to any point $(x, y)$ on the ellipse.
The distance squared $r^2 = X^2 + Y^2$ ranges from the square of the semi-minor axis to the square of the semi-major axis.
Here,$a^2 = \frac{25}{4} = 6.25$ and $b^2 = \frac{25}{9} \approx 2.77$.
Thus,$\min(X^2 + Y^2) = b^2 = \frac{25}{9}$ and $\max(X^2 + Y^2) = a^2 = \frac{25}{4}$.
The sum is $\frac{25}{9} + \frac{25}{4} = 25 \left( \frac{4 + 9}{36} \right) = \frac{25 \times 13}{36} = \frac{325}{36}$.

(B) The equation of the ellipse is $9x^2 + 16y^2 = 144$,which can be written as $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Here,$a^2 = 16$ and $b^2 = 9$.
The equation of a line with slope $m$ and $y$-intercept $c$ is $y = mx + c$. Given $c = 5$,the line is $y = mx + 5$.
For this line to have a common point with the ellipse,it must be either a secant or a tangent. The condition for the line $y = mx + c$ to be tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.
Substituting the values,we get $5^2 = 16m^2 + 9$.
$25 = 16m^2 + 9$ $\Rightarrow 16m^2 = 16$ $\Rightarrow m^2 = 1$.
Thus,$m = \pm 1$.
The positive slope is $m = 1$. For any slope $m > 1$,the line $y = mx + 5$ will intersect the ellipse at two points,and for $m < 1$,it will not intersect the ellipse. Therefore,the smallest positive slope for which the line has a common point with the ellipse is $1$.

(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Let the point $P$ on the ellipse be $(a \cos \theta, b \sin \theta)$.
The foci of the ellipse are $F_1(ae, 0)$ and $F_2(-ae, 0)$.
The centroid $(h, k)$ of $\triangle P F_1 F_2$ is given by:
$h = \frac{a \cos \theta + ae - ae}{3} = \frac{a \cos \theta}{3}$
$k = \frac{b \sin \theta + 0 + 0}{3} = \frac{b \sin \theta}{3}$
Thus,$\cos \theta = \frac{3h}{a}$ and $\sin \theta = \frac{3k}{b}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$\left(\frac{3h}{a}\right)^2 + \left(\frac{3k}{b}\right)^2 = 1$
$\frac{9h^2}{a^2} + \frac{9k^2}{b^2} = 1$
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{(a/3)^2} + \frac{y^2}{(b/3)^2} = 1$,which is an ellipse.

(B) The foci of the ellipse are $S'(5, 15)$ and $S(21, 15)$.
The distance between the foci is $2ae = \sqrt{(21-5)^2 + (15-15)^2} = 16$,so $ae = 8$.
The center of the ellipse is the midpoint of the foci: $(\frac{5+21}{2}, \frac{15+15}{2}) = (13, 15)$.
The $X$-axis $(y=0)$ is a tangent to the ellipse. The distance from the center $(13, 15)$ to the tangent line $y=0$ is $15$.
For an ellipse,the distance from the center to a tangent line is given by $\sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}$. Since the major axis is horizontal,the tangent is parallel to the major axis,so the distance is simply the semi-minor axis $b = 15$.
We know the relation $b^2 = a^2(1 - e^2) = a^2 - (ae)^2$.
Substituting the values $b = 15$ and $ae = 8$:
$15^2 = a^2 - 8^2$
$225 = a^2 - 64$
$a^2 = 289$
$a = 17$
The length of the major axis is $2a = 2 \times 17 = 34$.

(D) The regions are defined as:
$A_1: x^2 + 2y^2 \leq 1$ (an ellipse)
$A_2: |x|^3 + 2\sqrt{2}|y|^3 \leq 1$
$A_3: \max(|x|, \sqrt{2}|y|) \leq 1$ (a rectangle with vertices at $(\pm 1, \pm 1/\sqrt{2})$)
Let us test a point $(x, y)$ such that $|x| \leq 1$ and $\sqrt{2}|y| \leq 1$.
If $(x, y) \in A_3$,then $|x| \leq 1$ and $|y| \leq 1/\sqrt{2}$.
For $A_1$,we have $x^2 + 2y^2 \leq 1^2 + 2(1/\sqrt{2})^2 = 1 + 1 = 2$. This does not guarantee $A_3 \subset A_1$.
However,consider the boundary curves. The region $A_3$ is the rectangle $|x| \leq 1, |y| \leq 1/\sqrt{2}$.
The region $A_1$ is the ellipse $x^2 + 2y^2 \leq 1$.
The region $A_2$ is $|x|^3 + 2\sqrt{2}|y|^3 \leq 1$.
By comparing the growth of the powers,for $0 < |x|, |y| < 1$,we have $|x|^3 < |x|^2$ and $|y|^3 < |y|^2$.
Thus,$x^2 + 2y^2 \leq 1 \implies |x|^3 + 2\sqrt{2}|y|^3 \leq 1$ is not necessarily true,but checking the containment:
$A_1 \subset A_2 \subset A_3$ is the correct inclusion order,which means $A_3 \supset A_2 \supset A_1$.
Therefore,the correct option is $(d)$.

