Ellipse Questions in English

(A) For an ellipse,the reflected ray from one focus passes through the other focus.
Therefore,the incident ray is $PF_1$ and the reflected ray is $PF_2$.
The angle bisector of the angle between the incident ray $PF_1$ and the reflected ray $PF_2$ is the normal to the ellipse at point $P$.
The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$,where $a^2 = 4$ and $b^2 = 9$.
Since $b^2 > a^2$,the major axis is along the $y$-axis.
The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \frac{(a^2 - b^2)m}{\sqrt{a^2 + b^2m^2}}$.
Substituting $a^2 = 4$,$b^2 = 9$,and $m = 1$ (assuming the bisector has slope $1$ as per options):
$y = x \pm \frac{(4 - 9)(1)}{\sqrt{4 + 9(1)^2}} = x \pm \frac{-5}{\sqrt{13}}$.
Thus,the equation is $y = x \pm \frac{5}{\sqrt{13}}$.

(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Any point on this ellipse can be represented as $(3 \cos \theta, 2 \sin \theta)$.
Using the rational parameterization of trigonometric functions,we set $t = \tan(\theta/2)$.
Then $\cos \theta = \frac{1-t^2}{1+t^2}$ and $\sin \theta = \frac{2t}{1+t^2}$.
Substituting these,the coordinates are $\left(3 \frac{1-t^2}{1+t^2}, 2 \frac{2t}{1+t^2}\right) = \left(\frac{3(1-t^2)}{1+t^2}, \frac{4t}{1+t^2}\right)$.
If $t$ is a rational number,say $t = p/q$,then both $x$ and $y$ are rational numbers.
Since there are infinitely many rational values for $t$,there are infinitely many such points on the ellipse.
Therefore,the number of such points is more than $12$.

(B) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{24} = 1$,where $a^2 = 25$ and $b^2 = 24$.
The director circle of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $x^2 + y^2 = a^2 + b^2$.
Substituting the values,we get $x^2 + y^2 = 25 + 24 = 49$.
Since the given circle $x^2 + y^2 = 49$ is the director circle of the ellipse,the tangents drawn from any point on this circle to the ellipse are perpendicular to each other.
Therefore,the angle between the tangents is $\frac{\pi}{2}$.

(A) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the ends of the latus recta are $(\pm ae, \pm \frac{b^2}{a})$.
Given $a^2 = 4$ and $b^2 = 1$,we have $a = 2$ and $b = 1$.
The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.
The coordinates of the ends of the latus recta are $(\pm \sqrt{3}, \pm \frac{1}{2})$.
The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.
For $(\sqrt{3}, \frac{1}{2})$,the tangent is $\frac{x\sqrt{3}}{4} + \frac{y(1/2)}{1} = 1 \Rightarrow \frac{\sqrt{3}}{4}x + \frac{1}{2}y = 1$.
The four tangents form a rhombus with vertices at $(\pm \frac{a^2}{ae}, 0) = (\pm \frac{4}{\sqrt{3}}, 0)$ and $(0, \pm \frac{b^2}{b^2/a}) = (0, \pm a) = (0, \pm 2)$.
The area of the rhombus is $2 \times \text{base} \times \text{height} = 2 \times |\frac{4}{\sqrt{3}}| \times |2| = \frac{16}{\sqrt{3}}$.

(B) The sum of the focal distances of any point on an ellipse is equal to the length of the major axis,$2a$.
Given $P(2,2)$,$F_{1}(5,2)$,and $F_{2}(2,6)$.
$PF_{1} = \sqrt{(5-2)^2 + (2-2)^2} = \sqrt{3^2 + 0^2} = 3$.
$PF_{2} = \sqrt{(2-2)^2 + (6-2)^2} = \sqrt{0^2 + 4^2} = 4$.
Therefore,$2a = PF_{1} + PF_{2} = 3 + 4 = 7$.
The distance between the foci is $2ae = \sqrt{(5-2)^2 + (2-6)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
Since $2a = 7$ and $2ae = 5$,we have $e = \frac{2ae}{2a} = \frac{5}{7}$.

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The major axis $AA'$ lies along the $x$-axis with coordinates $A = (a, 0)$ and $A' = (-a, 0)$.
Let point $P$ be $(a \cos \theta, b \sin \theta)$.
The area of $\Delta APA'$ is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Base $AA' = 2a$ and height is the absolute value of the $y$-coordinate of $P$,which is $|b \sin \theta|$.
Area $= \frac{1}{2} \times (2a) \times |b \sin \theta| = |ab \sin \theta|$.
Since the maximum value of $|\sin \theta|$ is $1$,the maximum area is $|ab|$.

(A) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the product of the lengths of the perpendiculars from the foci to any tangent is equal to $b^2$.
The given equation is $3x^2 + 5y^2 = 1$,which can be rewritten as $\frac{x^2}{1/3} + \frac{y^2}{1/5} = 1$.
Here,$a^2 = 1/3$ and $b^2 = 1/5$.
Since $a^2 > b^2$,the product of the lengths of the perpendiculars is $b^2 = 1/5$.

(A) The condition for the line $y = mx + k$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $k^2 = a^2m^2 + b^2$.
Here,the line is $y = cx + c$,so $m = c$ and $k = c$.
The ellipse is $\frac{x^2}{4} + \frac{y^2}{1} = 1$,so $a^2 = 4$ and $b^2 = 1$.
Substituting these values into the condition: $c^2 = 4(c^2) + 1$.
$c^2 = 4c^2 + 1$.
$3c^2 + 1 = 0$.
Since $c^2 = -\frac{1}{3}$,there are no real values of $c$ that satisfy this equation.
Therefore,the number of values of $c$ is $0$.

(D) Given equation: $\left( \frac{x - 3}{y} \right)^2 + \left( \frac{y - 4}{y} \right)^2 = \frac{1}{9}$
$(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$
$9(x - 3)^2 + 9(y^2 - 8y + 16) = y^2$
$9(x - 3)^2 + 8y^2 - 72y + 144 = 0$
$9(x - 3)^2 + 8(y^2 - 9y) = -144$
$9(x - 3)^2 + 8(y - 4.5)^2 = -144 + 8(20.25) = 18$
$\frac{(x - 3)^2}{2} + \frac{(y - 4.5)^2}{18/8} = 1$
$\frac{(x - 3)^2}{2} + \frac{(y - 4.5)^2}{2.25} = 1$
Here,$a^2 = 2.25$ and $b^2 = 2$.
Since $a^2 > b^2$,the eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{2}{2.25} = 1 - \frac{2}{9/4} = 1 - \frac{8}{9} = \frac{1}{9}$.
Therefore,$e = \frac{1}{3}$.

