Ellipse Questions in English

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $S_1(ae, 0)$ and $S_2(-ae, 0)$. The ends of the major axis are $A(a, 0)$ and $A'(-a, 0)$. The ends of the minor axis are $B(0, b)$ and $B'(0, -b)$.
Given the distance from focus $S_1$ to the corresponding end of the major axis $A$ is $a - ae = 4 - \sqrt{7}$.
Given the distance from focus $S_1$ to the end of the minor axis $B$ is $\sqrt{(ae)^2 + b^2} = 4$. Since $b^2 = a^2(1 - e^2)$,we have $\sqrt{(ae)^2 + a^2 - a^2e^2} = 4$,which implies $a = 4$.
Substituting $a = 4$ into $a(1 - e) = 4 - \sqrt{7}$,we get $4(1 - e) = 4 - \sqrt{7}$,so $1 - e = 1 - \frac{\sqrt{7}}{4}$,which gives $e = \frac{\sqrt{7}}{4}$.
Then $b^2 = a^2(1 - e^2) = 16(1 - \frac{7}{16}) = 16(\frac{9}{16}) = 9$,so $b = 3$.
The distance $S_1S_2 = 2ae = 2(4)(\frac{\sqrt{7}}{4}) = 2\sqrt{7}$.
In $\triangle BS_1S_2$,$BS_1 = BS_2 = 4$ and $S_1S_2 = 2\sqrt{7}$.
Using the law of cosines,$\cos \theta = \frac{BS_1^2 + BS_2^2 - S_1S_2^2}{2(BS_1)(BS_2)} = \frac{4^2 + 4^2 - (2\sqrt{7})^2}{2(4)(4)} = \frac{16 + 16 - 28}{32} = \frac{4}{32} = \frac{1}{8}$.

(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since it passes through $(2,2)$,we have $\frac{4}{a^2} + \frac{4}{b^2} = 1$,which simplifies to $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$ ... $(i)$.
Since it passes through $(3,1)$,we have $\frac{9}{a^2} + \frac{1}{b^2} = 1$ ... $(ii)$.
Subtracting $(i)$ from $(ii)$,we get $\frac{8}{a^2} = 1 - \frac{1}{4} = \frac{3}{4}$.
Thus,$a^2 = \frac{32}{3}$,which implies $3a^2 = 32$.
Substituting $a^2$ into $(ii)$,we get $\frac{9}{32/3} + \frac{1}{b^2} = 1$,so $\frac{27}{32} + \frac{1}{b^2} = 1$.
This gives $\frac{1}{b^2} = 1 - \frac{27}{32} = \frac{5}{32}$,so $b^2 = \frac{32}{5}$,which implies $5b^2 = 32$.
Therefore,$3a^2 + 5b^2 = 32 + 32 = 64$.

(B) The foci are given as $F_1(-3, 0)$ and $F_2(9, 0)$.
Since the $y$-coordinates are the same,the major axis is horizontal.
The center $(h, k)$ is the midpoint of the foci: $h = \frac{-3+9}{2} = 3$ and $k = 0$.
The distance between the foci is $2ae = 9 - (-3) = 12$,so $ae = 6$.
Given eccentricity $e = \frac{1}{3}$,we have $a(\frac{1}{3}) = 6$,which gives $a = 18$.
Using the relation $b^2 = a^2(1-e^2)$,we get $b^2 = 18^2(1 - (\frac{1}{3})^2) = 324(1 - \frac{1}{9}) = 324(\frac{8}{9}) = 288$.
Thus,$b = \sqrt{288} = 12\sqrt{2}$.
The parametric equations for a shifted ellipse are $x = h + a \cos \theta$ and $y = k + b \sin \theta$.
Substituting the values,we get $x = 3 + 18 \cos \theta$ and $y = 12\sqrt{2} \sin \theta$.

(A) Given,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
Eccentricity $e = \frac{\sqrt{3}}{2}$ and length of latus rectum $L = \frac{2b^2}{a} = 1$.
We know that $b^2 = a^2(1 - e^2)$.
Substituting $e^2 = \frac{3}{4}$,we get $b^2 = a^2(1 - \frac{3}{4}) = \frac{a^2}{4}$,which implies $\frac{b^2}{a^2} = \frac{1}{4}$ or $b^2 = \frac{a^2}{4}$.
Substitute $b^2 = \frac{a^2}{4}$ into the latus rectum equation: $\frac{2(a^2/4)}{a} = 1$.
This simplifies to $\frac{a}{2} = 1$,so $a = 2$.
Then $b^2 = \frac{2^2}{4} = 1$,so $b = 1$.
The length of the major axis is $2a = 2(2) = 4$.
The length of the minor axis is $2b = 2(1) = 2$.
The sum of the lengths of the major and minor axes is $4 + 2 = 6$.