(C) Since the ellipse is tangent to both axes in the first quadrant,its center is $(a, b)$. The foci are $(a \pm ae, b)$ or $(a, b \pm ae)$. Given the geometry,the center is $(a, b)$ and the distance from the origin to the center is $\sqrt{a^2 + b^2}$.
For an ellipse tangent to the axes,$x_0 = a$ and $y_0 = b$. The foci are $(a \pm ae, b)$.
Given $\triangle OF_1F_2$ is isosceles with $\angle OF_1F_2 = 120^{\circ}$,we have $OF_1 = F_1F_2 = 2ae$.
Using the law of cosines in $\triangle OF_1F_2$ or coordinate geometry properties,we find $e = \frac{1}{2}$.

(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Any point $P$ on the ellipse can be represented as $(4 \cos \theta, 3 \sin \theta)$.
The equation of the tangent at $P$ is $\frac{x \cos \theta}{4} + \frac{y \sin \theta}{3} = 1$.
The coordinates of $A$ (where $y=0$) are $(4 \sec \theta, 0)$ and $B$ (where $x=0$) are $(0, 3 \operatorname{cosec} \theta)$.
The length $L$ of the segment $AB$ is given by $L^2 = 16 \sec^2 \theta + 9 \operatorname{cosec}^2 \theta$.
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$ and $\operatorname{cosec}^2 \theta = 1 + \cot^2 \theta$,we get $L^2 = 16(1 + \tan^2 \theta) + 9(1 + \cot^2 \theta) = 25 + 16 \tan^2 \theta + 9 \cot^2 \theta$.
By the $AM$-$GM$ inequality,$16 \tan^2 \theta + 9 \cot^2 \theta \geq 2 \sqrt{16 \tan^2 \theta \cdot 9 \cot^2 \theta} = 2 \cdot 4 \cdot 3 = 24$.
Thus,$L^2 \geq 25 + 24 = 49$,which implies $L \geq 7$.
The minimum length of the line segment $AB$ is $7$.

(B) The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos \theta, b \sin \theta)$ is $ax \sec \theta - by \operatorname{cosec} \theta = a^2 - b^2$.
Here $a^2 = 4$,so the equation is $2x \sec \theta - by \operatorname{cosec} \theta = 4 - b^2$.
The perpendicular distance $d$ from the origin $(0,0)$ is $d = \frac{|4 - b^2|}{\sqrt{4 \sec^2 \theta + b^2 \operatorname{cosec}^2 \theta}}$.
To maximize $d$,we minimize $f(\theta) = 4 \sec^2 \theta + b^2 \operatorname{cosec}^2 \theta = 4(1 + \tan^2 \theta) + b^2(1 + \cot^2 \theta) = (4 + b^2) + 4 \tan^2 \theta + b^2 \cot^2 \theta$.
Using $AM$-$GM$ inequality,$4 \tan^2 \theta + b^2 \cot^2 \theta \geq 2 \sqrt{4 \tan^2 \theta \cdot b^2 \cot^2 \theta} = 4b$.
The minimum value is $(4 + b^2 + 4b) = (2 + b)^2$.
Thus,$d_{max} = \frac{4 - b^2}{\sqrt{(2 + b)^2}} = \frac{(2 - b)(2 + b)}{2 + b} = 2 - b$.
Given $d_{max} = 1$,we have $2 - b = 1$,so $b = 1$.
The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(B) Given the directrix $x = \frac{a}{e} = 8$ and focus $(ae, 0) = (2, 0)$.
From these,$ae = 2$ and $\frac{a}{e} = 8$.
Multiplying gives $a^2 = 16$,so $a = 4$. Then $e = \frac{2}{4} = \frac{1}{2}$.
$b^2 = a^2(1 - e^2) = 16(1 - \frac{1}{4}) = 16(\frac{3}{4}) = 12$,so $b = 2\sqrt{3}$.
The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{12} = 1$.
The tangent at $P(4\cos\theta, 2\sqrt{3}\sin\theta)$ is $\frac{x\cos\theta}{4} + \frac{y\sin\theta}{2\sqrt{3}} = 1$.
Since it passes through $(0, 4\sqrt{3})$,we have $\frac{0}{4} + \frac{4\sqrt{3}\sin\theta}{2\sqrt{3}} = 1$,which gives $2\sin\theta = 1$,so $\sin\theta = \frac{1}{2}$.
Thus,$\theta = 30^\circ$. Point $P$ is $(4\cos 30^\circ, 2\sqrt{3}\sin 30^\circ) = (2\sqrt{3}, \sqrt{3})$.
The tangent equation is $\frac{x\cos 30^\circ}{4} + \frac{y\sin 30^\circ}{2\sqrt{3}} = 1$ $\Rightarrow \frac{x\sqrt{3}}{8} + \frac{y}{4\sqrt{3}} = 1$.
For $Q$,set $y = 0$: $\frac{x\sqrt{3}}{8} = 1 \Rightarrow x = \frac{8}{\sqrt{3}}$. So $Q = (\frac{8}{\sqrt{3}}, 0)$.
$PQ^2 = (2\sqrt{3} - \frac{8}{\sqrt{3}})^2 + (\sqrt{3} - 0)^2 = (\frac{6-8}{\sqrt{3}})^2 + 3 = \frac{4}{3} + 3 = \frac{13}{3}$.
$(3PQ)^2 = 9 \times PQ^2 = 9 \times \frac{13}{3} = 39$.