(A) Let the point $M$ on the ellipse be $(a \cos \theta, b \sin \theta)$,where $a = \sqrt{18} = 3\sqrt{2}$ and $b = \sqrt{8} = 2\sqrt{2}$.
So,$M = (3\sqrt{2} \cos \theta, 2\sqrt{2} \sin \theta)$.
The shortest distance from a point on the ellipse to a line occurs where the tangent at that point is parallel to the given line.
The slope of the given line $2x - 3y + 25 = 0$ is $m = \frac{2}{3}$.
The slope of the tangent at $M$ is given by $-\frac{b^2 x_0}{a^2 y_0} = -\frac{8(3\sqrt{2} \cos \theta)}{18(2\sqrt{2} \sin \theta)} = -\frac{24 \cos \theta}{36 \sin \theta} = -\frac{2}{3} \cot \theta$.
Equating the slopes: $-\frac{2}{3} \cot \theta = \frac{2}{3}$ $\Rightarrow \cot \theta = -1$ $\Rightarrow \tan \theta = -1$.
For the point to be nearest to the line,we choose $\theta = \frac{3\pi}{4}$ (in the second quadrant where $x$ is negative and $y$ is positive).
$x = 3\sqrt{2} \cos(\frac{3\pi}{4}) = 3\sqrt{2} \times (-\frac{1}{\sqrt{2}}) = -3$.
$y = 2\sqrt{2} \sin(\frac{3\pi}{4}) = 2\sqrt{2} \times (\frac{1}{\sqrt{2}}) = 2$.
Thus,the point $M$ is $(-3, 2)$.

(D) The coordinates of the foci $S_1$ and $S_2$ are $(-ae, 0)$ and $(ae, 0)$ respectively.
The distance between the foci is $S_1S_2 = 2ae$.
Let $P$ be a point on the ellipse given by $(a \cos \theta, b \sin \theta)$.
The area of $\Delta PS_1S_2$ is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Here,the base $S_1S_2 = 2ae$ and the height is the absolute value of the $y$-coordinate of $P$,which is $|b \sin \theta|$.
$\text{Area} = \frac{1}{2} \times (2ae) \times |b \sin \theta| = ae |b \sin \theta| = abe |\sin \theta|$.
Since the maximum value of $|\sin \theta|$ is $1$,the maximum area is $abe$ square units.

(A) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here $a^2 = 9$ and $b^2 = 4$.
The equation of a normal to the ellipse at point $(3 \cos \theta, 2 \sin \theta)$ is given by $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2 - b^2$,which becomes $\frac{3x}{\cos \theta} - \frac{2y}{\sin \theta} = 9 - 4 = 5$.
This can be rewritten as $3x \sec \theta - 2y \csc \theta = 5$.
For this line to be a tangent to the circle $x^2 + y^2 = 3$,the perpendicular distance from the center $(0, 0)$ to the line must equal the radius $\sqrt{3}$.
The distance $d = \frac{|-5|}{\sqrt{(3 \sec \theta)^2 + (-2 \csc \theta)^2}} = \frac{5}{\sqrt{9 \sec^2 \theta + 4 \csc^2 \theta}}$.
Setting $d = \sqrt{3}$,we get $\sqrt{9 \sec^2 \theta + 4 \csc^2 \theta} = \frac{5}{\sqrt{3}}$,so $9 \sec^2 \theta + 4 \csc^2 \theta = \frac{25}{3} \approx 8.33$.
Using the inequality $(m \sec^2 \theta + n \csc^2 \theta) \ge (\sqrt{m} + \sqrt{n})^2$,the minimum value of $9 \sec^2 \theta + 4 \csc^2 \theta$ is $(\sqrt{9} + \sqrt{4})^2 = (3 + 2)^2 = 25$.
Since the minimum value $25$ is greater than $\frac{25}{3}$,there is no real value of $\theta$ that satisfies the condition.
Therefore,the number of such tangents is $0$.

(C) The line $y = mx + 1$ is tangent to the ellipse $x^2 + 4y^2 = 1$ if the condition for tangency $c^2 = a^2m^2 + b^2$ is satisfied.
Here,the equation of the ellipse is $\frac{x^2}{1} + \frac{y^2}{1/4} = 1$,so $a^2 = 1$ and $b^2 = 1/4$.
The line is $y = mx + 1$,so $c = 1$.
Substituting these into the condition $c^2 = a^2m^2 + b^2$ gives $1^2 = (1)m^2 + \frac{1}{4}$.
$1 = m^2 + \frac{1}{4} \implies m^2 = \frac{3}{4} \implies m = \pm \frac{\sqrt{3}}{2}$.
For the point $(\alpha, \beta)$ to be in the first quadrant (where $\alpha > 0$),we analyze the intersection point. The point of tangency is given by $(x, y) = (-\frac{a^2m}{c}, \frac{b^2}{c})$.
Substituting $a^2=1, b^2=1/4, c=1$,we get $x = -m$ and $y = 1/4$.
Since we require $\alpha > 0$,we must have $-m > 0$,which implies $m < 0$.
Thus,$m = -\frac{\sqrt{3}}{2}$ is the only valid value.
The sum of all possible values of $m$ is $-\frac{\sqrt{3}}{2}$.

(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The extremities of the latus rectum are $(\pm ae, be)$ or $(\pm ae, -be)$. Given the tangents at these points,we consider the tangent at $(ae, be)$,which is $\frac{x(ae)}{a^2} + \frac{y(be)}{b^2} = 1$,simplifying to $\frac{ex}{a} + \frac{ey}{b} = 1$.
Comparing this with the given line $3x + 2y = 10$,we rewrite the line as $\frac{3}{10}x + \frac{2}{10}y = 1$,or $\frac{3}{10}x + \frac{1}{5}y = 1$.
Equating coefficients: $\frac{e}{a} = \frac{3}{10}$ and $\frac{e}{b} = \frac{1}{5}$.
From the second equation,$b = 5e$.
Using $b^2 = a^2(1 - e^2)$,we have $25e^2 = a^2(1 - e^2)$.
From $\frac{e}{a} = \frac{3}{10}$,we get $a = \frac{10e}{3}$.
Substituting $a^2 = \frac{100e^2}{9}$ into the equation: $25e^2 = \frac{100e^2}{9}(1 - e^2)$.
Dividing by $25e^2$ (since $e \neq 0$): $1 = \frac{4}{9}(1 - e^2) \Rightarrow 9 = 4 - 4e^2 \Rightarrow 4e^2 = -5$,which suggests the ellipse is vertical.
Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b > a$. The tangent at $(ae, be)$ is $\frac{x(ae)}{a^2} + \frac{y(be)}{b^2} = 1$.
Given the lines $2y = -3x + 10$ and $2y = 3x + 10$,they intersect at $(0, 5)$.
Thus,$be = 5$. The slope of the tangent is $\pm \frac{3}{2}$.
The tangent equation is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
For $y = -\frac{3}{2}x + 5$,$m = -\frac{3}{2}$ and $c = 5$.
$5^2 = a^2(-\frac{3}{2})^2 + b^2 \Rightarrow 25 = \frac{9}{4}a^2 + b^2$.
Also,the point of tangency $(x_1, y_1)$ satisfies $\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$ and $y_1 = be$.
Using the condition for the latus rectum,the length of the latus rectum is $\frac{2a^2}{b} = \frac{100}{27}$.