(D) Given the ellipses $S^{\prime} \equiv \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$.
Points $P(a \cos \theta, b \sin \theta)$ and $Q\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)$ lie on $S^{\prime}$.
Simplifying $Q$,we get $Q \equiv (-a \sin \theta, b \cos \theta)$.
Since $P$ lies on $S^{\prime}$,we have:
$\frac{a^2 \cos^2 \theta}{\alpha^2} + \frac{b^2 \sin^2 \theta}{\beta^2} = 1$ $(i)$
Since $Q$ lies on $S^{\prime}$,we have:
$\frac{a^2 \sin^2 \theta}{\alpha^2} + \frac{b^2 \cos^2 \theta}{\beta^2} = 1$ $(ii)$
Adding $(i)$ and $(ii)$:
$\frac{a^2}{\alpha^2}(\cos^2 \theta + \sin^2 \theta) + \frac{b^2}{\beta^2}(\sin^2 \theta + \cos^2 \theta) = 1 + 1$
$\frac{a^2}{\alpha^2} + \frac{b^2}{\beta^2} = 2$
$\frac{a^2 \beta^2 + b^2 \alpha^2}{\alpha^2 \beta^2} = 2$
Therefore,$\frac{1}{2}(a^2 \beta^2 + b^2 \alpha^2) = \alpha^2 \beta^2$.

(D) The given ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$ with center $(0,0)$.
To find the image of the ellipse in the line $x+y-10=0$,we find the image of the center $(0,0)$ in the line $x+y-10=0$.
Let the image be $(h,k)$. The line joining $(0,0)$ and $(h,k)$ is perpendicular to $x+y-10=0$,so its slope is $1$. Thus,$\frac{k-0}{h-0} = 1 \implies k=h$.
The midpoint $(\frac{h}{2}, \frac{k}{2})$ lies on $x+y-10=0$,so $\frac{h}{2}+\frac{k}{2}=10 \implies h+k=20$.
Substituting $k=h$,we get $2h=20 \implies h=10, k=10$.
The image of the center is $(10,10)$.
The shape and size of the ellipse remain unchanged,so the new equation is $\frac{(x-10)^2}{25}+\frac{(y-10)^2}{16}=1$.
Comparing this with the given equation $\frac{(x-10)^2}{16}+\frac{(y-10)^2}{25}=1$,we see the denominators are swapped.
Thus,Assertion $(A)$ is false.
Reason $(R)$ is a standard definition of the image of a curve,which is true.
Therefore,$(A)$ is false but $(R)$ is true.

(C) The given equation is $ax^2 + by^2 = 15$,which can be written as $\frac{x^2}{15/a} + \frac{y^2}{15/b} = 1$. Let $a'^2 = \frac{15}{a}$ and $b'^2 = \frac{15}{b}$.
For an ellipse,the distance between foci is $2a'e = 2 \Rightarrow a'e = 1$.
The distance between directrices is $\frac{2a'}{e} = 5 \Rightarrow \frac{a'}{e} = \frac{5}{2}$.
Multiplying these two equations: $(a'e) \times (a'/e) = 1 \times \frac{5}{2} \Rightarrow a'^2 = \frac{5}{2}$.
Since $a'^2 = \frac{15}{a}$,we have $\frac{15}{a} = \frac{5}{2} \Rightarrow a = 6$.
Using $a'e = 1$,we get $e^2 = \frac{1}{a'^2} = \frac{2}{5}$.
For an ellipse,$b'^2 = a'^2(1 - e^2) = \frac{5}{2}(1 - \frac{2}{5}) = \frac{5}{2} \times \frac{3}{5} = \frac{3}{2}$.
Since $b'^2 = \frac{15}{b}$,we have $\frac{15}{b} = \frac{3}{2} \Rightarrow b = 10$.
Therefore,$a + b = 6 + 10 = 16$.

(A) For Statement $I$: The equation $4x^2+y^2-8x-4y+4=0$ can be written as $4(x-1)^2-4 + (y-2)^2-4+4=0$,which simplifies to $4(x-1)^2+(y-2)^2=4$,or $\frac{(x-1)^2}{1} + \frac{(y-2)^2}{4} = 1$.
Here $a^2=1, b^2=4$,so $b > a$. The eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.
The directrices are $y-k = \pm \frac{b}{e}$,so $y-2 = \pm \frac{2}{\sqrt{3}/2} = \pm \frac{4}{\sqrt{3}}$.
Thus $y = 2 \pm \frac{4}{\sqrt{3}}$,which gives $y = \frac{6 \pm 4\sqrt{3}}{3}$,or $3y = 6 \pm 4\sqrt{3}$. Statement $I$ is true.
For Statement $II$: The equation $x^2+4y^2-4x-8y+4=0$ can be written as $(x-2)^2-4 + 4(y-1)^2-4+4=0$,which simplifies to $(x-2)^2+4(y-1)^2=4$,or $\frac{(x-2)^2}{4} + \frac{(y-1)^2}{1} = 1$.
Here $a^2=4, b^2=1$,so $a > b$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.
The latus rectum is $x-h = \pm ae$,so $x-2 = \pm 2(\frac{\sqrt{3}}{2}) = \pm \sqrt{3}$.
This gives $x = 2 \pm \sqrt{3}$. Statement $II$ claims the latus rectum is $y=2+\sqrt{3}$,which is incorrect. Statement $II$ is false.