(C) Let $p$ be the number of persons who speak both languages.
Given:
$\alpha + p = 75$ (Only English)
$\beta + p = 40$ (Only Hindi)
$\alpha + \beta + p = 100$ (Total persons)
Adding the first two equations: $\alpha + \beta + 2p = 115$.
Subtracting the third equation: $p = 115 - 100 = 15$.
Then,$\alpha = 75 - 15 = 60$ and $\beta = 40 - 15 = 25$.
The equation of the ellipse is $25(\beta^2 x^2 + \alpha^2 y^2) = \alpha^2 \beta^2$.
Dividing by $25 \alpha^2 \beta^2$: $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = \frac{1}{25}$.
This is $\frac{x^2}{(\alpha/5)^2} + \frac{y^2}{(\beta/5)^2} = 1$.
Here,$a = \frac{60}{5} = 12$ and $b = \frac{25}{5} = 5$.
Since $a > b$,the eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{144}} = \sqrt{\frac{119}{144}} = \frac{\sqrt{119}}{12}$.

(C) The ellipse is $E : \frac{x^2}{9} + \frac{y^2}{1} = 1$. The points of intersection with the positive axes are $A(3, 0)$ and $B(0, 1)$.
The major axis of $E$ lies along the $x$-axis with length $2a = 6$. Thus,the circle $C$ with the major axis as diameter has center $(0, 0)$ and radius $r = 3$. The equation of circle $C$ is $x^2 + y^2 = 9$.
The line passing through $A(3, 0)$ and $B(0, 1)$ has the equation $\frac{x}{3} + \frac{y}{1} = 1$,or $x + 3y = 3$. Substituting $x = 3 - 3y$ into the circle equation $x^2 + y^2 = 9$:
$(3 - 3y)^2 + y^2 = 9$
$9 - 18y + 9y^2 + y^2 = 9$
$10y^2 - 18y = 0$
$2y(5y - 9) = 0$
So,$y = 0$ (which corresponds to point $A$) or $y = \frac{9}{5}$.
If $y = \frac{9}{5}$,then $x = 3 - 3(\frac{9}{5}) = 3 - \frac{27}{5} = -\frac{12}{5}$. Thus,$P = (-\frac{12}{5}, \frac{9}{5})$.
The area of triangle $OAP$ with vertices $O(0, 0)$,$A(3, 0)$,and $P(-\frac{12}{5}, \frac{9}{5})$ is given by $\frac{1}{2} |x_O(y_A - y_P) + x_A(y_P - y_O) + x_P(y_O - y_A)| = \frac{1}{2} |0 + 3(\frac{9}{5} - 0) + (-\frac{12}{5})(0 - 0)| = \frac{1}{2} |\frac{27}{5}| = \frac{27}{10}$.
Here $m = 27$ and $n = 10$,which are coprime. Therefore,$m - n = 27 - 10 = 17$.