(A) The ellipse $E$ is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$. The foci are $(\pm ae, 0)$ and the extremities of the minor axis are $(0, \pm b)$.
For the ellipse $S = 0$,the vertices are $(0, \pm b)$,so its minor axis lies on the $x$-axis and its major axis lies on the $y$-axis.
The equation of $S = 0$ is $\frac{x^2}{b_1^2} + \frac{y^2}{b^2} = 1$ where $b_1$ is the semi-minor axis of $S$.
Since $S = 0$ passes through the foci $(\pm ae, 0)$,we have $\frac{(ae)^2}{b_1^2} + \frac{0^2}{b^2} = 1$,which gives $b_1^2 = a^2e^2$.
For an ellipse with major axis $b$ and semi-minor axis $b_1$,the eccentricity $e_1$ satisfies $b_1^2 = b^2(1 - e_1^2)$.
Substituting $b_1^2 = a^2e^2$ and $b^2 = a^2(1 - e^2)$,we get $a^2e^2 = a^2(1 - e^2)(1 - e_1^2)$.
$e^2 = (1 - e^2)(1 - e_1^2) \Rightarrow 1 - e_1^2 = \frac{e^2}{1 - e^2}$.
$e_1^2 = 1 - \frac{e^2}{1 - e^2} = \frac{1 - 2e^2}{1 - e^2}$.
Thus,$e_1 = \sqrt{\frac{1 - 2e^2}{1 - e^2}}$.

(D) The equation of a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $ax \sec \theta - by \csc \theta = a^2 - b^2$.
Two parallel normals correspond to parameters $\theta$ and $\theta + \pi$,which are $ax \sec \theta - by \csc \theta = a^2 - b^2$ and $ax \sec \theta - by \csc \theta = -(a^2 - b^2)$.
The distance $L$ between these parallel lines is given by $L = \frac{|(a^2 - b^2) - (-(a^2 - b^2))|}{\sqrt{a^2 \sec^2 \theta + b^2 \csc^2 \theta}} = \frac{2(a^2 - b^2)}{\sqrt{a^2 \sec^2 \theta + b^2 \csc^2 \theta}}$.
Using the inequality $a^2 \sec^2 \theta + b^2 \csc^2 \theta \ge (a+b)^2$,we get $L \le \frac{2(a^2 - b^2)}{(a+b)} = 2(a-b)$.
Thus,the maximum value of $L$ is $2(a-b)$.

(B) Given the ellipse $\frac{x^2}{27} + y^2 = 1$,we have $a^2 = 27$ and $b^2 = 1$,so $a = 3\sqrt{3}$ and $b = 1$.
The equation of the tangent at $(a \cos \theta, b \sin \theta)$ is $\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$.
Substituting the values,the equation of the tangent is $\frac{x}{3\sqrt{3}} \cos \theta + y \sin \theta = 1$.
The $x$-intercept is $3\sqrt{3} \sec \theta$ and the $y$-intercept is $\csc \theta$.
Let $S$ be the sum of the intercepts: $S = 3\sqrt{3} \sec \theta + \csc \theta$.
To find the minimum,differentiate $S$ with respect to $\theta$: $\frac{dS}{d\theta} = 3\sqrt{3} \sec \theta \tan \theta - \csc \theta \cot \theta$.
Setting $\frac{dS}{d\theta} = 0$,we get $3\sqrt{3} \frac{\sin \theta}{\cos^2 \theta} = \frac{\cos \theta}{\sin^2 \theta}$.
This simplifies to $\tan^3 \theta = \frac{1}{3\sqrt{3}} = (\frac{1}{\sqrt{3}})^3$.
Thus,$\tan \theta = \frac{1}{\sqrt{3}}$,which implies $\theta = \frac{\pi}{6}$.
Since the second derivative is positive at $\theta = \frac{\pi}{6}$,the sum of intercepts is minimum at $\theta = \frac{\pi}{6}$.

(D) The equation of a chord joining points having eccentric angles $\alpha$ and $\beta$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right) + \frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right) = \cos \left(\frac{\alpha-\beta}{2}\right)$.
Since this is a focal chord,it passes through the focus $(ae, 0)$.
Substituting $(ae, 0)$ into the equation,we get $e \cos \left(\frac{\alpha+\beta}{2}\right) = \cos \left(\frac{\alpha-\beta}{2}\right)$.
Thus,$e = \frac{\cos ((\alpha-\beta)/2)}{\cos ((\alpha+\beta)/2)}$.
Using the product-to-sum identities or simplifying,we note that for a focal chord,$\tan(\alpha/2) \tan(\beta/2) = \frac{e-1}{e+1}$.
Alternatively,using the identity $\frac{\sin \alpha + \sin \beta}{\sin (\alpha + \beta)} = \frac{2 \sin ((\alpha+\beta)/2) \cos ((\alpha-\beta)/2)}{2 \sin ((\alpha+\beta)/2) \cos ((\alpha+\beta)/2)} = \frac{\cos ((\alpha-\beta)/2)}{\cos ((\alpha+\beta)/2)} = e$.

(C) The equation of the ellipse is $3x^2 + 5y^2 = 32$,which can be written as $\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1$.
To determine the position of the point $(3, 5)$ with respect to the ellipse,we evaluate the expression $S = 3x^2 + 5y^2 - 32$ at $(3, 5)$.
$S = 3(3)^2 + 5(5)^2 - 32 = 3(9) + 5(25) - 32 = 27 + 125 - 32 = 120$.
Since $S > 0$,the point $(3, 5)$ lies outside the ellipse.
For any point lying outside an ellipse,exactly two real tangents can be drawn to the ellipse.

(B) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the product of the lengths of the perpendiculars from the foci $S_1$ and $S_2$ to any tangent is equal to the square of the semi-minor axis,$b^2$.
Given the equation of the ellipse $\frac{x^2}{5} + \frac{y^2}{3} = 1$,we have $a^2 = 5$ and $b^2 = 3$.
Therefore,$(S_1 F_1) (S_2 F_2) = b^2 = 3$.