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The length of the latus rectum is $\frac{2b^2}{a} = 6$,which implies $b^2 = 3a$.
The distance between a focus $(ae, 0)$ and its nearest vertex $(a, 0)$ is $a - ae = a(1 - e) = \frac{5}{3}$.
We know $b^2 = a^2(1 - e^2)$.
Substituting $b^2 = 3a$,we get $3a = a^2(1 - e^2)$,so $3 = a(1 - e^2) = a(1 - e)(1 + e)$.
Since $a(1 - e) = \frac{5}{3}$,we have $3 = \frac{5}{3}(1 + e)$.
This gives $1 + e = \frac{9}{5}$,so $e = \frac{9}{5} - 1 = \frac{4}{5}$.
Now,we check which equation is satisfied by $e = \frac{4}{5}$.
For option $A$: $25(\frac{4}{5})^2 - 40(\frac{4}{5}) + 16 = 25(\frac{16}{25}) - 32 + 16 = 16 - 32 + 16 = 0$.
Thus,$e$ satisfies $25e^2 - 40e + 16 = 0$.

(B) The given equation of the ellipse is $25x^2+16y^2-150x-64y-111=0$.
Rearranging the terms:
$25(x^2-6x) + 16(y^2-4y) = 111$
$25(x-3)^2 - 225 + 16(y-2)^2 - 64 = 111$
$25(x-3)^2 + 16(y-2)^2 = 400$
Dividing by $400$:
$\frac{(x-3)^2}{16} + \frac{(y-2)^2}{25} = 1$
Here,$a^2 = 16$ and $b^2 = 25$,so $a = 4$ and $b = 5$.
Since $b > a$,the major axis is vertical.
The length of the latus rectum $m = \frac{2a^2}{b} = \frac{2 \times 16}{5} = \frac{32}{5}$.
The length of the major axis $n = 2b = 2 \times 5 = 10$.
Thus,the ordered pair $(m, n) = \left(\frac{32}{5}, 10\right)$.

(D) By the definition of an ellipse,the distance from any point $P(x, y)$ on the ellipse to the focus $S(0, 0)$ is $e$ times the distance to the directrix $x = 4$.
$SP^2 = e^2 \times (\text{distance to directrix})^2$
$x^2 + y^2 = (\frac{1}{2})^2 (x - 4)^2$
$x^2 + y^2 = \frac{1}{4} (x^2 - 8x + 16)$
$4x^2 + 4y^2 = x^2 - 8x + 16$
$3x^2 + 8x + 4y^2 = 16$
Multiply by $3$ to complete the square for $x$:
$9x^2 + 24x + 12y^2 = 48$
$(3x + 4)^2 - 16 + 12y^2 = 48$
$(3x + 4)^2 + 12y^2 = 64$

(A) The given equation is $4(x-2y+1)^2 + 9(2x+y+2)^2 = 25$.
Divide by $25$: $\frac{(x-2y+1)^2}{25/4} + \frac{(2x+y+2)^2}{25/9} = 1$.
Let $X = \frac{x-2y+1}{\sqrt{1^2+(-2)^2}} = \frac{x-2y+1}{\sqrt{5}}$ and $Y = \frac{2x+y+2}{\sqrt{2^2+1^2}} = \frac{2x+y+2}{\sqrt{5}}$.
The equation becomes $\frac{5X^2}{25/4} + \frac{5Y^2}{25/9} = 1$,which simplifies to $\frac{X^2}{5/4} + \frac{Y^2}{5/9} = 1$.
Here $a^2 = 5/4$ and $b^2 = 5/9$,so $a = \sqrt{5}/2$ and $b = \sqrt{5}/3$.
The major axis corresponds to the term with the larger denominator,which is $X=0$,i.e.,$x-2y+1=0$.
The eccentricity $e = \sqrt{1 - b^2/a^2} = \sqrt{1 - (5/9)/(5/4)} = \sqrt{1 - 4/9} = \sqrt{5}/3$.
The length of the major axis is $2a = 2(\sqrt{5}/2) = \sqrt{5}$.
The centre is the intersection of $x-2y+1=0$ and $2x+y+2=0$,which is $(-4/5, -2/5)$.

(B) For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the coordinates of the vertices of an inscribed rectangle are $(\pm a \cos \theta, \pm b \sin \theta)$.
Area of the rectangle $A = (2a \cos \theta)(2b \sin \theta) = 2ab \sin 2\theta$.
For the area to be maximum,$\sin 2\theta = 1$,which implies $\theta = \frac{\pi}{4}$.
Given length $L = 2a \cos \theta = 2a \cos(\frac{\pi}{4}) = 2a(\frac{1}{\sqrt{2}}) = a\sqrt{2} = 8\sqrt{2}$,so $a = 8$.
Given breadth $B = 2b \sin \theta = 2b \sin(\frac{\pi}{4}) = 2b(\frac{1}{\sqrt{2}}) = b\sqrt{2} = 4\sqrt{2}$,so $b = 4$.
Thus,$\frac{b}{a} = \frac{4}{8} = \frac{1}{2}$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$.
The length of the latus rectum is $\frac{2a^2}{b}$ and the length of the minor axis is $2a$.
Given that the length of the latus rectum is $\frac{2}{3}$ times the length of the minor axis:
$\frac{2a^2}{b} = \frac{2}{3}(2a)$
$\frac{a^2}{b} = \frac{2a}{3}$
$\frac{a}{b} = \frac{2}{3}$
$\frac{a^2}{b^2} = \frac{4}{9}$
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}}$.
$e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.