(A) The equation of the ellipse is $kx^{2} + k^{2}y^{2} = 1$,which can be written as $\frac{x^{2}}{1/k} + \frac{y^{2}}{1/k^{2}} = 1$.
The end points on the axes are $(\pm 1/\sqrt{k}, 0)$ and $(0, \pm 1/k)$.
The equation of the chord in the first quadrant joining $(1/\sqrt{k}, 0)$ and $(0, 1/k)$ is $\frac{x}{1/\sqrt{k}} + \frac{y}{1/k} = 1$,which simplifies to $\sqrt{k}x + ky = 1$.
The radius $r_{k}$ of the circle $C_{k}$ is the perpendicular distance from the origin $(0,0)$ to this line:
$r_{k} = \frac{|0 + 0 - 1|}{\sqrt{(\sqrt{k})^{2} + k^{2}}} = \frac{1}{\sqrt{k + k^{2}}}$.
Therefore,$\frac{1}{r_{k}^{2}} = k + k^{2}$.
We need to calculate $\sum_{k=1}^{20} \frac{1}{r_{k}^{2}} = \sum_{k=1}^{20} (k + k^{2}) = \sum_{k=1}^{20} k + \sum_{k=1}^{20} k^{2}$.
Using the summation formulas $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$ for $n=20$:
$\sum_{k=1}^{20} k = \frac{20 \times 21}{2} = 210$.
$\sum_{k=1}^{20} k^{2} = \frac{20 \times 21 \times 41}{6} = 10 \times 7 \times 41 = 2870$.
Thus,$\sum_{k=1}^{20} \frac{1}{r_{k}^{2}} = 210 + 2870 = 3080$.

(C) The equation of the circle with centre $(2,0)$ and radius $r$ is $(x-2)^2 + y^2 = r^2$.
The equation of the ellipse is $x^2 + 4y^2 = 36$,which implies $y^2 = \frac{36-x^2}{4}$.
Substituting $y^2$ into the circle equation:
$(x-2)^2 + \frac{36-x^2}{4} = r^2$
$4(x^2 - 4x + 4) + 36 - x^2 = 4r^2$
$4x^2 - 16x + 16 + 36 - x^2 = 4r^2$
$3x^2 - 16x + 52 - 4r^2 = 0$.
For the circle to be inscribed,the quadratic equation must have a discriminant $D = 0$ for tangency:
$D = (-16)^2 - 4(3)(52 - 4r^2) = 0$
$256 - 12(52 - 4r^2) = 0$
$256 - 624 + 48r^2 = 0$
$48r^2 = 368$
$12r^2 = \frac{368}{4} = 92$.

(C) The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
Since $PQ$ and $RS$ are chords passing through the origin,$P, Q$ are symmetric about the origin,so $O$ is the midpoint of $PQ$ and $RS$.
Thus,$PQ = 2OP$ and $RS = 2OR$.
We have $\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4(OP)^2} + \frac{1}{4(OR)^2} = \frac{1}{4} \left( \frac{1}{(OP)^2} + \frac{1}{(OR)^2} \right)$.
Let $P = (2 \cos \alpha, 3 \sin \alpha)$ and $R = (2 \cos \theta, 3 \sin \theta)$.
Since $OP \perp OR$,the product of their slopes is $-1$: $\left( \frac{3 \sin \alpha}{2 \cos \alpha} \right) \left( \frac{3 \sin \theta}{2 \cos \theta} \right) = -1$ $\Rightarrow \tan \alpha \tan \theta = -\frac{4}{9}$.
Given $P = \left( \frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}} \right)$,we have $\cos \alpha = \frac{\sqrt{3}}{\sqrt{7}}$ and $\sin \alpha = \frac{2}{\sqrt{7}}$,so $\tan \alpha = \frac{2}{\sqrt{3}}$.
Then $\tan \theta = -\frac{4}{9} \times \frac{\sqrt{3}}{2} = -\frac{2 \sqrt{3}}{9}$.
Using $(OP)^2 = 4 \cos^2 \alpha + 9 \sin^2 \alpha = 4 \left( \frac{3}{7} \right) + 9 \left( \frac{4}{7} \right) = \frac{12+36}{7} = \frac{48}{7}$.
Using $(OR)^2 = 4 \cos^2 \theta + 9 \sin^2 \theta = \frac{4}{\sec^2 \theta} + \frac{9 \tan^2 \theta}{\sec^2 \theta} = \frac{4 + 9 \tan^2 \theta}{1 + \tan^2 \theta} = \frac{4 + 9(12/81)}{1 + 12/81} = \frac{4 + 4/3}{1 + 4/27} = \frac{16/3}{31/27} = \frac{16}{3} \times \frac{27}{31} = \frac{144}{31}$.
Substituting these: $\frac{1}{4} \left( \frac{7}{48} + \frac{31}{144} \right) = \frac{1}{4} \left( \frac{21+31}{144} \right) = \frac{52}{4 \times 144} = \frac{13}{144}$.
Thus $p=13, q=144$,so $p+q = 157$.