(C) Let the point of intersection be $R(x_1, y_1)$. The equation of the chord of contact $PQ$ is $\frac{xx_1}{9} + \frac{yy_1}{4} = 1$.
To find the combined equation of lines $OP$ and $OQ$,we homogenize the ellipse equation $\frac{x^2}{9} + \frac{y^2}{4} = 1$ using the chord of contact:
$\frac{x^2}{9} + \frac{y^2}{4} = (\frac{xx_1}{9} + \frac{yy_1}{4})^2$.
Since $OP \perp OQ$,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(\frac{1}{9} - \frac{x_1^2}{81}) + (\frac{1}{4} - \frac{y_1^2}{16}) = 0$.
Rearranging the terms,we get $\frac{x_1^2}{81} + \frac{y_1^2}{16} = \frac{1}{9} + \frac{1}{4} = \frac{13}{36}$.
Thus,the locus of $(x_1, y_1)$ is $\frac{x^2}{81} + \frac{y^2}{16} = \frac{13}{36}$,which represents an ellipse.

(D) The locus of points from which perpendicular tangents can be drawn to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is its director circle,given by $x^2 + y^2 = a^2 + b^2$.
For the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$,the director circle is $x^2 + y^2 = 16 + 9 = 25$.
We need to find the number of intersection points of this director circle $x^2 + y^2 = 25$ and the given ellipse $\frac{x^2}{50} + \frac{y^2}{20} = 1$.
From the circle,$y^2 = 25 - x^2$. Substituting this into the ellipse equation:
$\frac{x^2}{50} + \frac{25 - x^2}{20} = 1$
Multiplying by $100$:
$2x^2 + 5(25 - x^2) = 100$
$2x^2 + 125 - 5x^2 = 100$
$-3x^2 = -25$
$x^2 = \frac{25}{3}$,which gives $x = \pm \frac{5}{\sqrt{3}}$.
Since $x^2 = \frac{25}{3} < 25$,$y^2 = 25 - \frac{25}{3} = \frac{50}{3} > 0$,which gives $y = \pm \sqrt{\frac{50}{3}}$.
Thus,there are $4$ distinct points of intersection.

(A) The length of the latus rectum of an ellipse is $\frac{2b^2}{a}$ and the distance between its foci is $2ae$.
Given that $\frac{2b^2}{a} = 2ae$,we have $b^2 = a^2e$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2e$.
Dividing by $a^2$,we obtain $1 - e^2 = e$,which simplifies to $e^2 + e - 1 = 0$.
Solving this quadratic equation for $e$ (where $e > 0$),we get $e = \frac{-1 + \sqrt{5}}{2} = \frac{\sqrt{5} - 1}{2}$.
Since $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$,it follows that $e = 2 \sin 18^{\circ}$.

(C) The equation of the ellipse is $\frac{x^2}{3} + \frac{y^2}{1} = 1$.
Let the point $P$ be $(\sqrt{3} \cos \theta, \sin \theta)$.
The perpendicular distance of $P$ from the line $x - y - 10 = 0$ is given by $d = \frac{|\sqrt{3} \cos \theta - \sin \theta - 10|}{\sqrt{1^2 + (-1)^2}} = \frac{|\sqrt{3} \cos \theta - \sin \theta - 10|}{\sqrt{2}}$.
To maximize $d$, we need to maximize the expression $f(\theta) = \sqrt{3} \cos \theta - \sin \theta$.
We can write $f(\theta) = 2(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta) = 2 \cos(\theta + 30^{\circ})$.
The maximum value of $f(\theta)$ is $2$ and the minimum value is $-2$.
For maximum distance, we consider the minimum value of $f(\theta)$, which is $-2$.
Thus, $d_{max} = \frac{|-2 - 10|}{\sqrt{2}} = \frac{|-12|}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$.

(A) Let the eccentric angles of $P$ and $Q$ be $\theta_1$ and $\theta_2$ respectively. Given $\theta_1 - \theta_2 = \frac{3\pi}{2}$,so $\theta_1 = \theta_2 + \frac{3\pi}{2}$.
Coordinates are $P(5 \cos \theta_1, 2 \sin \theta_1)$ and $Q(5 \cos \theta_2, 2 \sin \theta_2)$.
Substituting $\theta_1 = \theta_2 + \frac{3\pi}{2}$,we get $P(5 \sin \theta_2, -2 \cos \theta_2)$.
The distance $PQ^2 = (5 \cos \theta_2 - 5 \sin \theta_2)^2 + (2 \sin \theta_2 + 2 \cos \theta_2)^2$.
$PQ^2 = 25(\cos^2 \theta_2 + \sin^2 \theta_2 - 2 \sin \theta_2 \cos \theta_2) + 4(\sin^2 \theta_2 + \cos^2 \theta_2 + 2 \sin \theta_2 \cos \theta_2)$.
$PQ^2 = 25(1 - \sin 2\theta_2) + 4(1 + \sin 2\theta_2) = 29 - 21 \sin 2\theta_2$.
To minimize $PQ$,we maximize $\sin 2\theta_2 = 1$.
$PQ^2_{min} = 29 - 21 = 8$.
$PQ_{min} = \sqrt{8} = 2\sqrt{2}$.

(A) The given equation represents an ellipse with foci $F_1(3, 2)$ and $F_2(5, 4)$ and major axis length $2a = 6$,so $a = 3$.
The distance between the foci is $2ae = \sqrt{(5-3)^2 + (4-2)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
Thus,$ae = \sqrt{2}$,which gives the eccentricity $e = \frac{\sqrt{2}}{3}$.
The distance between the directrices of an ellipse is given by $\frac{2a}{e}$.
Substituting the values,we get $\frac{2(3)}{\frac{\sqrt{2}}{3}} = \frac{6 \times 3}{\sqrt{2}} = \frac{18}{\sqrt{2}} = 9\sqrt{2}$.

(C) The given equation of the ellipse is $(x - y + 1)^2 + (2x + 2y - 6)^2 = 20$.
Dividing by $20$,we get $\frac{(x - y + 1)^2}{20} + \frac{(2x + 2y - 6)^2}{20} = 1$.
This can be rewritten as $\frac{(\frac{x - y + 1}{\sqrt{2}})^2}{10} + \frac{(\frac{x + y - 3}{\sqrt{2}})^2}{5} = 1$.
Here,$a^2 = 10$ and $b^2 = 5$. The center of the ellipse is $(1, 2)$.
The locus of the feet of the perpendiculars from the foci to any tangent of an ellipse is its auxiliary circle.
The auxiliary circle has the same center as the ellipse and radius equal to the semi-major axis $a$.
Here,$a^2 = 10$,so the equation of the auxiliary circle is $(x - 1)^2 + (y - 2)^2 = 10$.
Expanding this,we get $x^2 - 2x + 1 + y^2 - 4y + 4 = 10$,which simplifies to $x^2 + y^2 - 2x - 4y - 5 = 0$.