(A) Given,equation of ellipse $16 x^2+25 y^2=400$ $\Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1$.
Comparing to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we get $a=5$ and $b=4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Let $P(5 \cos \theta, 4 \sin \theta)$ be any point on the ellipse.
The focal distances are $P F_1 = a - ex = 5 - (\frac{3}{5})(5 \cos \theta) = 5 - 3 \cos \theta$ and $P F_2 = a + ex = 5 + 3 \cos \theta$.
Now,the product $P F_1 \cdot P F_2 = (5 - 3 \cos \theta)(5 + 3 \cos \theta) = 25 - 9 \cos^2 \theta$.
Since $0 \leq \cos^2 \theta \leq 1$,we have:
$25 - 9(1) \leq 25 - 9 \cos^2 \theta \leq 25 - 9(0)$
$16 \leq P F_1 \cdot P F_2 \leq 25$.
Thus,the value lies in the interval $[16, 25]$.

(D) The given equation of the ellipse is $9x^2+4y^2=36$.
Dividing both sides by $36$,we get $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (since $9 > 4$),we have $a^2 = 9$ and $b^2 = 4$,which gives $a = 3$ and $b = 2$.
For any point $P$ on an ellipse,the sum of its focal distances $PF_1 + PF_2$ is equal to the length of the major axis,which is $2a$.
Here,the major axis is along the $y$-axis because $a > b$.
Thus,the sum of the focal distances is $2a = 2 \times 3 = 6$.

(C) The length of the latus rectum is $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$.
The distance between the foci is $2ae = 4\sqrt{2}$,so $ae = 2\sqrt{2}$.
Squaring both sides,$a^2e^2 = 8$.
Using the relation $a^2e^2 = a^2 - b^2$,we have $a^2 - b^2 = 8$.
Substituting $b^2 = 2a$,we get $a^2 - 2a - 8 = 0$.
Solving the quadratic equation $(a - 4)(a + 2) = 0$,we get $a = 4$ (since $a > 0$).
Then $b^2 = 2(4) = 8$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{16} + \frac{y^2}{8} = 1$.
Multiplying by $16$,we get $x^2 + 2y^2 = 16$.

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since the ellipse passes through $(3 \sqrt{2}, \sqrt{10})$,we have $\frac{(3 \sqrt{2})^2}{a^2} + \frac{(\sqrt{10})^2}{b^2} = 1$,which simplifies to $\frac{18}{a^2} + \frac{10}{b^2} = 1$ $(i)$.
The foci are at $(\pm 4, 0)$,so $ae = 4$,which means $a^2 e^2 = 16$.
Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2 e^2$,we get $b^2 = a^2 - 16$.
Substituting $b^2$ into equation $(i)$: $\frac{18}{a^2} + \frac{10}{a^2 - 16} = 1$.
Multiplying by $a^2(a^2 - 16)$,we get $18(a^2 - 16) + 10a^2 = a^2(a^2 - 16)$.
$18a^2 - 288 + 10a^2 = a^4 - 16a^2$.
$a^4 - 44a^2 + 288 = 0$.
Factoring the quadratic in $a^2$: $(a^2 - 36)(a^2 - 8) = 0$.
Since $b^2 = a^2 - 16 > 0$,we must have $a^2 > 16$,so $a^2 = 36$.
Finally,$e^2 = \frac{16}{a^2} = \frac{16}{36} = \frac{4}{9}$.
Therefore,$e = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

(A) Since the foci of the given ellipse lie on the $Y$-axis,it is a vertical ellipse.
Let the required equation be $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,where $a^2 > b^2$.
The foci are $(0, \pm c) = (0, \pm 1)$,which implies $c = 1$.
The length of the major axis is $2a = \sqrt{5}$,so $a = \frac{\sqrt{5}}{2}$ and $a^2 = \frac{5}{4}$.
Using the relation $c^2 = a^2 - b^2$,we have $1 = \frac{5}{4} - b^2$.
Thus,$b^2 = \frac{5}{4} - 1 = \frac{1}{4}$.
Substituting $a^2$ and $b^2$ into the equation,we get $\frac{x^2}{1/4} + \frac{y^2}{5/4} = 1$.
This simplifies to $4x^2 + \frac{4y^2}{5} = 1$,or $20x^2 + 4y^2 = 5$.

