Ellipse Questions in English

(D) The equation of the chord of contact $QR$ for the point $P(8, 27)$ with respect to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is given by $T = 0$:
$\frac{8x}{4} + \frac{27y}{9} = 1 \Rightarrow 2x + 3y = 1$.
The equation of the pair of lines passing through the origin and the points $Q$ and $R$ is obtained by homogenizing the ellipse equation using the chord of contact:
$\frac{x^{2}}{4} + \frac{y^{2}}{9} = (2x + 3y)^{2}$.
$9x^{2} + 4y^{2} = 36(4x^{2} + 12xy + 9y^{2})$.
$9x^{2} + 4y^{2} = 144x^{2} + 432xy + 324y^{2}$.
$135x^{2} + 432xy + 320y^{2} = 0$.
Comparing this with $ax^{2} + 2hxy + by^{2} = 0$,we have $a = 135$,$2h = 432$ $(h = 216)$,and $b = 320$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^{2} - ab}}{a + b} \right|$.
$\tan \theta = \left| \frac{2\sqrt{216^{2} - 135 \cdot 320}}{135 + 320} \right| = \left| \frac{2\sqrt{46656 - 43200}}{455} \right| = \frac{2\sqrt{3456}}{455} = \frac{2 \cdot 24 \sqrt{6}}{455} = \frac{48 \sqrt{6}}{455}$.
Thus,$\theta = \tan^{-1} \left( \frac{48 \sqrt{6}}{455} \right)$,which is not among the given options.

(A) The given curve is $9x^{2} + 16y^{2} = 144$.
Dividing by $144$,we get $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.
Here,$a^{2} = 16$ and $b^{2} = 9$.
Let the equation of the tangent be $x + y = k$,which can be written as $y = -x + k$.
Comparing this with $y = mx + c$,we have $m = -1$ and $c = k$.
The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} + b^{2}$.
Substituting the values,we get $k^{2} = 16(-1)^{2} + 9$.
$k^{2} = 16 + 9 = 25$.
$k = \pm 5$.
Thus,the equations of the tangents are $x + y = 5$ and $x + y = -5$.

(C) The equation of the ellipse is $\frac{x^{2}}{32}+\frac{y^{2}}{18}=1$,where $a^{2}=32$ and $b^{2}=18$.
The equation of a tangent with slope $m = -\frac{3}{4}$ is $y = mx \pm \sqrt{a^{2}m^{2}+b^{2}}$.
Substituting the values: $y = -\frac{3}{4}x \pm \sqrt{32 \times (-\frac{3}{4})^{2} + 18}$.
$y = -\frac{3}{4}x \pm \sqrt{32 \times \frac{9}{16} + 18} = -\frac{3}{4}x \pm \sqrt{18 + 18} = -\frac{3}{4}x \pm 6$.
Considering the positive intercept,the equation is $y = -\frac{3}{4}x + 6$,which simplifies to $3x + 4y = 24$.
The intercepts are found by setting $y=0$ and $x=0$:
For $y=0$,$3x=24 \Rightarrow x=8$,so $A = (8, 0)$.
For $x=0$,$4y=24 \Rightarrow y=6$,so $B = (0, 6)$.
The area of $\Delta AOB = \frac{1}{2} \times |OA| \times |OB| = \frac{1}{2} \times 8 \times 6 = 24$ sq unit.

(A) The given equation of the ellipse is $5x^{2} + 9y^{2} = 45$. Dividing by $45$,we get $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$.
Here,$a^{2} = 9$ and $b^{2} = 5$.
The line $4x - 3y + k = 0$ can be written as $y = \frac{4}{3}x + \frac{k}{3}$.
Comparing this with the line $y = mx + c$,we have $m = \frac{4}{3}$ and $c = \frac{k}{3}$.
The condition for the line $y = mx + c$ to touch the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} + b^{2}$.
Substituting the values,we get $(\frac{k}{3})^{2} = 9(\frac{4}{3})^{2} + 5$.
$\frac{k^{2}}{9} = 9(\frac{16}{9}) + 5 = 16 + 5 = 21$.
$k^{2} = 9 \times 21 = 189$.
$k = \pm \sqrt{189} = \pm 3\sqrt{21}$.

(D) The equation of any normal to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at point $(a \cos \phi, b \sin \phi)$ is given by $ax \sec \phi - by \operatorname{cosec} \phi = a^{2} - b^{2} \quad \dots (i)$
Comparing this with the given line $x \cos \alpha + y \sin \alpha = p$,we have:
$\frac{a \sec \phi}{\cos \alpha} = \frac{-b \operatorname{cosec} \phi}{\sin \alpha} = \frac{a^{2} - b^{2}}{p}$
From this,we get $\sec \phi = \frac{(a^{2} - b^{2}) \cos \alpha}{ap}$ and $\operatorname{cosec} \phi = \frac{-(a^{2} - b^{2}) \sin \alpha}{bp}$.
Using the identity $\cos^{2} \phi + \sin^{2} \phi = 1$,we have $\frac{1}{\sec^{2} \phi} + \frac{1}{\operatorname{cosec}^{2} \phi} = 1$.
Substituting the values,we get $\frac{a^{2} p^{2}}{(a^{2} - b^{2})^{2} \cos^{2} \alpha} + \frac{b^{2} p^{2}}{(a^{2} - b^{2})^{2} \sin^{2} \alpha} = 1$.
Multiplying by $(a^{2} - b^{2})^{2}$,we get $p^{2} (a^{2} \sec^{2} \alpha + b^{2} \operatorname{cosec}^{2} \alpha) = (a^{2} - b^{2})^{2}$.

