Ellipse Questions in English

(A) The equation of a chord of a conic $S=0$ with mid-point $(x_1, y_1)$ is given by $T=S_1$.
Given equation of ellipse: $S = x^2+4y^2-2x+20y = 0$.
Mid-point $(x_1, y_1) = (2, -4)$.
$T = x(x_1) + 4y(y_1) - (x+x_1) + 10(y+y_1) = 0$.
Substituting $(x_1, y_1) = (2, -4)$:
$T = 2x + 4y(-4) - (x+2) + 10(y-4) = 2x - 16y - x - 2 + 10y - 40 = x - 6y - 42$.
$S_1 = (2)^2 + 4(-4)^2 - 2(2) + 20(-4) = 4 + 64 - 4 - 80 = -16$.
Equating $T = S_1$:
$x - 6y - 42 = -16$.
$x - 6y = 42 - 16$.
$x - 6y = 26$.

(C) Let the point $P$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $(a \cos \theta, b \sin \theta)$. Here $a^2=9$ and $b^2=4$,so $a=3, b=2$. Thus $P = (3 \cos \theta, 2 \sin \theta)$.
The auxiliary circle is $x^2 + y^2 = a^2 = 9$. The point $Q$ on the auxiliary circle corresponding to $P$ is $(3 \cos \theta, 3 \sin \theta)$.
The normal to the ellipse at $P(3 \cos \theta, 2 \sin \theta)$ is $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2 - b^2$,which is $\frac{3x}{\cos \theta} - \frac{2y}{\sin \theta} = 9 - 4 = 5$.
The normal to the circle at $Q(3 \cos \theta, 3 \sin \theta)$ passes through the center $(0,0)$ and $Q$,so its equation is $y = x \tan \theta$,or $x \sin \theta - y \cos \theta = 0$.
Let $R = (h, k)$. Since $R$ lies on both normals,we have $\frac{3h}{\cos \theta} - \frac{2k}{\sin \theta} = 5$ and $h \sin \theta = k \cos \theta$.
From $h \sin \theta = k \cos \theta$,we get $\sin \theta = \frac{k \cos \theta}{h}$. Substituting into the first equation: $\frac{3h}{\cos \theta} - \frac{2kh}{k \cos \theta} = 5 \implies \frac{3h-2h}{\cos \theta} = 5 \implies \cos \theta = \frac{h}{5}$.
Then $\sin \theta = \frac{k}{h} \cdot \frac{h}{5} = \frac{k}{5}$.
Using $\sin^2 \theta + \cos^2 \theta = 1$,we get $(\frac{k}{5})^2 + (\frac{h}{5})^2 = 1$,which simplifies to $h^2 + k^2 = 25$.
Thus,the locus of $R$ is $x^2 + y^2 = 25$.

(D) The coordinates of points $P$ and $Q$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $P = (a \cos \theta, b \sin \theta)$ and $Q = (a \cos(\frac{\pi}{2}+\theta), b \sin(\frac{\pi}{2}+\theta)) = (-a \sin \theta, b \cos \theta)$.
Let the midpoint of $PQ$ be $(x, y)$. Then,
$x = \frac{a(\cos \theta - \sin \theta)}{2} \Rightarrow \frac{2x}{a} = \cos \theta - \sin \theta$
$y = \frac{b(\sin \theta + \cos \theta)}{2} \Rightarrow \frac{2y}{b} = \sin \theta + \cos \theta$
Squaring and adding these equations:
$(\frac{2x}{a})^2 + (\frac{2y}{b})^2 = (\cos \theta - \sin \theta)^2 + (\sin \theta + \cos \theta)^2$
$\frac{4x^2}{a^2} + \frac{4y^2}{b^2} = (\cos^2 \theta + \sin^2 \theta - 2 \sin \theta \cos \theta) + (\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta) = 2$
$\frac{x^2}{a^2/2} + \frac{y^2}{b^2/2} = 1$
Comparing this with $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$,we get $\alpha = \frac{a}{\sqrt{2}}$ and $\beta = \frac{b}{\sqrt{2}}$.
Therefore,$\frac{a+b}{\alpha+\beta} = \frac{a+b}{\frac{a}{\sqrt{2}} + \frac{b}{\sqrt{2}}} = \frac{a+b}{\frac{1}{\sqrt{2}}(a+b)} = \sqrt{2}$.

(B) Given the equation of the conic:
$a^2 x^2 + b^2 y^2 = a^2(a^2 + b^2 - y^2)$
$\Rightarrow a^2 x^2 + b^2 y^2 + a^2 y^2 = a^2(a^2 + b^2)$
$\Rightarrow a^2 x^2 + y^2(a^2 + b^2) = a^2(a^2 + b^2)$
Dividing by $a^2(a^2 + b^2)$:
$\frac{x^2}{a^2 + b^2} + \frac{y^2}{a^2} = 1$
This is an ellipse with $A^2 = a^2 + b^2$ and $B^2 = a^2$. Since $A^2 > B^2$,the eccentricity $e$ is given by:
$e = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{a^2}{a^2 + b^2}} = \sqrt{\frac{b^2}{a^2 + b^2}} = \frac{b}{\sqrt{a^2 + b^2}}$
By the definition of a conic section,the distance from a point $P$ to the focus $S$ is $SP = e \cdot PM$,where $PM$ is the perpendicular distance to the directrix.
Thus,$PM = \frac{1}{e} SP$.
Comparing this with $PM = K SP$,we get $K = \frac{1}{e}$.
$K = \frac{\sqrt{a^2 + b^2}}{b}$.

(C) Given the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$,where $a^2=2$ and $b^2=1$.
The equation of the tangent at any point $(\sqrt{2}\cos \theta, \sin \theta)$ is given by $\frac{x \cos \theta}{\sqrt{2}} + y \sin \theta = 1$.
The tangent intersects the $x$-axis at point $A$ where $y=0$,so $A = (\frac{\sqrt{2}}{\cos \theta}, 0)$.
The tangent intersects the $y$-axis at point $B$ where $x=0$,so $B = (0, \frac{1}{\sin \theta})$.
Let $(h, k)$ be the midpoint of the segment $AB$. Then $h = \frac{\sqrt{2}}{2 \cos \theta}$ and $k = \frac{1}{2 \sin \theta}$.
From these,we have $\cos \theta = \frac{\sqrt{2}}{2h} = \frac{1}{\sqrt{2}h}$ and $\sin \theta = \frac{1}{2k}$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we get $(\frac{1}{2k})^2 + (\frac{1}{\sqrt{2}h})^2 = 1$.
This simplifies to $\frac{1}{4k^2} + \frac{1}{2h^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(D) The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
The equation of the tangent at point $(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1$.
The tangent meets the $x$-axis at $A\left(\frac{a}{\cos \theta}, 0\right)$ and the $y$-axis at $B\left(0, \frac{b}{\sin \theta}\right)$.
Let $(x, y)$ be the midpoint of $AB$. Then:
$x = \frac{a}{2 \cos \theta} \Rightarrow \cos \theta = \frac{a}{2x}$
$y = \frac{b}{2 \sin \theta} \Rightarrow \sin \theta = \frac{b}{2y}$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$\left(\frac{b}{2y}\right)^2 + \left(\frac{a}{2x}\right)^2 = 1$
$\frac{b^2}{4y^2} + \frac{a^2}{4x^2} = 1$
$\frac{a^2}{x^2} + \frac{b^2}{y^2} = 4$.
Thus,the locus of the midpoint is $\frac{a^2}{x^2} + \frac{b^2}{y^2} = 4$.