(B) By the definition of an ellipse,the distance from a point $(x, y)$ to the focus is $e$ times the distance from the point to the directrix: $\sqrt{(x-1)^2+(y-2)^2} = \frac{1}{2} \frac{|3x+4y-5|}{\sqrt{3^2+4^2}}$.
Squaring both sides,we get: $(x-1)^2+(y-2)^2 = \frac{1}{4} \frac{(3x+4y-5)^2}{25}$.
$100(x^2-2x+1+y^2-4y+4) = (3x+4y-5)^2$.
$100(x^2+y^2-2x-4y+5) = 9x^2+16y^2+25+24xy-30x-40y$.
$100x^2+100y^2-200x-400y+500 = 9x^2+16y^2+24xy-30x-40y+25$.
Rearranging terms: $91x^2+84y^2-24xy-170x-360y+475=0$.

(D) Let the equation of the required ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad (a>b)$.
According to the given information,the distance from each focus to the centre is $ae$,so the sum of distances is $ae + ae = 2ae = 8\sqrt{6} \Rightarrow ae = 4\sqrt{6}$.
Squaring both sides,$a^2e^2 = 16 \times 6 = 96$.
Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $a^2(1 - \frac{b^2}{a^2}) = 96$,which simplifies to $a^2 - b^2 = 96 \quad (i)$.
The smallest rectangle in which the ellipse is inscribed has dimensions $2a \times 2b$. Thus,$4ab = 80$ $\Rightarrow ab = 20$ $\Rightarrow a^2b^2 = 400 \quad (ii)$.
Using the identity $(a^2+b^2)^2 = (a^2-b^2)^2 + 4a^2b^2$,we get $(a^2+b^2)^2 = (96)^2 + 4(400) = 9216 + 1600 = 10816$.
Taking the square root,$a^2+b^2 = \sqrt{10816} = 104 \quad (iii)$.
Adding $(i)$ and $(iii)$,$2a^2 = 200 \Rightarrow a^2 = 100$.
Subtracting $(i)$ from $(iii)$,$2b^2 = 8 \Rightarrow b^2 = 4$.
Therefore,the equation of the ellipse is $\frac{x^2}{100}+\frac{y^2}{4}=1$.

(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
The coordinates of the foci are $A(-ae, 0)$ and $B(ae, 0)$,and the endpoint of the semi-minor axis is $T(0, b)$.
The slope of $AT$ is $m_1 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.
The slope of $BT$ is $m_2 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.
Since $\angle ATB = 90^{\circ}$,the product of the slopes is $-1$:
$\left(\frac{b}{ae}\right) \left(-\frac{b}{ae}\right) = -1$.
$\Rightarrow \frac{b^2}{a^2 e^2} = 1$ $\Rightarrow b^2 = a^2 e^2$.
We know that for an ellipse,$b^2 = a^2(1 - e^2)$.
Substituting $b^2 = a^2 e^2$ into the equation:
$a^2 e^2 = a^2(1 - e^2)$.
$e^2 = 1 - e^2
$ $\Rightarrow 2e^2 = 1
$ $\Rightarrow e^2 = \frac{1}{2}
$ $\Rightarrow e = \frac{1}{\sqrt{2}}$.

(B) Let the rectangle be $ABCD$ with dimensions $2a$ and $2b$. The center of the ellipse is the origin $O(0,0)$. The vertices of the rectangle are $(a, b), (-a, b), (-a, -b),$ and $(a, -b)$.
Let the angle between the diagonals be $2\theta$. From the geometry of the rectangle,$\tan \theta = \frac{b}{a}$.
The angle between the diagonals is given as $\tan(2\theta) = 4\sqrt{3}$.
Using the formula $\tan(2\theta) = \frac{2\tan \theta}{1-\tan^2 \theta}$,we have:
$4\sqrt{3} = \frac{2(b/a)}{1-(b/a)^2}$
$2\sqrt{3} = \frac{b/a}{1-(b/a)^2}$
Let $k = b/a$. Then $2\sqrt{3}(1-k^2) = k$,which implies $2\sqrt{3}k^2 + k - 2\sqrt{3} = 0$.
Solving the quadratic equation for $k$:
$k = \frac{-1 \pm \sqrt{1 - 4(2\sqrt{3})(-2\sqrt{3})}}{2(2\sqrt{3})} = \frac{-1 \pm \sqrt{1 + 48}}{4\sqrt{3}} = \frac{-1 \pm 7}{4\sqrt{3}}$.
Since $k = b/a > 0$,we take $k = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
The eccentricity $e$ of the ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - k^2}$.
$e = \sqrt{1 - (\frac{\sqrt{3}}{2})^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the correct option is $(b)$.

(A) Given,eccentricity $e = \frac{1}{2}$,centre is $(0, 0)$,and the equation of the directrix is $x = \frac{a}{e} = 4$.
Since $e = \frac{1}{2}$,we have $\frac{a}{1/2} = 4$,which implies $a = 2$.
Now,$b^2 = a^2(1 - e^2) = 2^2(1 - (\frac{1}{2})^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
The equation of the ellipse with centre at the origin and major axis along the $x$-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Multiplying by $12$,we get $3x^2 + 4y^2 = 12$.