(B) Let the coordinates of $P$ be $(a \cos \theta, b \sin \theta)$.
Since $CD$ is the semi-conjugate diameter,the coordinates of $D$ are $(a \cos(\theta + \frac{\pi}{2}), b \sin(\theta + \frac{\pi}{2})) = (-a \sin \theta, b \cos \theta)$.
Now,$CP^{2} = a^{2} \cos^{2} \theta + b^{2} \sin^{2} \theta$.
And $CD^{2} = (-a \sin \theta)^{2} + (b \cos \theta)^{2} = a^{2} \sin^{2} \theta + b^{2} \cos^{2} \theta$.
Adding these,we get $CP^{2} + CD^{2} = a^{2}(\cos^{2} \theta + \sin^{2} \theta) + b^{2}(\sin^{2} \theta + \cos^{2} \theta)$.
Since $\cos^{2} \theta + \sin^{2} \theta = 1$,we have $CP^{2} + CD^{2} = a^{2} + b^{2}$.

(A) The equation of the ellipse is $9x^2 + 16y^2 = 288$. Dividing by $288$,we get $\frac{x^2}{32} + \frac{y^2}{18} = 1$. Here $a^2 = 32$ and $b^2 = 18$.
The equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Since the tangent makes equal intercepts on the axes,its slope $m$ must be $\pm 1$. Given it makes equal intercepts,the line equation is $x + y = c$ or $x - y = c$. For equal intercepts on the axes,the slope is $-1$,so the tangent is $y = -x + c$,or $x + y = c$.
The condition for $y = mx + c$ to be a tangent is $c^2 = a^2m^2 + b^2$. Substituting $m = -1$,$a^2 = 32$,and $b^2 = 18$:
$c^2 = 32(-1)^2 + 18 = 32 + 18 = 50$.
Thus,$c = \pm 5\sqrt{2}$.
The intercepts are $A(c, 0)$ and $B(0, c)$. The area of $\triangle OAB$ is $\frac{1}{2} |c| |c| = \frac{1}{2} c^2 = \frac{1}{2} (50) = 25$ sq. units.

(C) The area of an ellipse given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by the formula $A = \pi ab$.
Given that the area is $\frac{\pi}{6}$,we have $\pi ab = \frac{\pi}{6}$,which implies $ab = \frac{1}{6}$.
Let us check the options:
For option $C$,the equation is $4x^2 + 9y^2 = 1$,which can be written as $\frac{x^2}{1/4} + \frac{y^2}{1/9} = 1$.
Here,$a^2 = \frac{1}{4} \implies a = \frac{1}{2}$ and $b^2 = \frac{1}{9} \implies b = \frac{1}{3}$.
The area is $\pi ab = \pi \times \frac{1}{2} \times \frac{1}{3} = \frac{\pi}{6}$.
Thus,option $C$ is correct.

(D) The given parametric equations are $x = 4 \cos \theta$ and $y = 3 \sin \theta$.
This represents an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a = 4$ and $b = 3$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values of $a$ and $b$, we get $A = \pi \times 4 \times 3 = 12 \pi$ sq. units.
Therefore, the correct option is $D$.

(D) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 4$ and $b^2 = 9$.
Thus,$a = 2$ and $b = 3$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values,we get $A = \pi \times 2 \times 3 = 6 \pi$.
Therefore,the correct option is $D$.

(A) The given equation of the ellipse is $4x^2 + 9y^2 = 1$.
Rewriting this in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we get:
$\frac{x^2}{(1/2)^2} + \frac{y^2}{(1/3)^2} = 1$.
Here,$a = \frac{1}{2}$ and $b = \frac{1}{3}$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values,we get $A = \pi \times \frac{1}{2} \times \frac{1}{3} = \frac{\pi}{6}$ sq. units.

(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 9$.
This implies $a = 4$ and $b = 3$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values of $a$ and $b$, we get $A = \pi \times 4 \times 3 = 12 \pi$.
Thus, the area of the region enclosed by the ellipse is $12 \pi$ square units.

(B) The equation of the ellipse is $\frac{x^{2}}{4}+y^{2}=1$,where $a^{2}=4$ and $b^{2}=1$.
The equation of the line is $y=mx+c$,where $m=4$.
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 $c^{2}=a^{2}m^{2}+b^{2}$.
Substituting the values,we get $c^{2}=4(4)^{2}+1$.
$c^{2}=4(16)+1=64+1=65$.
Therefore,$c=\pm \sqrt{65}$.
Since there are two possible values for $c$,the line touches the curve for $2$ values of $c$.