(A) The equation of the ellipse is $\frac{(x-1)^2}{a^2} + \frac{y^2}{b^2} = 1$.
The length of the latus rectum is $\frac{2b^2}{a} = \frac{1}{2}$,which implies $4b^2 = a$ (Equation $1$).
The minor axis endpoints are $(1, b)$ and $(1, -b)$,and the foci are $(1 \pm ae, 0)$. The angle subtended by the minor axis at a focus is $60^{\circ}$.
Considering the triangle formed by the focus $(1+ae, 0)$ and the endpoints of the minor axis $(1, b)$ and $(1, -b)$,the angle at the focus is $60^{\circ}$,so the half-angle is $30^{\circ}$.
Thus,$\tan(30^{\circ}) = \frac{b}{ae} = \frac{1}{\sqrt{3}}$,which implies $ae = b\sqrt{3}$.
Squaring both sides,$a^2e^2 = 3b^2$. Since $a^2e^2 = a^2 - b^2$,we have $a^2 - b^2 = 3b^2$,so $a^2 = 4b^2$ (Equation $2$).
From Equation $1$,$b^2 = \frac{a}{4}$. Substituting this into Equation $2$,$a^2 = 4(\frac{a}{4}) = a$.
Since $a > 0$,we get $a = 1$. Then $b^2 = \frac{1}{4}$,so $b = \frac{1}{2}$.
The length of the major axis is $2a = 2(1) = 2$,and the length of the minor axis is $2b = 2(\frac{1}{2}) = 1$.
The square of the sum of the lengths is $(2a + 2b)^2 = (2 + 1)^2 = 3^2 = 9$.

(A) The equation of the chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with mid-point $(x_1, y_1)$ is given by $T=S_1$,where $T = \frac{xx_1}{a^2}+\frac{yy_1}{b^2}$ and $S_1 = \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$.
Given $a^2=25, b^2=16$ and $(x_1, y_1) = (1, \frac{2}{5})$.
$T = \frac{x(1)}{25}+\frac{y(2/5)}{16} = \frac{x}{25}+\frac{y}{40}$.
$S_1 = \frac{1^2}{25}+\frac{(2/5)^2}{16}-1 = \frac{1}{25}+\frac{4/25}{16}-1 = \frac{1}{25}+\frac{1}{100}-1 = \frac{4+1-100}{100} = -\frac{95}{100} = -\frac{19}{20}$.
So,$\frac{x}{25}+\frac{y}{40} = -\frac{19}{20}$.
Multiplying by $200$,we get $8x+5y = -190$,or $y = \frac{-8x-190}{5}$.
Substituting this into the ellipse equation $\frac{x^2}{25}+\frac{y^2}{16}=1$ leads to a quadratic in $x$. The length of the chord is given by $L = \frac{2b}{a^2} \sqrt{a^2x_1^2+b^2y_1^2} \sqrt{a^2+b^2-x_1^2-y_1^2}$ (or using the distance formula between intersection points).
Calculating the distance,we obtain $L = \frac{\sqrt{1691}}{5}$.

(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The length of the minor axis is $2b$.
The distance between the foci is $2ae$.
According to the problem,$2b = \frac{1}{2} (2ae) = ae$.
Thus,$\frac{b}{a} = \frac{e}{2}$.
We know that for an ellipse,$b^2 = a^2(1 - e^2)$,which implies $\frac{b^2}{a^2} = 1 - e^2$.
Substituting $\frac{b}{a} = \frac{e}{2}$,we get $(\frac{e}{2})^2 = 1 - e^2$.
$\frac{e^2}{4} = 1 - e^2$.
$e^2 + \frac{e^2}{4} = 1$.
$\frac{5e^2}{4} = 1$.
$e^2 = \frac{4}{5}$.
$e = \frac{2}{\sqrt{5}}$.