(B) The given equation of the ellipse is $3x^2 + 4y^2 = 12$,which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Here,$a^2 = 4$ and $b^2 = 3$,so $a = 2$ and $b = \sqrt{3}$.
The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$.
The coordinates of the end of the latus rectum in the third quadrant are $(-ae, -\frac{b^2}{a}) = (-2 \times \frac{1}{2}, -\frac{3}{2}) = (-1, -\frac{3}{2})$.
The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.
Substituting $(x_1, y_1) = (-1, -\frac{3}{2})$,$a^2 = 4$,and $b^2 = 3$:
$\frac{4x}{-1} - \frac{3y}{-3/2} = 4 - 3$ $\Rightarrow -4x + 2y = 1$ $\Rightarrow y = 2x + \frac{1}{2}$.
Substituting $y = 2x + \frac{1}{2}$ into the ellipse equation $3x^2 + 4y^2 = 12$:
$3x^2 + 4(2x + \frac{1}{2})^2 = 12$ $\Rightarrow 3x^2 + 4(4x^2 + 2x + \frac{1}{4}) = 12$ $\Rightarrow 3x^2 + 16x^2 + 8x + 1 = 12$.
$19x^2 + 8x - 11 = 0$.
Factoring the quadratic: $(x + 1)(19x - 11) = 0$.
The roots are $x = -1$ (the point $L_1^{\prime}$) and $x = \frac{11}{19}$ (the point $P$).
Thus,$a = \frac{11}{19}$.

(D) Given the ellipse equation $\frac{x^2}{4} + y^2 = 1$ and the line $y = x + 1$.
Substitute $y = x + 1$ into the ellipse equation:
$\frac{x^2}{4} + (x + 1)^2 = 1$
$\frac{x^2}{4} + x^2 + 2x + 1 = 1$
$\frac{5x^2}{4} + 2x = 0$
$x(\frac{5x}{4} + 2) = 0$
So,$x_1 = 0$ and $x_2 = -\frac{8}{5}$.
Corresponding $y$ values are $y_1 = 0 + 1 = 1$ and $y_2 = -\frac{8}{5} + 1 = -\frac{3}{5}$.
The points of intersection are $P(0, 1)$ and $Q(-\frac{8}{5}, -\frac{3}{5})$.
The length of the chord $PQ$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Length $= \sqrt{(-\frac{8}{5} - 0)^2 + (-\frac{3}{5} - 1)^2}$
$= \sqrt{(-\frac{8}{5})^2 + (-\frac{8}{5})^2} = \sqrt{\frac{64}{25} + \frac{64}{25}} = \sqrt{\frac{128}{25}} = \frac{8\sqrt{2}}{5}$.

(B) For the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$,we have $a^2=9$ and $b^2=4$.
Thus,$a=3$ and $b=2$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
The focus $S$ on the positive $X$-axis is $(ae, 0) = (3 \times \frac{\sqrt{5}}{3}, 0) = (\sqrt{5}, 0)$.
$A$ point $P$ on the ellipse is $(3 \cos \theta, 2 \sin \theta)$.
The distance $SP$ is given by the focal distance formula $SP = a - ex$ for a focus on the positive $X$-axis,where $x = 3 \cos \theta$.
So,$SP = 3 - (\frac{\sqrt{5}}{3})(3 \cos \theta) = 3 - \sqrt{5} \cos \theta$.
Given $SP = 1$,we have $1 = 3 - \sqrt{5} \cos \theta$.
$\sqrt{5} \cos \theta = 2$.
$\cos \theta = \frac{2}{\sqrt{5}}$.

(C) The given curve is $4x^2 + 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
This is an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a^2 = 9$ $(a = 3)$ and $b^2 = 4$ $(b = 2)$.
The parametric coordinates of any point on the ellipse are $(a \cos \theta, b \sin \theta) = (3 \cos \theta, 2 \sin \theta)$.
The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $\theta$ is given by $ax \sec \theta - by \operatorname{cosec} \theta = a^2 - b^2$.
Substituting $a = 3$,$b = 2$,and $\theta = \frac{7\pi}{4}$:
$3x \sec(\frac{7\pi}{4}) - 2y \operatorname{cosec}(\frac{7\pi}{4}) = 9 - 4$
Since $\sec(\frac{7\pi}{4}) = \sec(2\pi - \frac{\pi}{4}) = \sec(\frac{\pi}{4}) = \sqrt{2}$ and $\operatorname{cosec}(\frac{7\pi}{4}) = \operatorname{cosec}(2\pi - \frac{\pi}{4}) = -\operatorname{cosec}(\frac{\pi}{4}) = -\sqrt{2}$:
$3x(\sqrt{2}) - 2y(-\sqrt{2}) = 5$
$3\sqrt{2}x + 2\sqrt{2}y = 5$
Thus,the equation of the normal is $3\sqrt{2}x + 2\sqrt{2}y - 5 = 0$.

(A) We know that two conics $\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1$ and $\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=1$ intersect orthogonally if and only if $a_1^2-b_1^2 = a_2^2-b_2^2$,which can be rearranged as $a_1^2-a_2^2 = b_1^2-b_2^2$.
Given the curves are $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$.
Applying the condition for orthogonal intersection:
$a^2-25 = b^2-16$
Rearranging the terms to find $a^2-b^2$:
$a^2-b^2 = 25-16$
$a^2-b^2 = 9$
Thus,the value is $9$.