(B) The given equation is $\frac{x^{2}}{2-\lambda} - \frac{y^{2}}{\lambda-5} = 1$.
For this to represent an ellipse,the equation must be of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where $a^{2} > 0$ and $b^{2} > 0$.
Rewriting the equation: $\frac{x^{2}}{2-\lambda} + \frac{y^{2}}{5-\lambda} = 1$.
For this to be an ellipse,both denominators must be positive:
$2 - \lambda > 0 \implies \lambda < 2$
$5 - \lambda > 0 \implies \lambda < 5$
Taking the intersection of these two conditions,we get $\lambda < 2$.

(C) Given equation: $x^2+3y^2=12$
Divide by $12$: $\frac{x^2}{12} + \frac{y^2}{4} = 1$
This is an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 12$ and $b^2 = 4$.
Thus,$a = \sqrt{12} = 2\sqrt{3}$ and $b = 2$.
The length of the latus rectum is given by $\frac{2b^2}{a}$.
Length $= \frac{2 \times 4}{2\sqrt{3}} = \frac{4}{\sqrt{3}}$ units.

(C) Given equation of the ellipse is $9x^{2} + 25y^{2} = 225$.
Dividing both sides by $225$,we get:
$\frac{9x^{2}}{225} + \frac{25y^{2}}{225} = 1$
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
Comparing this with the standard form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we have $a^{2} = 25$ and $b^{2} = 9$.
Thus,$a = 5$ and $b = 3$.
The eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$.
$e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

(C) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$. Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a^{2}=16$ and $b^{2}=4$,so $a=4$ and $b=2$.
Let $\theta$ be the eccentric angle of the point $P(2, \sqrt{3})$.
The parametric coordinates of any point on the ellipse are given by $x = a \cos \theta$ and $y = b \sin \theta$.
Substituting the values,we have $x = 4 \cos \theta$ and $y = 2 \sin \theta$.
Given the point $(2, \sqrt{3})$,we equate:
$2 = 4 \cos \theta \Rightarrow \cos \theta = \frac{2}{4} = \frac{1}{2}$
$\sqrt{3} = 2 \sin \theta \Rightarrow \sin \theta = \frac{\sqrt{3}}{2}$
Since both $\sin \theta = \frac{\sqrt{3}}{2}$ and $\cos \theta = \frac{1}{2}$ are satisfied at $\theta = \frac{\pi}{3}$,the eccentric angle is $\frac{\pi}{3}$.

(D) The given equation of the ellipse is $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.
Comparing this with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a^{2}=36$,which implies $a=6$.
By the definition of an ellipse,the sum of the focal distances of any point $P$ on the ellipse is equal to the length of the major axis.
Therefore,$PS + PS^{\prime} = 2a$.
Substituting the value of $a$,we get $PS + PS^{\prime} = 2 \times 6 = 12$.

(D) The equation of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $x^{2}+y^{2}=a^{2}$.
Area of the auxiliary circle $= \pi a^{2}$.
Area of the ellipse $= \pi ab$.
According to the problem,the area of the auxiliary circle is twice the area of the ellipse:
$\pi a^{2} = 2(\pi ab)$
$a^{2} = 2ab$
$a = 2b \Rightarrow b = \frac{a}{2}$.
The eccentricity $e$ of the ellipse is given by:
$e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 - \frac{(a/2)^{2}}{a^{2}}}$
$e = \sqrt{1 - \frac{a^{2}/4}{a^{2}}}$
$e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(D) The given curve is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The point $P$ is $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$.
Taking the derivative,$\frac{2x}{a^{2}} + \frac{2y}{b^{2}}\frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{b^{2}x}{a^{2}y}$.
At $P$,the slope of the tangent $m_{t} = -\frac{b^{2}(a/\sqrt{2})}{a^{2}(b/\sqrt{2})} = -\frac{b}{a}$.
The equation of the tangent at $P$ is $y - \frac{b}{\sqrt{2}} = -\frac{b}{a}\left(x - \frac{a}{\sqrt{2}}\right)$,which simplifies to $bx + ay = \sqrt{2}ab$.
For the $x$-axis $(y=0)$,$x = \sqrt{2}a$. So,the first vertex is $A(\sqrt{2}a, 0)$.
The slope of the normal $m_{n} = \frac{a}{b}$.
The equation of the normal at $P$ is $y - \frac{b}{\sqrt{2}} = \frac{a}{b}\left(x - \frac{a}{\sqrt{2}}\right)$,which simplifies to $ax - by = \frac{a^{2}-b^{2}}{\sqrt{2}}$.
For the $x$-axis $(y=0)$,$x = \frac{a^{2}-b^{2}}{a\sqrt{2}}$. So,the second vertex is $B\left(\frac{a^{2}-b^{2}}{a\sqrt{2}}, 0\right)$.
The third vertex is $P\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \left|\sqrt{2}a - \frac{a^{2}-b^{2}}{a\sqrt{2}}\right| \times \frac{b}{\sqrt{2}}$.
$= \frac{1}{2} \times \left|\frac{2a^{2}-a^{2}+b^{2}}{a\sqrt{2}}\right| \times \frac{b}{\sqrt{2}} = \frac{1}{2} \times \frac{a^{2}+b^{2}}{a\sqrt{2}} \times \frac{b}{\sqrt{2}} = \frac{b(a^{2}+b^{2})}{4a}$.