(C) The ellipse $S$ is $\frac{x^2}{4}+\frac{y^2}{9}=1$. Its vertices are $(0, \pm 3)$. Let $A = (0, 3)$.
For the ellipse $S^{\prime} \equiv \frac{x^2}{9}+\frac{y^2}{4}=1$,the major axis is along the $x$-axis,$a^2=9, b^2=4$. The eccentricity $e = \sqrt{1-\frac{4}{9}} = \frac{\sqrt{5}}{3}$. The foci are $(\pm ae, 0) = (\pm \sqrt{5}, 0)$.
Let $F = (\sqrt{5}, 0)$. Point $P$ divides $\overline{OF}$ in the ratio $2:1$. Thus,$P = \left(\frac{2(\sqrt{5})+1(0)}{2+1}, 0\right) = \left(\frac{2\sqrt{5}}{3}, 0\right)$.
The line passing through $A(0, 3)$ and $P\left(\frac{2\sqrt{5}}{3}, 0\right)$ has the equation $y - 3 = \frac{0-3}{\frac{2\sqrt{5}}{3}-0}(x-0)$,which simplifies to $y = -\frac{9}{2\sqrt{5}}x + 3$.
Substituting $y = 3 - \frac{9}{2\sqrt{5}}x$ into $\frac{x^2}{4} + \frac{y^2}{9} = 1$,we get $\frac{x^2}{4} + \frac{(3 - \frac{9}{2\sqrt{5}}x)^2}{9} = 1$.
$\frac{x^2}{4} + \frac{9(1 - \frac{3}{2\sqrt{5}}x)^2}{9} = 1$ $\Rightarrow \frac{x^2}{4} + 1 - \frac{3}{\sqrt{5}}x + \frac{9}{20}x^2 = 1$.
$\frac{5x^2 + 9x^2}{20} = \frac{3}{\sqrt{5}}x$ $\Rightarrow \frac{14x^2}{20} = \frac{3}{\sqrt{5}}x$ $\Rightarrow x = 0$ or $x = \frac{3}{\sqrt{5}} \times \frac{20}{14} = \frac{30}{7\sqrt{5}}$.
For $x = \frac{30}{7\sqrt{5}}$,$y = 3 - \frac{9}{2\sqrt{5}}(\frac{30}{7\sqrt{5}}) = 3 - \frac{270}{70} = 3 - \frac{27}{7} = -\frac{6}{7}$.
The chord length is $\sqrt{(\frac{30}{7\sqrt{5}} - 0)^2 + (-\frac{6}{7} - 3)^2} = \sqrt{\frac{900}{49 \times 5} + (-\frac{27}{7})^2} = \sqrt{\frac{180}{49} + \frac{729}{49}} = \sqrt{\frac{909}{49}} = \frac{3\sqrt{101}}{7}$.
Thus,$k=7$.

(A) Given,the major axis of the ellipse lies on the $X$-axis,so its equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a^2 > b^2$.
Since it passes through $(-3, 1)$,we have $\frac{9}{a^2} + \frac{1}{b^2} = 1$ $(i)$.
The eccentricity is $e = \sqrt{\frac{2}{5}}$,so $e^2 = \frac{2}{5}$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = a^2(1 - \frac{2}{5}) = a^2(\frac{3}{5})$,which implies $b^2 = \frac{3a^2}{5}$ (ii).
Substituting (ii) into $(i)$: $\frac{9}{a^2} + \frac{5}{3a^2} = 1$.
Multiplying by $3a^2$: $27 + 5 = 3a^2$,so $3a^2 = 32$,which gives $a^2 = \frac{32}{3}$.
Then $b^2 = \frac{3}{5} \times \frac{32}{3} = \frac{32}{5}$.
The equation of the ellipse is $\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1$,which simplifies to $3x^2 + 5y^2 = 32$.

(D) Given equation: $4x^2 + y^2 - 8x + 2y + 1 = 0$
Completing the square: $4(x^2 - 2x) + (y^2 + 2y) + 1 = 0$
$4(x-1)^2 - 4 + (y+1)^2 - 1 + 1 = 0$
$4(x-1)^2 + (y+1)^2 = 4$
Dividing by $4$: $\frac{(x-1)^2}{1} + \frac{(y+1)^2}{4} = 1$
Here,$a^2 = 1$ and $b^2 = 4$. Since $b^2 > a^2$,the major axis is vertical.
Eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.
Center $(h, k) = (1, -1)$.
Foci are given by $(h, k \pm be) = (1, -1 \pm 2 \cdot \frac{\sqrt{3}}{2}) = (1, -1 \pm \sqrt{3})$.
Directrices are given by $y = k \pm \frac{b}{e} = -1 \pm \frac{2}{\sqrt{3}/2} = -1 \pm \frac{4}{\sqrt{3}}$.
For the focus $(1, -1-\sqrt{3})$,the corresponding directrix is $y = -1 - \frac{4}{\sqrt{3}}$ $\Rightarrow \sqrt{3}y = -\sqrt{3} - 4$ $\Rightarrow \sqrt{3}y + \sqrt{3} + 4 = 0$.