(B) Let the foci be $F_1 = (ae, 0)$ and $F_2 = (-ae, 0)$,and the extremity of the minor axis be $B = (0, b)$.
The lines $BF_1$ and $BF_2$ are inclined at an angle of $60^{\circ}$ to each other,so the angle each makes with the $y$-axis is $30^{\circ}$.
Thus,the slope of $BF_1$ is $\tan(90^{\circ} - 30^{\circ}) = \tan 60^{\circ} = \sqrt{3}$.
The slope of $BF_1$ is also given by $\frac{b - 0}{0 - ae} = -\frac{b}{ae}$.
Since the angle is $60^{\circ}$ between the lines,the angle between $BF_1$ and the $y$-axis is $30^{\circ}$,so $\frac{ae}{b} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
Therefore,$\frac{b}{ae} = \sqrt{3}$,which implies $b = ae\sqrt{3}$.
Using the relation $e^2 = 1 - \frac{b^2}{a^2}$,we substitute $b^2 = 3a^2e^2$:
$e^2 = 1 - \frac{3a^2e^2}{a^2} = 1 - 3e^2$.
$4e^2 = 1 \Rightarrow e^2 = \frac{1}{4}$.
Since $e > 0$,we have $e = \frac{1}{2}$.

(C) The given equation of the ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$.
Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a^{2}=25$ and $b^{2}=9$,so $a=5$ and $b=3$.
The ends of the minor axis are $B(0, 3)$ and $B^{\prime}(0, -3)$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
The foci $S$ and $S^{\prime}$ are $(\pm ae, 0) = (\pm 5 \times \frac{4}{5}, 0) = (\pm 4, 0)$.
Thus,$S(4, 0)$ and $S^{\prime}(-4, 0)$.
The rhombus $SBS^{\prime}B^{\prime}$ has diagonals $SS^{\prime}$ and $BB^{\prime}$.
The length of diagonal $SS^{\prime} = 4 - (-4) = 8$.
The length of diagonal $BB^{\prime} = 3 - (-3) = 6$.
The area of a rhombus is $\frac{1}{2} \times \text{product of diagonals} = \frac{1}{2} \times 8 \times 6 = 24 \text{ sq. unit}$.

(C) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Foci are $S(ae, 0)$ and $T(-ae, 0)$.
$B(0, b)$ is the end point of the minor axis.
Since $\triangle STB$ is an equilateral triangle,the distance $SB = ST = TB$.
$ST = ae - (-ae) = 2ae$.
$SB = \sqrt{(ae-0)^{2} + (0-b)^{2}} = \sqrt{a^{2}e^{2} + b^{2}}$.
Since $SB = ST$,we have $SB^{2} = ST^{2}$.
$a^{2}e^{2} + b^{2} = (2ae)^{2} = 4a^{2}e^{2}$.
$b^{2} = 3a^{2}e^{2}$.
Using the relation $b^{2} = a^{2}(1-e^{2})$,we get:
$a^{2}(1-e^{2}) = 3a^{2}e^{2}$.
$1 - e^{2} = 3e^{2}$.
$4e^{2} = 1$.
$e^{2} = \frac{1}{4}$.
$e = \frac{1}{2}$ (since eccentricity $e > 0$).

(C) The given equation of the ellipse is $3x^{2} + 4y^{2} = 48$.
Dividing by $48$,we get $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$.
Here,$a^{2} = 16$ and $b^{2} = 12$,so $a = 4$ and $b = 2\sqrt{3}$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{12}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2}$.
The coordinates of the extremity of the latus rectum $P$ in the first quadrant are $(ae, \frac{b^{2}}{a}) = (4 \times \frac{1}{2}, \frac{12}{4}) = (2, 3)$.
Let the eccentric angle of $P$ be $\theta$. Then $P = (a \cos \theta, b \sin \theta) = (4 \cos \theta, 2\sqrt{3} \sin \theta)$.
Equating the coordinates: $4 \cos \theta = 2 \Rightarrow \cos \theta = \frac{1}{2}$ and $2\sqrt{3} \sin \theta = 3 \Rightarrow \sin \theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Since $\cos \theta = \frac{1}{2}$ and $\sin \theta = \frac{\sqrt{3}}{2}$,we have $\theta = \frac{\pi}{3}$.

(C) The given equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$.
Here,$a^{2}=144$ and $b^{2}=25$.
The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{25}{144}}=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}$.
The foci of the ellipse are $(\pm ae, 0) = (\pm 12 \times \frac{\sqrt{119}}{12}, 0) = (\pm \sqrt{119}, 0)$.
The circle has its center at $(0, \sqrt{2})$ and passes through the foci $(\pm \sqrt{119}, 0)$.
The radius $r$ of the circle is the distance between the center $(0, \sqrt{2})$ and one of the foci $(\sqrt{119}, 0)$.
$r = \sqrt{(\sqrt{119}-0)^{2} + (0-\sqrt{2})^{2}}$
$r = \sqrt{119 + 2} = \sqrt{121} = 11$.

(C) Given that the distance between the foci of an ellipse is equal to the length of the latus rectum.
The distance between the foci is $2ae$ and the length of the latus rectum is $\frac{2b^2}{a}$.
Therefore,$2ae = \frac{2b^2}{a} \implies ae = \frac{b^2}{a} \implies b^2 = a^2e$.
We know that for an ellipse,$b^2 = a^2(1 - e^2)$.
Substituting $b^2 = a^2e$ into the equation,we get $a^2e = a^2(1 - e^2)$.
Dividing by $a^2$ (since $a \neq 0$),we get $e = 1 - e^2$.
Rearranging gives the quadratic equation $e^2 + e - 1 = 0$.
Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.
Since eccentricity $e$ must be positive $(0 < e < 1)$,we take the positive root: $e = \frac{\sqrt{5} - 1}{2}$.

(B) Let $P(\sqrt{10} \cos \theta, \sqrt{8} \sin \theta)$ be the point on the ellipse $\frac{x^{2}}{10}+\frac{y^{2}}{8}=1$.
Given that the distance of $P$ from the centre $(0,0)$ is $3$ units.
So,$OP^2 = 3^2 = 9$.
$10 \cos^2 \theta + 8 \sin^2 \theta = 9$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we can write $9 = 9(\sin^2 \theta + \cos^2 \theta)$.
$10 \cos^2 \theta + 8 \sin^2 \theta = 9 \sin^2 \theta + 9 \cos^2 \theta$.
Rearranging the terms,we get $\cos^2 \theta = \sin^2 \theta$.
$\tan^2 \theta = 1$.
Since the point is in the first quadrant,$\theta = \frac{\pi}{4}$.