(D) The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.
Given the line $x \cos \alpha + y \sin \alpha = 4$,we can rewrite it as $y = -x \cot \alpha + 4 \operatorname{cosec} \alpha$.
Here,$m = -\cot \alpha$,$c = 4 \operatorname{cosec} \alpha$,$a^2 = 25$,and $b^2 = 9$.
Substituting these into the condition $c^2 = a^2m^2 + b^2$:
$16 \operatorname{cosec}^2 \alpha = 25 \cot^2 \alpha + 9$.
Using $\operatorname{cosec}^2 \alpha = 1 + \cot^2 \alpha$,we get $16(1 + \cot^2 \alpha) = 25 \cot^2 \alpha + 9$.
$16 + 16 \cot^2 \alpha = 25 \cot^2 \alpha + 9$.
$9 \cot^2 \alpha = 7 \implies \cot^2 \alpha = 7/9$.
Thus,$\tan^2 \alpha = 9/7$,which gives $\tan \alpha = 3/\sqrt{7}$.
Therefore,$\alpha = \tan^{-1}(3/\sqrt{7})$.

(D) The equation of a pair of tangents from $(\alpha, \beta)$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by the condition $SS_{1}=T^{2}$.
The tangents are perpendicular if the sum of the coefficients of $x^{2}$ and $y^{2}$ in the combined equation is zero.
Expanding the condition,we get:
$\left(\frac{1}{a^{2}}\right)\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right) - \frac{\alpha^{2}}{a^{4}} + \left(\frac{1}{b^{2}}\right)\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right) - \frac{\beta^{2}}{b^{4}} = 0$
Simplifying this expression:
$\frac{\beta^{2}}{a^{2}b^{2}} - \frac{1}{a^{2}} + \frac{\alpha^{2}}{a^{2}b^{2}} - \frac{1}{b^{2}} = 0$
$\alpha^{2} + \beta^{2} = a^{2}b^{2}\left(\frac{1}{a^{2}} + \frac{1}{b^{2}}\right) = a^{2} + b^{2}$
Thus,the locus of the point $(\alpha, \beta)$ is $x^{2} + y^{2} = a^{2} + b^{2}$,which is known as the director circle.

(C) The equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$.
Let the point $P$ on the ellipse be $(a \cos \theta, b \sin \theta)$.
The area of $\triangle S P S^{\prime}$ is given by the determinant formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$\text{Area} = \frac{1}{2} |ae(b \sin \theta - 0) + a \cos \theta(0 - 0) + (-ae)(0 - b \sin \theta)|$
$\text{Area} = \frac{1}{2} |aeb \sin \theta + aeb \sin \theta| = |abe \sin \theta|$.
Since the maximum value of $\sin \theta$ is $1$,the maximum area of $\triangle S P S^{\prime}$ is $abe$.

(C) By definition,an ellipse is the set of all points in a plane such that the sum of their distances from two fixed points (called foci) is a constant.
Therefore,the locus of such a point is an ellipse.

(A) The given equation of the ellipse is $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.
Comparing this with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we have $a^{2}=36$ and $b^{2}=16$.
The eccentricity $e$ of an ellipse is given by the formula $e=\sqrt{1-\frac{b^{2}}{a^{2}}}$.
Substituting the values,we get $e=\sqrt{1-\frac{16}{36}}$.
$e=\sqrt{\frac{36-16}{36}}=\sqrt{\frac{20}{36}}$.
$e=\frac{\sqrt{20}}{6}=\frac{2 \sqrt{5}}{6}$.

(A) The area of an ellipse given by the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is calculated as $\pi |ab|$.
Given the equation $\frac{x^{2}}{25}+\frac{y^{2}}{\lambda^{2}}=1$,we identify $a^{2} = 25$ and $b^{2} = \lambda^{2}$.
Thus,$a = 5$ and $b = |\lambda|$.
The area is given as $20 \pi$ sq units.
Substituting these values into the area formula: $\pi \times 5 \times |\lambda| = 20 \pi$.
Dividing both sides by $5 \pi$,we get $|\lambda| = 4$.
Therefore,$\lambda = \pm 4$.