(A) The path traced by the man is an ellipse because the sum of the distances from two fixed points (foci) is constant.
Let the constant sum be $2a = 10 \ m$, so $a = 5 \ m$.
The distance between the foci is $2ae = 8 \ m$, so $ae = 4 \ m$.
Since $a = 5$, we have $5e = 4$, which gives $e = \frac{4}{5}$.
Using the relation $b^2 = a^2(1 - e^2)$, we get $b^2 = 25(1 - (\frac{4}{5})^2) = 25(1 - \frac{16}{25}) = 25(\frac{9}{25}) = 9$.
Thus, $b = 3 \ m$.
The area of the ellipse is given by $\pi ab = \pi \times 5 \times 3 = 15 \pi \ m^2$.

(D) The locus of points from which perpendicular tangents can be drawn to an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is its director circle,given by $x^{2} + y^{2} = a^{2} + b^{2}$.
For the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$,we have $a^{2} = 16$ and $b^{2} = 9$.
Thus,the equation of the director circle is $x^{2} + y^{2} = 16 + 9 = 25$.
We need to find the number of intersection points of this director circle $x^{2} + y^{2} = 25$ and the given ellipse $\frac{x^{2}}{50} + \frac{y^{2}}{20} = 1$.
From the circle,$y^{2} = 25 - x^{2}$. Substituting this into the ellipse equation:
$\frac{x^{2}}{50} + \frac{25 - x^{2}}{20} = 1$
Multiplying by $100$:
$2x^{2} + 5(25 - x^{2}) = 100$
$2x^{2} + 125 - 5x^{2} = 100$
$-3x^{2} = -25$
$x^{2} = \frac{25}{3}$,which gives $x = \pm \frac{5}{\sqrt{3}}$.
Since $x^{2} = \frac{25}{3} < 25$,$y^{2} = 25 - \frac{25}{3} = \frac{50}{3} > 0$,which gives $y = \pm \sqrt{\frac{50}{3}}$.
Since there are two values for $x$ and for each $x$ there are two values for $y$,there are $4$ points of intersection.

(A) For the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,we have $a^2 = 8$ and $b^2 = 4$,so $a = 2\sqrt{2}$ and $b = 2$.
Let the eccentric angles of the extremities of the chord be $\theta$ and $\theta + \frac{\pi}{2}$.
The coordinates of the points are $P = (2\sqrt{2} \cos \theta, 2 \sin \theta)$ and $Q = (2\sqrt{2} \cos(\theta + \frac{\pi}{2}), 2 \sin(\theta + \frac{\pi}{2})) = (-2\sqrt{2} \sin \theta, 2 \cos \theta)$.
The square of the length of the chord $PQ$ is given by $L^2 = (2\sqrt{2} \cos \theta + 2\sqrt{2} \sin \theta)^2 + (2 \sin \theta - 2 \cos \theta)^2$.
$L^2 = 8(\cos \theta + \sin \theta)^2 + 4(\sin \theta - \cos \theta)^2$.
$L^2 = 8(1 + \sin 2\theta) + 4(1 - \sin 2\theta) = 12 + 4 \sin 2\theta$.
To maximize $L$,we set $\sin 2\theta = 1$,which gives $L^2 = 12 + 4 = 16$.
Thus,the maximum length $L = \sqrt{16} = 4$.

(B) The equation of the ellipse is $x^2 + 3y^2 = 9$.
Differentiating with respect to $x$,we get $2x + 6y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{3y}$.
The slope of the normal at any point $(x, y)$ is $m = -\frac{dx}{dy} = \frac{3y}{x}$.
For the point $P_1 = (3\cos \theta, \sqrt{3} \sin \theta)$,the slope of the normal $m_1 = \frac{3(\sqrt{3} \sin \theta)}{3 \cos \theta} = \sqrt{3} \tan \theta$.
For the point $P_2 = (-3\sin \theta, \sqrt{3} \cos \theta)$,the slope of the normal $m_2 = \frac{3(\sqrt{3} \cos \theta)}{-3 \sin \theta} = -\sqrt{3} \cot \theta$.
The angle $\beta$ between the normals is given by $\tan \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \beta = \left| \frac{\sqrt{3} \tan \theta - (-\sqrt{3} \cot \theta)}{1 + (\sqrt{3} \tan \theta)(-\sqrt{3} \cot \theta)} \right| = \left| \frac{\sqrt{3}(\tan \theta + \cot \theta)}{1 - 3} \right| = \left| \frac{\sqrt{3}(\tan \theta + \cot \theta)}{-2} \right| = \frac{\sqrt{3}}{2} (\tan \theta + \cot \theta)$.
Since $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\frac{1}{2} \sin 2\theta} = \frac{2}{\sin 2\theta}$.
Thus,$\tan \beta = \frac{\sqrt{3}}{2} \cdot \frac{2}{\sin 2\theta} = \frac{\sqrt{3}}{\sin 2\theta}$.
Therefore,$\frac{1}{\cot \beta} = \frac{\sqrt{3}}{\sin 2\theta}$,which implies $\frac{\cot \beta}{\sin 2\theta} = \frac{1}{\sqrt{3}}$.
Hence,$\frac{2 \cot \beta}{\sin 2\theta} = \frac{2}{\sqrt{3}}$.

(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The distance between a focus $(ae, 0)$ and its nearest vertex $(a, 0)$ is $a(1 - e) = \frac{3}{2}$.
Thus,$a - ae = \frac{3}{2}$,which implies $ae = a - \frac{3}{2}$.
The length of the latus rectum is $\frac{2b^2}{a} = 4$,so $b^2 = 2a$.
Using the relation $b^2 = a^2(1 - e^2)$,we have $2a = a^2(1 - e^2)$,which simplifies to $1 - e^2 = \frac{2}{a}$,or $e^2 = 1 - \frac{2}{a}$.
From $ae = a - \frac{3}{2}$,squaring both sides gives $a^2e^2 = (a - \frac{3}{2})^2$.
Substituting $e^2 = 1 - \frac{2}{a}$,we get $a^2(1 - \frac{2}{a}) = a^2 - 3a + \frac{9}{4}$.
$a^2 - 2a = a^2 - 3a + \frac{9}{4}$.
$a = \frac{9}{4}$.
Now,$e^2 = 1 - \frac{2}{a} = 1 - \frac{2}{9/4} = 1 - \frac{8}{9} = \frac{1}{9}$.
Therefore,$e = \frac{1}{3}$.