(B) Given the equation of the ellipse: $16x^2 + 25y^2 = 400$.
Divide both sides by $400$ to get the standard form: $\frac{16x^2}{400} + \frac{25y^2}{400} = 1$,which simplifies to $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 16$.
Thus,$a = 5$ and $b = 4$.
The length of the latus rectum of an ellipse is given by the formula $\frac{2b^2}{a}$.
Substituting the values: $\text{Length} = \frac{2 \times 16}{5} = \frac{32}{5}$ units.

(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $S(ae, 0)$ and $S'(-ae, 0)$,and the extremity of the minor axis is $B(0, b)$.
Given that the angle $\angle SBS' = 90^{\circ}$.
Since the triangle $\triangle SBS'$ is isosceles with $SB = S'B$,the altitude from $B$ to $SS'$ bisects the angle $\angle SBS'$.
Thus,$\angle OBS = 45^{\circ}$.
In the right-angled triangle $\triangle OBS$,we have $\tan(45^{\circ}) = \frac{OB}{OS} = \frac{b}{ae}$.
Since $\tan(45^{\circ}) = 1$,we get $1 = \frac{b}{ae}$,which implies $b = ae$.
Using the relation $b^2 = a^2(1 - e^2)$,we substitute $b^2 = a^2e^2$:
$a^2e^2 = a^2(1 - e^2)$
$e^2 = 1 - e^2$
$2e^2 = 1$
$e^2 = \frac{1}{2}$
$e = \frac{1}{\sqrt{2}}$.

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.
The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$.
The extremity of the minor axis is $B(0, b)$.
The slope of $SB$ is $m_1 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.
The slope of $S^{\prime}B$ is $m_2 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.
Since $\angle SBS^{\prime} = 90^{\circ}$,the product of the slopes is $-1$,so $m_1 \times m_2 = -1$.
$(-\frac{b}{ae}) \times (\frac{b}{ae}) = -1$.
$\frac{b^2}{a^2 e^2} = 1 \Rightarrow b^2 = a^2 e^2$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2 e^2$.
$1 - e^2 = e^2$ $\Rightarrow 2e^2 = 1$ $\Rightarrow e^2 = \frac{1}{2}$.
Since eccentricity $e > 0$,we have $e = \frac{1}{\sqrt{2}}$.

(B) The given equation of the ellipse is $x^{2}+2y^{2}=4$.
Dividing by $4$,we get $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$.
Comparing this with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we have $a^{2}=4$,so $a=2$.
The auxiliary circle of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by $x^{2}+y^{2}=a^{2}$.
Here,the equation of the auxiliary circle is $x^{2}+y^{2}=4$.
The coordinates of a point on the auxiliary circle corresponding to an eccentric angle $\theta$ are given by $(a \cos \theta, a \sin \theta)$.
Given $\theta = 60^{\circ}$ and $a=2$,the coordinates are $(2 \cos 60^{\circ}, 2 \sin 60^{\circ})$.
Substituting the values,we get $(2 \times \frac{1}{2}, 2 \times \frac{\sqrt{3}}{2}) = (1, \sqrt{3})$.
Thus,the correct option is $B$.

(A) The given equation of the ellipse is $16 x^{2}+25 y^{2}+32 x-100 y=284$.
Rearranging the terms,we get $16(x^{2}+2 x)+25(y^{2}-4 y)=284$.
Completing the square,$16(x^{2}+2 x+1)+25(y^{2}-4 y+4)=284+16+100$.
$16(x+1)^{2}+25(y-2)^{2}=400$.
Dividing by $400$,we get $\frac{(x+1)^{2}}{25}+\frac{(y-2)^{2}}{16}=1$.
Here,$a^{2}=25$. The auxiliary circle of an ellipse $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$ is $(x-h)^{2}+(y-k)^{2}=a^{2}$.
Thus,the equation of the auxiliary circle is $(x+1)^{2}+(y-2)^{2}=25$.
Expanding this,$x^{2}+2 x+1+y^{2}-4 y+4=25$.
Therefore,$x^{2}+y^{2}+2 x-4 y-20=0$.

(A) Let the equation of the chord $AB$ be $lx + my = 1$.
Homogenizing the equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the help of the chord equation $lx + my = 1$,we get:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = (lx + my)^2$
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = l^2x^2 + m^2y^2 + 2lmxy$
$x^2(\frac{1}{a^2} - l^2) + y^2(\frac{1}{b^2} - m^2) - 2lmxy = 0$
Since $AB$ subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero.
$(\frac{1}{a^2} - l^2) + (\frac{1}{b^2} - m^2) = 0$
$l^2 + m^2 = \frac{1}{a^2} + \frac{1}{b^2}$
Now,the distance of the chord $lx + my = 1$ from the origin is $p = \frac{1}{\sqrt{l^2 + m^2}}$.
If $A$ and $B$ are points on the ellipse,the coordinates can be represented such that $OA^2$ and $OB^2$ relate to the chord properties. For a chord subtending a right angle at the origin,the expression $\frac{1}{OA^2} + \frac{1}{OB^2}$ is constant and equal to $\frac{1}{a^2} + \frac{1}{b^2}$.

(A) Given the equation of the ellipse: $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $y = 2t^2$.
Substitute $y = 2t^2$ into the ellipse equation:
$\frac{x^2}{9} + \frac{(2t^2)^2}{4} = 1$
$\frac{x^2}{9} + \frac{4t^4}{4} = 1$
$\frac{x^2}{9} + t^4 = 1$
$x^2 = 9(1 - t^4)$
For the intersection points to be real,$x^2$ must be greater than or equal to $0$:
$9(1 - t^4) \geq 0$
$1 - t^4 \geq 0$
$t^4 \leq 1$
$(t^2 - 1)(t^2 + 1) \leq 0$
Since $t^2 + 1 > 0$ for all real $t$,we must have $t^2 - 1 \leq 0$.
$t^2 \leq 1$
$|t| \leq 1$

(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos \theta, b \sin \theta)$ is given by $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
To find the $x$-intercept $T$,set $y = 0$: $\frac{x \cos \theta}{a} = 1 \implies x = \frac{a}{\cos \theta}$. Thus,$OT = \frac{a}{\cos \theta}$.
To find the $y$-intercept $T_1$,set $x = 0$: $\frac{y \sin \theta}{b} = 1 \implies y = \frac{b}{\sin \theta}$. Thus,$OT_1 = \frac{b}{\sin \theta}$.
The product $(OT)(OT_1) = \frac{a}{\cos \theta} \cdot \frac{b}{\sin \theta} = \frac{ab}{\sin \theta \cos \theta} = \frac{2ab}{\sin 2\theta}$.
Since $0 < \theta < \frac{\pi}{2}$,the maximum value of $\sin 2\theta$ is $1$ (at $\theta = \frac{\pi}{4}$),which makes the expression $\frac{2ab}{\sin 2\theta}$ reach its minimum value.
Therefore,the minimum value is $2ab$.