(A) The $4^{th}$ roots of unity are $1, -1, i, -i$. The real roots are $\alpha=1, \beta=-1$ and the other roots are $\gamma=i, \delta=-i$.
For the first conic $|z-1|+|z+1|=4$,this represents an ellipse with foci at $(\pm 1, 0)$ and major axis $2a=4$,so $a=2$. The distance between foci is $2ae_1=2$,so $e_1=\frac{1}{2}$.
For the second conic $|z-i|+|z+i|=6$,this represents an ellipse with foci at $(0, \pm 1)$ and major axis $2a=6$,so $a=3$. The distance between foci is $2ae_2=2$,so $e_2=\frac{1}{3}$.
The sum of the eccentricities is $e_1+e_2 = \frac{1}{2}+\frac{1}{3} = \frac{5}{6}$.

(B) Let the coordinates of the moving point be $(x, y)$.
According to the definition of an ellipse,the sum of the distances of a point from two fixed points (foci) is constant $(2a)$.
The given condition is $\sqrt{(x - ae)^2 + y^2} + \sqrt{(x + ae)^2 + y^2} = 2a$.
Squaring and simplifying this equation:
$\sqrt{(x - ae)^2 + y^2} = 2a - \sqrt{(x + ae)^2 + y^2}$.
Squaring both sides:
$(x - ae)^2 + y^2 = 4a^2 + (x + ae)^2 + y^2 - 4a\sqrt{(x + ae)^2 + y^2}$.
$x^2 - 2aex + a^2e^2 + y^2 = 4a^2 + x^2 + 2aex + a^2e^2 + y^2 - 4a\sqrt{(x + ae)^2 + y^2}$.
$-4aex - 4a^2 = -4a\sqrt{(x + ae)^2 + y^2}$.
$ex + a = \sqrt{(x + ae)^2 + y^2}$.
Squaring again:
$e^2x^2 + 2aex + a^2 = x^2 + 2aex + a^2e^2 + y^2$.
$x^2(1 - e^2) + y^2 = a^2(1 - e^2)$.
Dividing by $a^2(1 - e^2)$:
$\frac{x^2(1 - e^2)}{a^2(1 - e^2)} + \frac{y^2}{a^2(1 - e^2)} = 1$.
Since $b^2 = a^2(1 - e^2)$,we get $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

(D) Given the equation: $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$.
Let $P = (x, y)$,$A = (2, 0)$,and $B = (-2, 0)$.
The equation represents the sum of distances from point $P$ to points $A$ and $B$,which is $PA + PB = 4$.
The distance between $A(2, 0)$ and $B(-2, 0)$ is $AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$.
Since $PA + PB = AB$,the point $P$ must lie on the line segment connecting $A$ and $B$.
Given the condition $-2 < x < 2$ and $y=0$,this represents the line segment on the $x$-axis between $x = -2$ and $x = 2$.
Thus,the equation represents a line segment.
Hence,option $D$ is correct.

(C) Let $P(x, y)$ be any point on the locus. According to the problem,the sum of distances from $P$ to $F_1(0, 2)$ and $F_2(0, -2)$ is $6$.
$\sqrt{x^2 + (y - 2)^2} + \sqrt{x^2 + (y + 2)^2} = 6$
$\sqrt{x^2 + (y - 2)^2} = 6 - \sqrt{x^2 + (y + 2)^2}$
Squaring both sides:
$x^2 + y^2 - 4y + 4 = 36 + x^2 + y^2 + 4y + 4 - 12\sqrt{x^2 + (y + 2)^2}$
$-8y - 36 = -12\sqrt{x^2 + (y + 2)^2}$
$2y + 9 = 3\sqrt{x^2 + (y + 2)^2}$
Squaring again:
$4y^2 + 36y + 81 = 9(x^2 + y^2 + 4y + 4)$
$4y^2 + 36y + 81 = 9x^2 + 9y^2 + 36y + 36$
$9x^2 + 5y^2 = 45$

(B) The given equation is $9x^2 + 25y^2 - 90x - 150y + 225 = 0$.
Rearranging the terms,we get $9(x^2 - 10x) + 25(y^2 - 6y) = -225$.
Completing the square,$9(x^2 - 10x + 25) + 25(y^2 - 6y + 9) = -225 + 225 + 225$.
$9(x - 5)^2 + 25(y - 3)^2 = 225$.
Dividing by $225$,we get $\frac{(x - 5)^2}{25} + \frac{(y - 3)^2}{9} = 1$.
This is the equation of an ellipse with $a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.
The length of the latus rectum is given by $\frac{2b^2}{a} = \frac{2 \times 9}{5} = \frac{18}{5}$.

(C) Given the equation of the ellipse: $x^2+2y^2-4x+12y+14=0$
Rearranging the terms: $(x^2-4x) + 2(y^2+6y) = -14$
Completing the square for $x$ and $y$: $(x^2-4x+4) + 2(y^2+6y+9) = -14+4+18$
$(x-2)^2 + 2(y+3)^2 = 8$
Dividing by $8$: $\frac{(x-2)^2}{8} + \frac{(y+3)^2}{4} = 1$
Comparing this with the standard form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$,the center $(h, k)$ is $(2, -3)$.