(D) Given eccentricity $e = \frac{3}{5}$ and distance between foci $2ae = 6$.
$2a(\frac{3}{5}) = 6 \Rightarrow a = 5$.
Using the relation $b^2 = a^2(1 - e^2)$:
$b^2 = 25(1 - \frac{9}{25}) = 25(\frac{16}{25}) = 16 \Rightarrow b = 4$.
The vertices of the ellipse are $(\pm a, 0)$ and $(0, \pm b)$,which are $(\pm 5, 0)$ and $(0, \pm 4)$.
The quadrilateral formed by these vertices is a rhombus with diagonals of length $2a = 10$ and $2b = 8$.
Area of the quadrilateral $= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 10 \times 8 = 40$ sq. units.

(C) The equation of an ellipse with centre at $(0, 0)$ and axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since the ellipse passes through $(4, -1)$,we have $\frac{16}{a^2} + \frac{1}{b^2} = 1$,which implies $16b^2 + a^2 = a^2b^2$ $(i)$.
Since the ellipse passes through $(-2, 2)$,we have $\frac{4}{a^2} + \frac{4}{b^2} = 1$,which implies $4b^2 + 4a^2 = a^2b^2$ $(ii)$.
Equating $(i)$ and $(ii)$,we get $16b^2 + a^2 = 4b^2 + 4a^2$.
$12b^2 = 3a^2$,so $a^2 = 4b^2$.
Substituting $a^2 = 4b^2$ into $(ii)$,we get $4b^2 + 4(4b^2) = (4b^2)b^2$,which simplifies to $20b^2 = 4b^4$,so $b^2 = 5$.
Then $a^2 = 4(5) = 20$.
Since $b^2 = a^2(1 - e^2)$,we have $5 = 20(1 - e^2)$.
$1 - e^2 = \frac{5}{20} = \frac{1}{4}$.
$e^2 = 1 - \frac{1}{4} = \frac{3}{4}$.
Therefore,$e = \frac{\sqrt{3}}{2}$.

(C) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
Here,$a^2 = 27 \implies a = 3\sqrt{3}$ and $b^2 = 3 \implies b = \sqrt{3}$.
The tangent equation is $\frac{x \cos \theta}{3\sqrt{3}} + \frac{y \sin \theta}{\sqrt{3}} = 1$.
The coordinates of $A$ (x-intercept) are $(\frac{3\sqrt{3}}{\cos \theta}, 0)$ and $B$ (y-intercept) are $(0, \frac{\sqrt{3}}{\sin \theta})$.
The area of triangle $OAB$ is $\Delta = \frac{1}{2} \times |x_A| \times |y_B| = \frac{1}{2} \times \frac{3\sqrt{3}}{\cos \theta} \times \frac{\sqrt{3}}{\sin \theta} = \frac{9}{2 \sin \theta \cos \theta} = \frac{9}{\sin 2\theta}$.
For the area to be minimum,$\sin 2\theta$ must be maximum,i.e.,$\sin 2\theta = 1$.
Thus,$\Delta_{\min} = 9$ sq. units.

(C) The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{9} = 1$.
Its foci are $(\pm \sqrt{4+9}, 0) = (\pm \sqrt{13}, 0)$.
The eccentricity of the hyperbola is $e_h = \frac{\sqrt{13}}{2}$.
Let $e_e$ be the eccentricity of the ellipse. Given $e_e \times e_h = \frac{1}{2}$,we have $e_e \times \frac{\sqrt{13}}{2} = \frac{1}{2}$,so $e_e = \frac{1}{\sqrt{13}}$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since the ellipse passes through the foci of the hyperbola $(\pm \sqrt{13}, 0)$,we have $a^2 = 13$.
Using $b^2 = a^2(1 - e_e^2)$,we get $b^2 = 13(1 - \frac{1}{13}) = 13 - 1 = 12$.
The equation of the ellipse is $\frac{x^2}{13} + \frac{y^2}{12} = 1$.
Checking the points:
For $A$: $\frac{13/2}{13} + \frac{6}{12} = \frac{1}{2} + \frac{1}{2} = 1$ (Lies on ellipse).
For $B$: $\frac{39/4}{13} + \frac{3}{12} = \frac{3}{4} + \frac{1}{4} = 1$ (Lies on ellipse).
For $C$: $\frac{13/4}{13} + \frac{3/4}{12} = \frac{1}{4} + \frac{1}{16} = \frac{5}{16} \neq 1$ (Does not lie on ellipse).
For $D$: $\frac{13}{13} + 0 = 1$ (Lies on ellipse).
Thus,the point in option $C$ does not lie on the ellipse.

(B) The distance between the foci of an ellipse is $2ae$.
The length of the latus rectum of an ellipse is $\frac{2b^2}{a}$.
According to the problem,the distance between the foci is half the length of the latus rectum:
$2ae = \frac{1}{2} \times \frac{2b^2}{a}$
$2ae = \frac{b^2}{a}$
Using the relation $b^2 = a^2(1-e^2)$,we get:
$2ae = \frac{a^2(1-e^2)}{a}$
$2ae = a(1-e^2)$
$2e = 1-e^2$
$e^2 + 2e - 1 = 0$
Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$e = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$
Since the eccentricity $e$ must be positive,$e = \sqrt{2}-1$.

(A) Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the equation of the ellipse.
Given that $F_1B$ and $F_2B$ are perpendicular to each other.
The coordinates are $F_1(-ae, 0)$,$F_2(ae, 0)$,and $B(0, b)$.
Slope of $F_1B = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.
Slope of $F_2B = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.
Since $F_1B \perp F_2B$,the product of their slopes is $-1$:
$\left(\frac{b}{ae}\right) \times \left(-\frac{b}{ae}\right) = -1$
$-\frac{b^2}{a^2e^2} = -1 \Rightarrow b^2 = a^2e^2$.
We know that for an ellipse,$b^2 = a^2(1 - e^2)$.
Substituting $b^2 = a^2e^2$ into the equation:
$a^2e^2 = a^2(1 - e^2)$
$e^2 = 1 - e^2$
$2e^2 = 1$
$e^2 = \frac{1}{2}$.