(C) Let the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $a > b$.
Let $P(a \cos \theta, b \sin \theta)$ be a point on the ellipse.
The equation of the tangent at $P$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
The directrix is $x = \frac{a}{e}$.
Substituting $x = \frac{a}{e}$ into the tangent equation:
$\frac{(\frac{a}{e}) \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \implies \frac{\cos \theta}{e} + \frac{y \sin \theta}{b} = 1$.
$y = \frac{b(1 - \frac{\cos \theta}{e})}{\sin \theta} = \frac{b(e - \cos \theta)}{e \sin \theta}$.
Let $S(ae, 0)$ be the focus. The slope of $SP$ is $m_1 = \frac{b \sin \theta}{a \cos \theta - ae} = \frac{b \sin \theta}{a(\cos \theta - e)}$.
The slope of the line segment from the focus to the intersection point $Q(\frac{a}{e}, \frac{b(e - \cos \theta)}{e \sin \theta})$ is $m_2 = \frac{\frac{b(e - \cos \theta)}{e \sin \theta} - 0}{\frac{a}{e} - ae} = \frac{b(e - \cos \theta)}{e \sin \theta} \times \frac{e}{a(1 - e^{2})} = \frac{b(e - \cos \theta)}{a \sin \theta (1 - e^{2})}$.
Since $b^{2} = a^{2}(1 - e^{2})$,we have $1 - e^{2} = \frac{b^{2}}{a^{2}}$.
$m_2 = \frac{b(e - \cos \theta)}{a \sin \theta (\frac{b^{2}}{a^{2}})} = \frac{a(e - \cos \theta)}{b \sin \theta}$.
Note that $m_1 \times m_2 = \frac{b \sin \theta}{a(\cos \theta - e)} \times \frac{a(e - \cos \theta)}{b \sin \theta} = -1$.
Therefore,the angle subtended is $\frac{\pi}{2}$.

(B, D) The given equation of the ellipse is $4x^{2} + 9y^{2} = 1$.
Differentiating with respect to $x$,we get $8x + 18yy' = 0$,which implies $y' = -\frac{8x}{18y} = -\frac{4x}{9y}$.
The slope of the line $8x = 9y$ is $y = \frac{8}{9}x$,so the slope $m = \frac{8}{9}$.
Since the tangents are parallel to the line,their slopes must be equal: $-\frac{4x}{9y} = \frac{8}{9}$.
This simplifies to $-4x = 8y$,or $x = -2y$.
Substituting $x = -2y$ into the ellipse equation: $4(-2y)^{2} + 9y^{2} = 1$.
$4(4y^{2}) + 9y^{2} = 1$ $\Rightarrow 16y^{2} + 9y^{2} = 1$ $\Rightarrow 25y^{2} = 1$.
Thus,$y^{2} = \frac{1}{25}$,which gives $y = \pm \frac{1}{5}$.
If $y = \frac{1}{5}$,then $x = -2(\frac{1}{5}) = -\frac{2}{5}$.
If $y = -\frac{1}{5}$,then $x = -2(-\frac{1}{5}) = \frac{2}{5}$.
The required points are $\left(-\frac{2}{5}, \frac{1}{5}\right)$ and $\left(\frac{2}{5}, -\frac{1}{5}\right)$.

(C) For the ellipse $E_1: 3x^2 + 5y^2 = 32$,check the position of the point $(3,5)$ by substituting it into the equation: $3(3)^2 + 5(5)^2 = 3(9) + 5(25) = 27 + 125 = 152$. Since $152 > 32$,the point $(3,5)$ lies outside the ellipse $E_1$. Thus,$2$ tangents can be drawn to $E_1$.
For the ellipse $E_2: 25x^2 + 9y^2 = 450$,check the position of the point $(3,5)$: $25(3)^2 + 9(5)^2 = 25(9) + 9(25) = 225 + 225 = 450$. Since the result equals $450$,the point $(3,5)$ lies on the ellipse $E_2$. Thus,only $1$ tangent can be drawn to $E_2$.
The total number of tangents is $2 + 1 = 3$.

(C) The equation of the line is $y=x+\lambda$.
On comparing it with $y=mx+c$,we get $m=1$ and $c=\lambda$.
The equation of the ellipse is $2x^{2}+3y^{2}=1$,which can be written as $\frac{x^{2}}{1/2} + \frac{y^{2}}{1/3} = 1$.
On comparing it with $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we get $a^{2}=\frac{1}{2}$ and $b^{2}=\frac{1}{3}$.
If the line touches the ellipse,the condition for tangency is $c^{2}=a^{2}m^{2}+b^{2}$.
Substituting the values,we get $\lambda^{2} = \frac{1}{2}(1)^{2} + \frac{1}{3} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.
Therefore,$\lambda = \pm \sqrt{\frac{5}{6}}$.
Given the options,the correct value is $\lambda = \sqrt{\frac{5}{6}}$.

(A) Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$,we have $a^{2}=9$ and $b^{2}=5$,so $a=3$ and $b=\sqrt{5}$.
The eccentricity $e$ is given by $e = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
The foci are at $(\pm ae, 0) = (\pm 3 \times \frac{2}{3}, 0) = (\pm 2, 0)$.
The ends of the latus recta are at $(\pm 2, \pm \frac{b^{2}}{a}) = (\pm 2, \pm \frac{5}{3})$.
Consider the point $P(2, \frac{5}{3})$ in the first quadrant. The equation of the tangent at $P$ is $\frac{x(2)}{9} + \frac{y(5/3)}{5} = 1$,which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$,or $2x + 3y = 9$.
This line intersects the $x$-axis at $A(\frac{9}{2}, 0)$ and the $y$-axis at $B(0, 3)$.
The quadrilateral formed by the four tangents is symmetric about both axes. The area of the quadrilateral is $4 \times \text{Area}(\Delta OAB) = 4 \times (\frac{1}{2} \times OA \times OB) = 4 \times \frac{1}{2} \times \frac{9}{2} \times 3 = 27 \text{ sq units}$.