(C) The given equation of the ellipse is $7x^2 + 16y^2 = 112$.
Dividing by $112$,we get $\frac{x^2}{16} + \frac{y^2}{7} = 1$.
Here,$a^2 = 16$ and $b^2 = 7$.
The eccentricity $e$ of the ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.
$e = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.
By the definition of an ellipse,the ratio of the distance of a point $P$ from the focus $S$ to its distance from the directrix $L$ is equal to the eccentricity $e$.
Therefore,$\frac{SP}{PM} = e = \frac{3}{4}$.

(D) Let the point be $(x, y) = (2 \cos \theta-3, 3 \sin \theta-4)$.
From this,we have $\cos \theta = \frac{x+3}{2}$ and $\sin \theta = \frac{y+4}{3}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we substitute the expressions:
$(\frac{x+3}{2})^2 + (\frac{y+4}{3})^2 = 1$
$\frac{x^2+6x+9}{4} + \frac{y^2+8y+16}{9} = 1$
Multiplying by $36$ to clear the denominators:
$9(x^2+6x+9) + 4(y^2+8y+16) = 36$
$9x^2 + 54x + 81 + 4y^2 + 32y + 64 = 36$
$9x^2 + 4y^2 + 54x + 32y + 145 - 36 = 0$
$9x^2 + 4y^2 + 54x + 32y + 109 = 0$.

(A) $1$. For an ellipse,the distance from the focus $(x_1, y_1)$ to the directrix $ax + by + c = 0$ is given by $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.
$2$. Here,focus is $(1, -2)$ and directrix is $3x + 4y - 15 = 0$.
$3$. $d = \frac{|3(1) + 4(-2) - 15|}{\sqrt{3^2 + 4^2}} = \frac{|3 - 8 - 15|}{5} = \frac{|-20|}{5} = 4$.
$4$. The formula for the distance from focus to directrix is $\frac{a}{e} - ae = \frac{a(1 - e^2)}{e}$. Thus,Reason $(R)$ is true.
$5$. The length of the latus rectum is $\frac{2b^2}{a} = 4$,so $b^2 = 2a$.
$6$. We know $b^2 = a^2(1 - e^2)$,so $2a = a^2(1 - e^2) \implies 2 = a(1 - e^2)$.
$7$. From step $3$ and $4$,$\frac{a(1 - e^2)}{e} = 4$. Substituting $a(1 - e^2) = 2$,we get $\frac{2}{e} = 4 \implies e = \frac{1}{2}$.
$8$. Since both Assertion $(A)$ and Reason $(R)$ are true and $(R)$ provides the correct formula used to derive $e$ in $(A)$,$(A)$ and $(R)$ are true and $(R)$ is the correct explanation to $(A)$.

(A) The given equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ with $b < 2$.
Here,$a^2 = 4$,so $a = 2$.
The foci are $F(c, 0)$ and $F'(-c, 0)$,and the end of the minor axis is $B(0, b)$.
The area of $\triangle FBF'$ is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2c) \times b = bc$.
Given $bc = \sqrt{3}$,so $b^2 c^2 = 3$,which implies $c^2 = \frac{3}{b^2}$.
For an ellipse,$b^2 = a^2(1 - e^2) = a^2 - a^2 e^2 = a^2 - c^2$.
Thus,$b^2 = 4 - c^2$,or $c^2 = 4 - b^2$.
Substituting $c^2 = \frac{3}{b^2}$ into $c^2 = 4 - b^2$:
$\frac{3}{b^2} = 4 - b^2$ $\Rightarrow 3 = 4b^2 - b^4$ $\Rightarrow b^4 - 4b^2 + 3 = 0$.
Let $t = b^2$,then $t^2 - 4t + 3 = 0 \Rightarrow (t - 1)(t - 3) = 0$.
So,$b^2 = 1$ or $b^2 = 3$.
Case $1$: $b^2 = 1$. Then $c^2 = 4 - 1 = 3$,so $c = \sqrt{3}$.
Eccentricity $e = \frac{c}{a} = \frac{\sqrt{3}}{2}$.
Case $2$: $b^2 = 3$. Then $c^2 = 4 - 3 = 1$,so $c = 1$.
Eccentricity $e = \frac{c}{a} = \frac{1}{2}$.
Thus,the eccentricity is $\frac{\sqrt{3}}{2}$ or $\frac{1}{2}$.

(D) Given the equation of the ellipse: $16x^2 + 25y^2 = 400$.
Dividing both sides by $400$,we get: $\frac{16x^2}{400} + \frac{25y^2}{400} = 1$,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 is given by the formula $\frac{2b^2}{a}$.
Substituting the values: $\frac{2 \times 16}{5} = \frac{32}{5}$.