(D) Given equation: $25x^2 + 100x + 4y^2 - 4y + 100 = 0$
Complete the squares: $25(x^2 + 4x + 4) + 4(y^2 - y + 1/4) = -100 + 100 + 1$
$25(x + 2)^2 + 4(y - 1/2)^2 = 1$
Divide by $1$: $\frac{(x + 2)^2}{(1/5)^2} + \frac{(y - 1/2)^2}{(1/2)^2} = 1$
Here $a^2 = 1/25$ and $b^2 = 1/4$. Since $b > a$,the major axis is vertical $(x = -2)$.
Eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1/25}{1/4}} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$.
The foci are $(h, k \pm be)$,where $(h, k) = (-2, 1/2)$.
Foci $= (-2, 1/2 \pm (1/2)(\sqrt{21}/5)) = (-2, 1/2 \pm \sqrt{21}/10) = (-2, \frac{5 \pm \sqrt{21}}{10})$.

(A) The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$,and the end of the minor axis is $B(0, b)$.
Since $\triangle SBS^{\prime}$ is an isosceles right-angled triangle with the right angle at $B$,we have $SB = S^{\prime}B$ and $SB^2 + S^{\prime}B^2 = SS^{\prime 2}$.
Given $S(ae, 0)$,$S^{\prime}(-ae, 0)$,and $B(0, b)$,the distance $SB = \sqrt{(ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.
Similarly,$S^{\prime}B = \sqrt{(-ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.
The distance $SS^{\prime} = 2ae$.
Applying the Pythagorean theorem: $SB^2 + S^{\prime}B^2 = SS^{\prime 2}$.
$(a^2e^2 + b^2) + (a^2e^2 + b^2) = (2ae)^2$.
$2(a^2e^2 + b^2) = 4a^2e^2$.
$a^2e^2 + b^2 = 2a^2e^2$.
$b^2 = a^2e^2$.
Since $b^2 = a^2(1-e^2)$,we have $a^2(1-e^2) = a^2e^2$.
$1-e^2 = e^2$.
$2e^2 = 1$.
$e^2 = \frac{1}{2}$.
$e = \frac{1}{\sqrt{2}}$.

(A) Given the equation of the ellipse: $\frac{x^2}{25}+\frac{y^2}{16}=1$.
Here,$a^2=25 \implies a=5$ and $b^2=16 \implies b=4$.
The eccentricity $e$ is given by $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
By the definition of an ellipse,the sum of focal distances of any point $P$ on the ellipse is $PS+PS^{\prime} = 2a$.
Thus,$PS+PS^{\prime} = 2 \times 5 = 10$.
Given $SP=8$,we have $8+PS^{\prime} = 10 \implies PS^{\prime} = 2$.
The distance between the foci $SS^{\prime}$ is $2ae = 2 \times 5 \times \frac{3}{5} = 6$.
Checking the options:
$4+S^{\prime}P = 4+2 = 6$.
Therefore,$SS^{\prime} = 4+S^{\prime}P$.

(A) The given equation of the ellipse is $\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1$.
Comparing this with the standard form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$,we have $h=3, k=2, a^2=25, b^2=16$.
Thus,$a=5$ and $b=4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
$(i)$ The equation of the major axis is $y = k$,so $y = 2$. This matches $(q)$.
$(ii)$ The equations of the directrices are $x = h \pm \frac{a}{e}$.
$x = 3 \pm \frac{5}{3/5} = 3 \pm \frac{25}{3}$.
$x = 3 + \frac{25}{3} = \frac{34}{3} \Rightarrow 3x = 34$.
$x = 3 - \frac{25}{3} = -\frac{16}{3} \Rightarrow 3x = -16$.
Thus,$3x = 34$ matches $(p)$.
$(iii)$ The equations of the latera recta are $x = h \pm ae$.
$x = 3 \pm (5 \times \frac{3}{5}) = 3 \pm 3$.
$x = 3 + 3 = 6$ or $x = 3 - 3 = 0$.
Thus,$x = 6$ matches $(s)$.
Therefore,the correct matches are $(i) \rightarrow (q)$,$(ii) \rightarrow (p)$,$(iii) \rightarrow (s)$.

(P-6, Q-1, R-3) The given equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here,$a^2=25 \Rightarrow a=5$ and $b^2=16 \Rightarrow b=4$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
$(P)$ The directrix corresponding to the focus $(-ae, 0) = (-3, 0)$ is $x = -\frac{a}{e} = -\frac{5}{3/5} = -\frac{25}{3}$,which gives $3x + 25 = 0$. This matches $(6)$.
$(Q)$ The tangent at the vertex $(0, 4)$ is $y = 4$. This matches $(1)$.
$(R)$ The latus rectum through the focus $(ae, 0) = (3, 0)$ is $x = ae = 5 \times \frac{3}{5} = 3$. This matches $(3)$.