(C) Let the two perpendicular lines be the coordinate axes. Let the ends of the line segment be $A(0, m)$ and $B(n, 0)$.
The length of the line segment is $AB = \sqrt{m^2 + n^2} = a + b$,so $m^2 + n^2 = (a + b)^2$.
Let $P(h, k)$ be a point on the line segment $AB$ that divides it into parts of lengths $a$ and $b$. By the section formula,$P$ divides $AB$ in the ratio $b : a$ (since $AP = a$ and $PB = b$).
Thus,$h = \frac{b(0) + a(n)}{a + b} = \frac{an}{a + b} \Rightarrow n = \frac{(a + b)h}{a}$.
And $k = \frac{b(m) + a(0)}{a + b} = \frac{bm}{a + b} \Rightarrow m = \frac{(a + b)k}{b}$.
Substituting these into $m^2 + n^2 = (a + b)^2$,we get:
$\left(\frac{(a + b)k}{b}\right)^2 + \left(\frac{(a + b)h}{a}\right)^2 = (a + b)^2$.
Dividing by $(a + b)^2$,we obtain $\frac{k^2}{b^2} + \frac{h^2}{a^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is an ellipse.

(B) Let the focus be $F(0, 0)$ and the directrix be $x = 4$. Let the end of the minor axis be $B(h, k)$.
By the definition of an ellipse,the distance from the focus to the point $B$ is $e$ times the distance from $B$ to the directrix.
$BF = e \cdot BM$
$BF = \sqrt{h^2 + k^2}$ and $BM = |4 - h|$.
So,$\sqrt{h^2 + k^2} = e(4 - h)$.
Also,for an ellipse,the distance from the center to the focus is $ae$ and the distance from the center to the directrix is $a/e$. Since the focus is at $(0,0)$ and the directrix is $x=4$,the distance between them is $a/e - ae = 4$.
For the end of the minor axis,the distance from the center to the focus is $ae$,so $h = -ae$ and $k = b$.
From $a/e - ae = 4$,we have $a(1 - e^2) = 4e$,which implies $b^2/a = 4e$,so $b^2 = 4ae$.
Substituting $h = -ae$ and $k = b$,we get $k^2 = 4(-h) = -4h$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = -4x$,which is a parabola.

(C) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The positive end of the minor axis is $P(0, b)$.
Let $(h, k)$ be the midpoint of a chord passing through $(0, b)$ and some point $(x_1, y_1)$ on the ellipse.
Then,$h = \frac{x_1 + 0}{2} \Rightarrow x_1 = 2h$ and $k = \frac{y_1 + b}{2} \Rightarrow y_1 = 2k - b$.
Since $(x_1, y_1)$ lies on the ellipse,we have $\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$.
Substituting the values of $x_1$ and $y_1$,we get $\frac{(2h)^2}{a^2} + \frac{(2k - b)^2}{b^2} = 1$.
This simplifies to $\frac{4h^2}{a^2} + \frac{4(k - b/2)^2}{b^2} = 1$,which can be rewritten as $\frac{h^2}{(a/2)^2} + \frac{(k - b/2)^2}{(b/2)^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{(a/2)^2} + \frac{(y - b/2)^2}{(b/2)^2} = 1$,which represents an ellipse.

(D) Let the point on the ellipse be $P(x_0, y_0)$. The equation of the tangent at $P$ is $\frac{x x_0}{2} + y y_0 = 1$.
The intercepts of this tangent on the $x$-axis and $y$-axis are found by setting $y=0$ and $x=0$ respectively.
For $y=0$,$x = \frac{2}{x_0}$. For $x=0$,$y = \frac{1}{y_0}$.
Let the midpoint of the intercepted portion be $(h, k)$. Then $h = \frac{1}{x_0}$ and $k = \frac{1}{2 y_0}$.
This gives $x_0 = \frac{1}{h}$ and $y_0 = \frac{1}{2 k}$.
Since $(x_0, y_0)$ lies on the ellipse $\frac{x_0^2}{2} + y_0^2 = 1$,we substitute the values:
$\frac{(1/h)^2}{2} + (1/2k)^2 = 1$
$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(C) Given the equation of the ellipse is $x^{2}+4y^{2}=4$.
Dividing by $4$,we get $\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$.
The positive end of the minor axis is $B(0, 1)$.
Let the mid-point of the chord $BP$ be $M(h, k)$,where $P(x, y)$ is a point on the ellipse.
Then,$(h, k) = \left(\frac{0+x}{2}, \frac{1+y}{2}\right)$.
This implies $h = \frac{x}{2} \Rightarrow x = 2h$ and $k = \frac{1+y}{2} \Rightarrow y = 2k-1$.
Since $P(x, y)$ lies on the ellipse $x^{2}+4y^{2}=4$,substituting the values of $x$ and $y$ gives:
$(2h)^{2} + 4(2k-1)^{2} = 4$
$4h^{2} + 4(4k^{2} - 4k + 1) = 4$
$4h^{2} + 16k^{2} - 16k + 4 = 4$
$4h^{2} + 16k^{2} - 16k = 0$
Dividing by $4$,we get $h^{2} + 4k^{2} - 4k = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2} + 4(y^{2} - y) = 0$.
Completing the square: $x^{2} + 4(y^{2} - y + \frac{1}{4}) = 4(\frac{1}{4}) = 1$.
$x^{2} + 4(y - \frac{1}{2})^{2} = 1$.
Dividing by $1$,we get $\frac{x^{2}}{1} + \frac{(y - 1/2)^{2}}{1/4} = 1$.
This is an ellipse with centre $(0, 1/2)$,semi-major axis $a=1$,and semi-minor axis $b=1/2$.

(D) Given the ellipse equation is $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$.
Comparing with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we have $a^{2}=9$ and $b^{2}=1$.
The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$.
The foci are $(\pm ae, 0) = (\pm 3 \times \frac{2\sqrt{2}}{3}, 0) = (\pm 2\sqrt{2}, 0)$.
Let $P(h, k)$ be a point such that the foci $F_{1}(2\sqrt{2}, 0)$ and $F_{2}(-2\sqrt{2}, 0)$ subtend a right angle at $P$.
Thus,the product of the slopes of $PF_{1}$ and $PF_{2}$ is $-1$.
$\frac{k-0}{h-2\sqrt{2}} \times \frac{k-0}{h+2\sqrt{2}} = -1$.
$\frac{k^{2}}{h^{2}-(2\sqrt{2})^{2}} = -1$.
$k^{2} = -(h^{2}-8)$.
$h^{2}+k^{2} = 8$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}=8$.