(D) The definition of an ellipse states that $SP + S^{\prime}P = 2a$. Given $SP + S^{\prime}P = 2$,we have $2a = 2$,so $a = 1$.
The distance between the foci $SS^{\prime}$ is given by $\sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$.
Since $SS^{\prime} = 2ae$,we have $2ae = \sqrt{2}$,which implies $e = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
The center of the ellipse is the midpoint of the foci $S$ and $S^{\prime}$,which is $(\frac{1+0}{2}, \frac{0+1}{2}) = (\frac{1}{2}, \frac{1}{2})$.
The major axis lies along the line passing through $S(1,0)$ and $S^{\prime}(0,1)$. The slope of this line is $m = \frac{1-0}{0-1} = -1$.
The unit vector along the major axis is $\vec{u} = \frac{S-S^{\prime}}{|S-S^{\prime}|} = \frac{(1, -1)}{\sqrt{2}} = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.
The endpoints of the major axis are at a distance $a=1$ from the center $C(\frac{1}{2}, \frac{1}{2})$ along the direction of the major axis.
Thus,$A, A^{\prime} = C \pm a\vec{u} = (\frac{1}{2} \pm \frac{1}{\sqrt{2}}, \frac{1}{2} \mp \frac{1}{\sqrt{2}})$.
Therefore,$x_1 = \frac{1}{2} + \frac{1}{\sqrt{2}}$ and $x_2 = \frac{1}{2} - \frac{1}{\sqrt{2}}$.
Summing these,$x_1 + x_2 = (\frac{1}{2} + \frac{1}{\sqrt{2}}) + (\frac{1}{2} - \frac{1}{\sqrt{2}}) = 1$.

(D) The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$.
Here,$a^2=25$ and $b^2=16$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The foci are $(\pm ae, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$.
Let the focal chord pass through the focus $F(3,0)$ and the point $A(0,3)$ on the minor axis.
The slope of the chord $AF$ is $m = \frac{3-0}{0-3} = -1$.
The equation of the chord is $y - 0 = -1(x - 3)$,which simplifies to $x + y - 3 = 0$.
The perpendicular distance $d$ from the centre $(0,0)$ to the line $x + y - 3 = 0$ is given by $d = \frac{|0 + 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}}$.

(B) The given equation of the ellipse is $x^2+3y^2=6$,which can be rewritten as $\frac{x^2}{6}+\frac{y^2}{2}=1$.
Let the eccentric angle of the point be $\theta$. The coordinates of any point on the ellipse are given by $(a \cos \theta, b \sin \theta)$,where $a^2=6$ and $b^2=2$.
Thus,the coordinates are $(\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)$.
The distance of this point from the centre $(0,0)$ is $2$ units.
So,$(\sqrt{6} \cos \theta)^2 + (\sqrt{2} \sin \theta)^2 = 2^2$.
$6 \cos^2 \theta + 2 \sin^2 \theta = 4$.
$6 \cos^2 \theta + 2(1 - \cos^2 \theta) = 4$.
$4 \cos^2 \theta + 2 = 4$ $\Rightarrow 4 \cos^2 \theta = 2$ $\Rightarrow \cos^2 \theta = \frac{1}{2}$.
$\cos \theta = \pm \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{\pi}{4}$ or $\frac{3\pi}{4}$.

(D) Given the ellipse equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
Since $b^2 > a^2$ $(9 > 4)$,this is a vertical ellipse with $a^2 = 4$ and $b^2 = 9$.
The eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$,so $4 = 9(1 - e^2)$ $\Rightarrow 1 - e^2 = \frac{4}{9}$ $\Rightarrow e^2 = \frac{5}{9}$ $\Rightarrow e = \frac{\sqrt{5}}{3}$.
The foci are $(0, \pm be) = (0, \pm 3 \times \frac{\sqrt{5}}{3}) = (0, \pm \sqrt{5})$.
Let $S = (0, \sqrt{5})$ and $S' = (0, -\sqrt{5})$. Let $P = \left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$.
The focal distances are $PS$ and $PS'$.
$PS = \sqrt{\left(\frac{4}{\sqrt{5}} - 0\right)^2 + \left(\frac{3}{\sqrt{5}} - \sqrt{5}\right)^2} = \sqrt{\frac{16}{5} + \left(\frac{3-5}{\sqrt{5}}\right)^2} = \sqrt{\frac{16}{5} + \frac{4}{5}} = \sqrt{\frac{20}{5}} = 2$.
$PS' = \sqrt{\left(\frac{4}{\sqrt{5}} - 0\right)^2 + \left(\frac{3}{\sqrt{5}} + \sqrt{5}\right)^2} = \sqrt{\frac{16}{5} + \left(\frac{3+5}{\sqrt{5}}\right)^2} = \sqrt{\frac{16}{5} + \frac{64}{5}} = \sqrt{\frac{80}{5}} = \sqrt{16} = 4$.
Thus,the focal distances are $4$ and $2$.