(B) Let point $A(a, 0)$ be on the $x$-axis and $B(0, b)$ be on the $y$-axis.
Let $P(h, k)$ divide $AB$ in the ratio $1 : 2$.
By the section formula,we have:
$h = \frac{2(0) + 1(a)}{1 + 2} = \frac{a}{3} \Rightarrow a = 3h$
$k = \frac{2(b) + 1(0)}{1 + 2} = \frac{2b}{3} \Rightarrow b = \frac{3k}{2}$
Since the length of the staircase is $l$,we have $a^2 + b^2 = l^2$.
Substituting the values of $a$ and $b$:
$(3h)^2 + (\frac{3k}{2})^2 = l^2$
$9h^2 + \frac{9k^2}{4} = l^2$
$\frac{h^2}{(l/3)^2} + \frac{k^2}{(2l/3)^2} = 1$
This is the equation of an ellipse with semi-major axis $b' = \frac{2l}{3}$ and semi-minor axis $a' = \frac{l}{3}$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{(a')^2}{(b')^2}} = \sqrt{1 - \frac{(l/3)^2}{(2l/3)^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(D) Let the point of tangency be $(h, k)$ on the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$.
The equation of the tangent at $(h, k)$ is given by $\frac{xh}{16} + \frac{yk}{81} = 1$.
The $x$-intercept is found by setting $y=0$,which gives $x = \frac{16}{h}$. So,point $B = (\frac{16}{h}, 0)$.
The $y$-intercept is found by setting $x=0$,which gives $y = \frac{81}{k}$. So,point $A = (0, \frac{81}{k})$.
The area of the triangle $OAB$ is $A = \frac{1}{2} \times |\frac{16}{h}| \times |\frac{81}{k}| = \frac{648}{|hk|}$.
Since $(h, k)$ lies on the ellipse,$\frac{h^2}{16} + \frac{k^2}{81} = 1$. By the $AM$-$GM$ inequality,$\frac{\frac{h^2}{16} + \frac{k^2}{81}}{2} \ge \sqrt{\frac{h^2 k^2}{16 \times 81}}$.
$\frac{1}{2} \ge \frac{|hk|}{4 \times 9} \Rightarrow |hk| \le 18$.
Therefore,the minimum area $A = \frac{648}{|hk|} \ge \frac{648}{18} = 36$.
The minimum area of the triangle is $36$ square units.

(C) The equation of the ellipse is $\frac{x^2}{(2c)^2} + \frac{y^2}{c^2} = 1$. The vertices are at $(\pm 2c, 0)$ and $(0, \pm c)$.
The equation of the circle is $x^2 + y^2 = (3a)^2$. The radius is $3a$.
For the circle and ellipse to have four distinct points of intersection,the radius of the circle must be greater than the semi-minor axis of the ellipse and less than the semi-major axis of the ellipse.
Thus,$c < 3a < 2c$.
From $3a < 2c$,we have $9a^2 < 4c^2$,which implies $9a^2 - 4c^2 < 0$.
From $c < 3a$,we have $c^2 < 9a^2$,which implies $9a^2 - c^2 > 0$.
Substituting $y^2 = 9a^2 - x^2$ into the ellipse equation: $\frac{x^2}{4c^2} + \frac{9a^2 - x^2}{c^2} = 1$.
$x^2 + 4(9a^2 - x^2) = 4c^2$ $\Rightarrow x^2 + 36a^2 - 4x^2 = 4c^2$ $\Rightarrow 3x^2 = 36a^2 - 4c^2$ $\Rightarrow x^2 = 12a^2 - \frac{4}{3}c^2$.
For two distinct values of $x^2$ (which give four points),we need $0 < x^2 < 9a^2$ (since $x^2 < r^2$).
$0 < 12a^2 - \frac{4}{3}c^2 < 9a^2$.
$12a^2 - \frac{4}{3}c^2 > 0$ $\Rightarrow 36a^2 > 4c^2$ $\Rightarrow 9a^2 > c^2$ $\Rightarrow 3a > c$.
$12a^2 - \frac{4}{3}c^2 < 9a^2$ $\Rightarrow 3a^2 < \frac{4}{3}c^2$ $\Rightarrow 9a^2 < 4c^2$ $\Rightarrow 3a < 2c$.
Combining these,$c < 3a < 2c$. The condition $9ac - 9a^2 - 2c^2 > 0$ is derived from the intersection analysis.

(C) Given the equations of the ellipses:
$E_1: \frac{x^2}{3} + \frac{y^2}{2} = 1$
Here,$a^2 = 3$ and $b^2 = 2$. The eccentricity $e_1$ is given by $e_1 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2}{3}} = \frac{1}{\sqrt{3}}$.
$E_2: \frac{x^2}{16} + \frac{y^2}{b^2} = 1$
Assuming $16 > b^2$,the eccentricity $e_2$ is $e_2 = \sqrt{1 - \frac{b^2}{16}} = \frac{\sqrt{16 - b^2}}{4}$.
Given $e_1 \times e_2 = \frac{1}{2}$,we have:
$\frac{1}{\sqrt{3}} \times \frac{\sqrt{16 - b^2}}{4} = \frac{1}{2}$
$\frac{\sqrt{16 - b^2}}{4\sqrt{3}} = \frac{1}{2}$
$\sqrt{16 - b^2} = 2\sqrt{3} = \sqrt{12}$
$16 - b^2 = 12$ $\Rightarrow b^2 = 4$ $\Rightarrow b = 2$.
The length of the minor axis of $E_2$ is $2b = 2 \times 2 = 4$.

(A) The given ellipse is $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Here,$a^2 = 9$ and $b^2 = 4$.
The normal is parallel to the line $4x - 2y - 5 = 0$,so the slope of the normal is $m_n = \frac{-4}{-2} = 2$.
The slope of the tangent at that point is $m_t = -\frac{1}{m_n} = -\frac{1}{2}$.
The equation of the tangent with slope $m$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
The point of contact is $\left( -\frac{a^2m}{c}, \frac{b^2}{c} \right)$ where $c = \pm \sqrt{a^2m^2 + b^2}$.
Here $m = -\frac{1}{2}$,so $c = \sqrt{9(-\frac{1}{2})^2 + 4} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
The point of contact is $\left( -\frac{9(-1/2)}{5/2}, \frac{4}{5/2} \right) = \left( \frac{9/2}{5/2}, \frac{8}{5} \right) = \left( \frac{9}{5}, \frac{8}{5} \right)$.
Since the normal is parallel to $4x - 2y - 5 = 0$,the point is $\left( \frac{9}{5}, \frac{8}{5} \right)$.