(B) The equation of the ellipse is $x^{2}+9y^{2}=9$,which can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$.
First,find the intersection points:
$1$. Intersection of $x+y=1$ and $3y=x+3$:
Substitute $x=1-y$ into $3y=x+3$: $3y=(1-y)+3$ $\Rightarrow 4y=4$ $\Rightarrow y=1$. Then $x=0$. So,point $P$ is $(0, 1)$.
$2$. Intersection of $x+y=1$ and $x^{2}+9y^{2}=9$:
Substitute $x=1-y$ into the ellipse equation: $(1-y)^{2}+9y^{2}=9$ $\Rightarrow 1-2y+y^{2}+9y^{2}=9$ $\Rightarrow 10y^{2}-2y-8=0$ $\Rightarrow 5y^{2}-y-4=0$ $\Rightarrow (5y+4)(y-1)=0$. Thus $y=1$ (gives $P(0,1)$) or $y=-4/5$. If $y=-4/5$,$x=1-(-4/5)=9/5$. So,point $R$ is $(9/5, -4/5)$.
$3$. Intersection of $3y=x+3$ and $x^{2}+9y^{2}=9$:
Substitute $y=(x+3)/3$ into the ellipse equation: $x^{2}+9((x+3)/3)^{2}=9$ $\Rightarrow x^{2}+(x+3)^{2}=9$ $\Rightarrow x^{2}+x^{2}+6x+9=9$ $\Rightarrow 2x^{2}+6x=0$ $\Rightarrow 2x(x+3)=0$. Thus $x=0$ (gives $P(0,1)$) or $x=-3$. If $x=-3$,$y=(-3+3)/3=0$. So,point $Q$ is $(-3, 0)$.
The vertices of $\triangle PQR$ are $P(0, 1)$,$Q(-3, 0)$,and $R(9/5, -4/5)$.
The area of $\triangle PQR = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
$= \frac{1}{2} |0(0 - (-4/5)) + (-3)(-4/5 - 1) + (9/5)(1 - 0)|$
$= \frac{1}{2} |0 + (-3)(-9/5) + 9/5|$
$= \frac{1}{2} |27/5 + 9/5| = \frac{1}{2} |36/5| = \frac{18}{5}$.

(D) Given that $f$ is a strictly decreasing function on $\mathbb{R}$.
For the equation $\frac{x^2}{f(a^2+5a+3)} + \frac{y^2}{f(a+15)} = 1$ to represent an ellipse with the major axis along the $y$-axis,the denominator of the $y^2$ term must be greater than the denominator of the $x^2$ term.
Thus,$f(a+15) > f(a^2+5a+3)$.
Since $f$ is a strictly decreasing function,$f(x_1) > f(x_2)$ implies $x_1 < x_2$.
Therefore,$a+15 < a^2+5a+3$.
Rearranging the inequality,we get $a^2+4a-12 > 0$.
Factoring the quadratic expression,we have $(a+6)(a-2) > 0$.
Solving this inequality,we find that $a < -6$ or $a > 2$.
Thus,the interval for $a$ is $(-\infty, -6) \cup (2, \infty)$.

(B) Given center $C = (1, -2)$,focus $F_1 = (3, -2)$,and vertex $A_1 = (5, -2)$.
Since the $y$-coordinates are the same,the major axis is horizontal.
The distance from the center to the vertex is $a = |5 - 1| = 4$.
The distance from the center to the focus is $ae = |3 - 1| = 2$.
Thus,$e = \frac{2}{4} = \frac{1}{2}$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 4^2(1 - (\frac{1}{2})^2) = 16(1 - \frac{1}{4}) = 16(\frac{3}{4}) = 12$.
The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 \times 12}{4} = \frac{24}{4} = 6$.

(B) For $E_1$,the distance between foci is $2ae = 8$. Given $e = \frac{4}{5}$,we have $2a(\frac{4}{5}) = 8 \Rightarrow a = 5$.
Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 25(1 - \frac{16}{25}) = 25(\frac{9}{25}) = 9$.
The length of the latus rectum $\ell_1 = \frac{2b^2}{a} = \frac{2 \times 9}{5} = \frac{18}{5}$.
For $E_2$,$A < B$,so the eccentricity formula is $A^2 = B^2(1 - e^2) = B^2(1 - \frac{16}{25}) = \frac{9}{25}B^2$,which gives $A = \frac{3}{5}B$.
The length of the latus rectum $\ell_2 = \frac{2A^2}{B} = \frac{2(9/25)B^2}{B} = \frac{18}{25}B$.
Given $2\ell_1^2 = 9\ell_2$,we have $2(\frac{18}{5})^2 = 9(\frac{18}{25}B) \Rightarrow 2 \times \frac{324}{25} = \frac{162}{25}B$.
Solving for $B$,$B = \frac{2 \times 324}{162} = 4$.
The distance between the foci of $E_2$ is $2Be = 2 \times 4 \times \frac{4}{5} = \frac{32}{5}$.

(C) Let the point $(h, k)$ on the circle $x^2 + y^2 = 4$ be represented as $(2 \cos \theta, 2 \sin \theta)$.
Let the point $(x, y) = (2h + 1, 3k + 2)$.
Substituting $h = 2 \cos \theta$ and $k = 2 \sin \theta$,we get $x = 4 \cos \theta + 1$ and $y = 6 \sin \theta + 2$.
Rearranging,we have $\cos \theta = \frac{x - 1}{4}$ and $\sin \theta = \frac{y - 2}{6}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we get $(\frac{x - 1}{4})^2 + (\frac{y - 2}{6})^2 = 1$.
This is the equation of an ellipse with semi-major axis $a = 6$ and semi-minor axis $b = 4$.
The eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{16}{36} = 1 - \frac{4}{9} = \frac{5}{9}$.
Therefore,$\frac{5}{e^2} = \frac{5}{5/9} = 9$.

(C) The function is $f(t) = -t^{2} + 2t - \frac{3}{4}$.
To find the maximum value,we complete the square: $f(t) = -(t^{2} - 2t + 1) + 1 - \frac{3}{4} = -(t-1)^{2} + \frac{1}{4}$.
The maximum value is $e = \frac{1}{4}$,so $e^{2} = \frac{1}{16}$.
For an ellipse,$e^{2} = 1 - \frac{b^{2}}{a^{2}}$,so $\frac{b^{2}}{a^{2}} = 1 - \frac{1}{16} = \frac{15}{16} \Rightarrow b^{2} = \frac{15}{16}a^{2}$.
The length of the latus rectum is $\frac{2b^{2}}{a} = 30$,so $b^{2} = 15a$.
Equating the expressions for $b^{2}$: $\frac{15}{16}a^{2} = 15a$.
Since $a \neq 0$,we have $\frac{a}{16} = 1 \Rightarrow a = 16$.
Then $b^{2} = 15(16) = 240$.
Thus,$a^{2} + b^{2} = 16^{2} + 240 = 256 + 240 = 496$.