(D) Let the foci be $S_1(-ae, 0)$ and $S_2(ae, 0)$,and the end of the minor axis be $A(0, b)$.
The angle $\angle S_1 A S_2 = 90^{\circ}$.
In $\Delta S_1 A S_2$,since $\angle S_1 A S_2 = 90^{\circ}$,the median $AO$ to the hypotenuse $S_1 S_2$ is half the length of the hypotenuse.
$AO = \frac{1}{2} S_1 S_2$
$b = \frac{1}{2} (2ae) = ae$
Since $b^2 = a^2(1 - e^2)$,we have:
$(ae)^2 = a^2(1 - e^2)$
$a^2 e^2 = a^2 - a^2 e^2$
$2a^2 e^2 = a^2$
$e^2 = \frac{1}{2}$
$e = \frac{1}{\sqrt{2}}$

(C) Given the equation of the ellipse: $\frac{x^2}{36}+\frac{y^2}{20}=1$.
Comparing this with the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we have $a^2=36$ and $b^2=20$,which gives $a=6$.
The eccentricity $e$ is calculated as $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{20}{36}} = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3}$.
The equations of the directrices for an ellipse with $a > b$ are $x = \pm \frac{a}{e}$.
Substituting the values,we get $x = \pm \frac{6}{2/3} = \pm 9$.
The distance between the directrices is the difference between these two values: $|9 - (-9)| = 18$.

(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
Length of the latus rectum $= \frac{2b^2}{a}$.
Length of the minor axis $= 2b$.
According to the given condition,the latus rectum is equal to half of the minor axis:
$\frac{2b^2}{a} = \frac{1}{2} (2b) = b$.
Dividing both sides by $b$ (since $b \neq 0$),we get $\frac{2b}{a} = 1$,which implies $\frac{b}{a} = \frac{1}{2}$.
Squaring both sides,we get $\frac{b^2}{a^2} = \frac{1}{4}$.
The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting the value of $\frac{b^2}{a^2}$:
$e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(A) Given,length of major axis $2a = 6 \Rightarrow a = 3$ and length of minor axis $2b = 2 \Rightarrow b = 1$.
The major axis is along the line $x-y+1=0$. The minor axis is perpendicular to the major axis and passes through the center $(5,6)$.
The slope of the major axis is $m_1 = 1$. Therefore,the slope of the minor axis is $m_2 = -1$.
The equation of the minor axis is $y - 6 = -1(x - 5) \Rightarrow x + y - 11 = 0$.
The equation of the ellipse with center $(h,k) = (5,6)$ is given by $\frac{(\text{distance from major axis})^2}{b^2} + \frac{(\text{distance from minor axis})^2}{a^2} = 1$.
Here,the distance from the major axis $x-y+1=0$ is $\frac{|x-y+1|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y+1|}{\sqrt{2}}$.
The distance from the minor axis $x+y-11=0$ is $\frac{|x+y-11|}{\sqrt{1^2+1^2}} = \frac{|x+y-11|}{\sqrt{2}}$.
Substituting these into the standard form $\frac{d_1^2}{b^2} + \frac{d_2^2}{a^2} = 1$:
$\frac{(\frac{x-y+1}{\sqrt{2}})^2}{1^2} + \frac{(\frac{x+y-11}{\sqrt{2}})^2}{3^2} = 1$
$\frac{(x-y+1)^2}{2} + \frac{(x+y-11)^2}{18} = 1$
Multiplying by $18$:
$9(x-y+1)^2 + (x+y-11)^2 = 18$.

(A) The given equation of the ellipse is $9x^2 + 25y^2 = 225$.
Dividing both sides by $225$,we get $\frac{9x^2}{225} + \frac{25y^2}{225} = 1$,which simplifies to $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.
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 coordinates of the foci are $(\pm ae, 0)$.
Substituting the values,we get $(\pm 5 \times \frac{4}{5}, 0) = (\pm 4, 0)$.

(D) For the inscribed ellipse $E_1: \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$,the rectangle $R$ has vertices at $(\pm 3, \pm 2)$.
Since $E_2: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ circumscribes the rectangle $R$,it must pass through the vertices of $R$,such as $(3, 2)$.
Substituting $(3, 2)$ into the equation of $E_2$: $\frac{3^2}{a^2} + \frac{2^2}{b^2} = 1 \Rightarrow \frac{9}{a^2} + \frac{4}{b^2} = 1$.
Given that $E_2$ passes through $(0, 4)$,we have $\frac{0^2}{a^2} + \frac{4^2}{b^2} = 1$,which gives $b^2 = 16$,so $b = 4$.
Substituting $b^2 = 16$ into the equation $\frac{9}{a^2} + \frac{4}{16} = 1$:
$\frac{9}{a^2} + \frac{1}{4} = 1$ $\Rightarrow \frac{9}{a^2} = \frac{3}{4}$ $\Rightarrow a^2 = 12$ $\Rightarrow a = 2\sqrt{3}$.
Thus,$a = 2\sqrt{3}$ and $b = 4$.

(D) Consider the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
The distance between its foci is $2c = 6$,which implies $c = 3$.
The length of the minor axis is $2b = 8$,which implies $b = 4$.
For an ellipse,the relationship between $a, b,$ and $c$ is $a^2 = b^2 + c^2$.
Substituting the values,we get $a^2 = 4^2 + 3^2 = 16 + 9 = 25$,so $a = 5$.
The eccentricity $e$ is given by $e = \frac{c}{a}$.
Therefore,$e = \frac{3}{5}$.