(D) The slope of the chord joining $(0, 1)$ and $(2, 0)$ is $m = \frac{0 - 1}{2 - 0} = -\frac{1}{2}$.
Since the tangents at $P_1$ and $P_2$ are parallel to this chord,their slope is $m = -\frac{1}{2}$.
The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Here $a^2 = 4, b^2 = 1$,and $m = -\frac{1}{2}$.
Thus,$y = -\frac{1}{2}x \pm \sqrt{4(-\frac{1}{2})^2 + 1} = -\frac{1}{2}x \pm \sqrt{1 + 1} = -\frac{1}{2}x \pm \sqrt{2}$.
Substituting $y = -\frac{1}{2}x + \sqrt{2}$ into the ellipse equation $\frac{x^2}{4} + y^2 = 1$:
$\frac{x^2}{4} + (-\frac{x}{2} + \sqrt{2})^2 = 1$ $\Rightarrow \frac{x^2}{4} + \frac{x^2}{4} - \sqrt{2}x + 2 = 1$ $\Rightarrow \frac{x^2}{2} - \sqrt{2}x + 1 = 0$ $\Rightarrow x^2 - 2\sqrt{2}x + 2 = 0$ $\Rightarrow (x - \sqrt{2})^2 = 0$.
So,$x = \sqrt{2}$,which gives $y = -\frac{1}{2}(\sqrt{2}) + \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Point $P_1 = (\sqrt{2}, \frac{1}{\sqrt{2}})$.
Similarly,for $y = -\frac{1}{2}x - \sqrt{2}$,we get $x = -\sqrt{2}$ and $y = -\frac{1}{\sqrt{2}}$. Point $P_2 = (-\sqrt{2}, -\frac{1}{\sqrt{2}})$.
The distance $P_1P_2 = \sqrt{(\sqrt{2} - (-\sqrt{2}))^2 + (\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}))^2} = \sqrt{(2\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{8 + 2} = \sqrt{10}$.

(C) The given ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$. Here $a^2 = 2$ and $b^2 = 1$.
The equation of the tangent to the ellipse at point $(a \cos \theta, b \sin \theta)$ is $\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$.
The tangent intersects 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 $P(h, k)$ be the midpoint of the segment $AB$. Then:
$h = \frac{a}{2 \cos \theta} \Rightarrow \cos \theta = \frac{a}{2h}$
$k = \frac{b}{2 \sin \theta} \Rightarrow \sin \theta = \frac{b}{2k}$
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$\left( \frac{a}{2h} \right)^2 + \left( \frac{b}{2k} \right)^2 = 1$
$\frac{a^2}{4h^2} + \frac{b^2}{4k^2} = 1$
Substituting $a^2 = 2$ and $b^2 = 1$:
$\frac{2}{4h^2} + \frac{1}{4k^2} = 1$
$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The length of the latus rectum is $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.
The distance between the foci is $2ae$ and the length of the minor axis is $2b$. Given $2ae = 2b$,we have $ae = b$,so $a^2e^2 = b^2$.
Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$,we substitute $a^2e^2 = b^2$ to get $b^2 = a^2 - b^2$,or $2b^2 = a^2$.
Substituting $b^2 = 4a$ into $2b^2 = a^2$,we get $2(4a) = a^2$,so $a^2 = 8a$. Since $a \neq 0$,$a = 8$.
Then $b^2 = 4(8) = 32$,so $b = \sqrt{32} = 4\sqrt{2}$.
The equation of the ellipse is $\frac{x^2}{64} + \frac{y^2}{32} = 1$.
Testing the points,for $(4\sqrt{3}, 2\sqrt{2})$: $\frac{(4\sqrt{3})^2}{64} + \frac{(2\sqrt{2})^2}{32} = \frac{48}{64} + \frac{8}{32} = \frac{3}{4} + \frac{1}{4} = 1$. Thus,the point lies on the ellipse.

(A) Let the equation of 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 $B(0, b)$.
Since $\Delta S'BS$ is a right-angled triangle at $B$,the product of the slopes of $BS$ and $BS'$ is $-1$.
Slope of $BS = \frac{b-0}{0-ae} = -\frac{b}{ae}$.
Slope of $BS' = \frac{b-0}{0-(-ae)} = \frac{b}{ae}$.
Since the angle at $B$ is $90^\circ$,the product of slopes is $-1$,so $(-\frac{b}{ae}) \times (\frac{b}{ae}) = -1$,which implies $b^2 = a^2e^2$.
We know that for an ellipse,$b^2 = a^2(1-e^2)$,so $a^2e^2 = a^2 - a^2e^2$,which gives $a^2e^2 = \frac{a^2}{2}$,so $b^2 = \frac{a^2}{2}$.
The area of $\Delta S'BS = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2ae) \times b = aeb = 8$.
Squaring both sides,$a^2e^2b^2 = 64$. Substituting $a^2e^2 = b^2$,we get $b^4 = 64$,so $b^2 = 8$.
Since $b^2 = \frac{a^2}{2}$,we have $8 = \frac{a^2}{2}$,so $a^2 = 16$,which means $a = 4$.
The length of the latus rectum is $\frac{2b^2}{a} = \frac{2(8)}{4} = 4$.

(C) Let $P = (h, k)$. The perimeter of $\Delta AOP$ is $AP + OP + AO = 4$.
Given $O(0, 0)$ and $A(0, 1)$,we have $AO = 1$.
So,$AP + OP = 4 - 1 = 3$.
Using the distance formula,$\sqrt{h^2 + (k - 1)^2} + \sqrt{h^2 + k^2} = 3$.
$\sqrt{h^2 + (k - 1)^2} = 3 - \sqrt{h^2 + k^2}$.
Squaring both sides:
$h^2 + k^2 - 2k + 1 = 9 + h^2 + k^2 - 6\sqrt{h^2 + k^2}$.
$-2k - 8 = -6\sqrt{h^2 + k^2}$.
$k + 4 = 3\sqrt{h^2 + k^2}$.
Squaring again:
$k^2 + 8k + 16 = 9(h^2 + k^2)$.
$k^2 + 8k + 16 = 9h^2 + 9k^2$.
$9h^2 + 8k^2 - 8k - 16 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2 + 8y^2 - 8y = 16$.

(A) The equation of the ellipse is $4x^2 + y^2 = 8$.
Differentiating with respect to $x$,we get $8x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{4x}{y}$.
The slope of the tangent at $(1, 2)$ is $m_1 = -\frac{4(1)}{2} = -2$.
Let the slope of the tangent at $(a, b)$ be $m_2 = -\frac{4a}{b}$.
Since the tangents are perpendicular,$m_1 \times m_2 = -1$,so $(-2) \times (-\frac{4a}{b}) = -1$,which gives $\frac{8a}{b} = -1$,or $b = -8a$.
Since $(a, b)$ lies on the ellipse,$4a^2 + b^2 = 8$.
Substituting $b = -8a$,we get $4a^2 + (-8a)^2 = 8$,which simplifies to $4a^2 + 64a^2 = 8$.
Thus,$68a^2 = 8$,so $a^2 = \frac{8}{68} = \frac{2}{17}$.