(D) The equation of the line is $\alpha x+4y-\sqrt{7}=0$. This line touches the ellipse $3x^{2}+4y^{2}=1$,which can be written as $\frac{x^{2}}{1/3} + \frac{y^{2}}{1/4} = 1$.
Comparing with the condition of tangency $c^{2}=a^{2}m^{2}+b^{2}$,where $m = -\frac{\alpha}{4}$ and $c = \frac{\sqrt{7}}{4}$:
$(\frac{\sqrt{7}}{4})^{2} = (\frac{1}{3})(\frac{-\alpha}{4})^{2} + \frac{1}{4} \implies \frac{7}{16} = \frac{\alpha^{2}}{48} + \frac{1}{4} \implies \frac{\alpha^{2}}{48} = \frac{3}{16} \implies \alpha^{2} = 9 \implies \alpha = \pm 3$.
Since $P$ is in the first quadrant,the tangent is $3x+4y=\sqrt{7}$.
The point of contact $P(x_{1}, y_{1})$ for the tangent $3x+4y=\sqrt{7}$ is given by $\frac{3x}{x_{1}} = \frac{4y}{y_{1}} = \frac{1}{1/7}$ is not correct,rather using $3xx_{1}+4yy_{1}=1$,we get $3x_{1} = \frac{3}{\sqrt{7}}$ and $4y_{1} = \frac{4}{\sqrt{7}}$,so $P = (\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}})$.
For the ellipse $\frac{x^{2}}{1/3} + \frac{y^{2}}{1/4} = 1$,$a^{2} = 1/3$ and $b^{2} = 1/4$. Eccentricity $e = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{1/4}{1/3}} = \sqrt{1-\frac{3}{4}} = \frac{1}{2}$.
The focal distances are $a \pm ex_{1} = \frac{1}{\sqrt{3}} \pm \frac{1}{2}(\frac{1}{\sqrt{7}}) = \frac{1}{\sqrt{3}} \pm \frac{1}{2\sqrt{7}}$.
Thus,one of the focal distances is $\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}}$.

(D) Given the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$,we have $a^2=25$ and $b^2=9$,so $a=5$ and $b=3$.
Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}$.
The foci are $S(ae, 0) = (4, 0)$ and $S^{\prime}(-ae, 0) = (-4, 0)$.
Since $P(\alpha, \beta)$ lies on the ellipse,$SP+S^{\prime}P=2a=10$.
Given $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$.
Using $(SP+S^{\prime}P)^2 = (SP)^2+(S^{\prime}P)^2+2SP \cdot S^{\prime}P$,we substitute $(SP)^2+(S^{\prime}P)^2 = (SP+S^{\prime}P)^2-2SP \cdot S^{\prime}P$.
So,$(SP+S^{\prime}P)^2-3SP \cdot S^{\prime}P=37$.
$10^2-3SP \cdot S^{\prime}P=37$ $\Rightarrow 100-3SP \cdot S^{\prime}P=37$ $\Rightarrow 3SP \cdot S^{\prime}P=63$ $\Rightarrow SP \cdot S^{\prime}P=21$.
For any point $P(\alpha, \beta)$ on the ellipse,$SP = a-e\alpha$ and $S^{\prime}P = a+e\alpha$.
$SP \cdot S^{\prime}P = a^2-e^2\alpha^2 = 25-\frac{16}{25}\alpha^2 = 21$.
$\frac{16}{25}\alpha^2 = 4 \Rightarrow \alpha^2 = \frac{100}{16} = \frac{25}{4}$.
From $\frac{\alpha^2}{25}+\frac{\beta^2}{9}=1$,we have $\frac{25/4}{25}+\frac{\beta^2}{9}=1$ $\Rightarrow \frac{1}{4}+\frac{\beta^2}{9}=1$ $\Rightarrow \frac{\beta^2}{9}=\frac{3}{4}$ $\Rightarrow \beta^2=\frac{27}{4}$.
Thus,$\alpha^2+\beta^2 = \frac{25}{4}+\frac{27}{4} = \frac{52}{4} = 13$.

(B) The equation of the line is $y = x + 1$. Substituting this into the ellipse equation $\frac{x^2}{2} + y^2 = 1$:
$\frac{x^2}{2} + (x+1)^2 = 1$
$\frac{x^2}{2} + x^2 + 2x + 1 = 1$
$\frac{3x^2}{2} + 2x = 0$
$x(\frac{3x}{2} + 2) = 0$
So,$x = 0$ or $x = -\frac{4}{3}$.
If $x = 0$,then $y = 1$,so $A = (0, 1)$.
If $x = -\frac{4}{3}$,then $y = -\frac{4}{3} + 1 = -\frac{1}{3}$,so $B = (-\frac{4}{3}, -\frac{1}{3})$.
The center of the ellipse is $O(0, 0)$.
The slope of $OA$ is $m_1 = \frac{1-0}{0-0} = \infty$ (vertical line,angle is $\frac{\pi}{2}$).
The slope of $OB$ is $m_2 = \frac{-1/3 - 0}{-4/3 - 0} = \frac{1}{4}$.
The angle $\theta$ that $OB$ makes with the negative $x$-axis is $\tan^{-1}(\frac{1}{4})$.
The angle $\angle AOB$ is the angle between the $y$-axis and the line $OB$,which is $\frac{\pi}{2} + \tan^{-1}(\frac{1}{4})$.

(A) Given the directrix $x = a/e = 9$ and eccentricity $e = 1/3$,we find $a = 9 \times (1/3) = 3$.
The focus $P$ is at $(ae, 0) = (3 \times 1/3, 0) = (1, 0)$,so $\alpha = 1$.
The equation of the ellipse is $x^2/a^2 + y^2/b^2 = 1$,where $b^2 = a^2(1 - e^2) = 9(1 - 1/9) = 8$.
Thus,the ellipse is $x^2/9 + y^2/8 = 1$.
The locus of the midpoint $(h, k)$ of a chord passing through a point $(x_0, y_0)$ for an ellipse $x^2/a^2 + y^2/b^2 = 1$ is given by $T = S_1$,which is $xh/a^2 + yk/b^2 = h^2/a^2 + k^2/b^2$.
Since the chord passes through $(1, 0)$,the equation of the chord with midpoint $(h, k)$ is $xh/9 + yk/8 = h^2/9 + k^2/8$.
Substituting $(x, y) = (1, 0)$ into this equation,we get $h/9 = h^2/9 + k^2/8$.
Multiplying by $72$,we get $8h = 8h^2 + 9k^2$,which simplifies to $9k^2 = 8h(1 - h)$.
Replacing $(h, k)$ with $(x, y)$,the locus is $9y^2 = 8x(1 - x)$.