(D) The given equation of the ellipse is $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$.
To find the centre of the ellipse,we set the linear expressions inside the squares to zero:
$x+y-3=0 \quad (i)$
$x-y+1=0 \quad (ii)$
Adding equations $(i)$ and $(ii)$:
$(x+y-3) + (x-y+1) = 0 + 0$
$2x - 2 = 0 \Rightarrow x = 1$
Substituting $x=1$ into equation $(i)$:
$1+y-3=0 \Rightarrow y=2$
Thus,the centre of the ellipse is $(1,2)$.

(B) The $y$-coordinate of the foci is $0$,so the major axis is on the $X$-axis. \\
Given $ae = 4$. \\
Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. \\
Since $b^2 = a^2(1 - e^2) = a^2 - (ae)^2 = a^2 - 16$,we substitute the point $(4\sqrt{2}, 2\sqrt{6})$: \\
$\frac{(4\sqrt{2})^2}{a^2} + \frac{(2\sqrt{6})^2}{a^2 - 16} = 1$ \\
$\frac{32}{a^2} + \frac{24}{a^2 - 16} = 1$ \\
$32(a^2 - 16) + 24a^2 = a^2(a^2 - 16)$ \\
$32a^2 - 512 + 24a^2 = a^4 - 16a^2$ \\
$a^4 - 72a^2 + 512 = 0$ \\
$(a^2 - 64)(a^2 - 8) = 0$ \\
Since $a > ae$,$a^2$ must be greater than $16$,so $a^2 = 64$ and $a = 8$. \\
Using $ae = 4$,we get $8e = 4$,which implies $e = \frac{1}{2}$.

(D) The given line is $2x + 5y = 12$ and the ellipse is $4x^2 + 5y^2 = 20$.
Substituting $x = \frac{12 - 5y}{2}$ into the ellipse equation:
$4\left(\frac{12 - 5y}{2}\right)^2 + 5y^2 = 20$
$(12 - 5y)^2 + 5y^2 = 20$
$144 - 120y + 25y^2 + 5y^2 = 20$
$30y^2 - 120y + 124 = 0$
$15y^2 - 60y + 62 = 0$
The discriminant $D = (-60)^2 - 4(15)(62) = 3600 - 3720 = -120$.
Since $D < 0$,the line does not intersect the ellipse at any real points.
Therefore,the condition of intersecting in two distinct points is not satisfied,and the answer is None of these.

(A) Given equation of the ellipse: $x^2+4 y^2+2 x+16 y+13=0$
Completing the square for $x$ and $y$ terms:
$(x^2+2x+1) + 4(y^2+4y+4) + 13 - 1 - 16 = 0$
$(x+1)^2 + 4(y+2)^2 = 4$
Dividing by $4$:
$\frac{(x+1)^2}{4} + \frac{(y+2)^2}{1} = 1$
Comparing with the standard form $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$,we get $a^2 = 4$ and $b^2 = 1$.
Since $a^2 > b^2$,the eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$.
$1 = 4(1 - e^2)$
$1 - e^2 = \frac{1}{4}$
$e^2 = 1 - \frac{1}{4} = \frac{3}{4}$
$e = \frac{\sqrt{3}}{2}$

(C) Given that the distance between the foci is $2ae = 6$,so $ae = 3$.
Given that the length of the minor axis is $2b = 8$,so $b = 4$.
We know the relation $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$.
Substituting $b = 4$ and $ae = 3$,we get $16 = a^2 - (3)^2$.
$16 = a^2 - 9$ $\Rightarrow a^2 = 25$ $\Rightarrow a = 5$.
Now,the eccentricity $e = \frac{ae}{a} = \frac{3}{5}$.

(C) The given equation of the ellipse is $2x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{3} + \frac{y^2}{2} = 1$.
Two points $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate with respect to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if $\frac{x_1 x_2}{a^2} + \frac{y_1 y_2}{b^2} = 1$.
Here,$(x_1, y_1) = (1, 2)$,$(x_2, y_2) = (k, -1)$,$a^2 = 3$,and $b^2 = 2$.
Substituting these values into the condition:
$\frac{(1)(k)}{3} + \frac{(2)(-1)}{2} = 1$
$\frac{k}{3} - 1 = 1$
$\frac{k}{3} = 2$
$k = 6$.

(A) The given equation is $36x^2 + 144y^2 - 36x - 96y - 119 = 0$.
Rearranging the terms,we get $36(x^2 - x) + 144(y^2 - \frac{2}{3}y) = 119$.
Completing the square,we have $36(x^2 - x + \frac{1}{4}) + 144(y^2 - \frac{2}{3}y + \frac{1}{9}) = 119 + 9 + 16$.
This simplifies to $36(x - \frac{1}{2})^2 + 144(y - \frac{1}{3})^2 = 144$.
Dividing by $144$,we get $\frac{(x - 1/2)^2}{4} + \frac{(y - 1/3)^2}{1} = 1$.
This is an ellipse with $a^2 = 4$ and $b^2 = 1$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(B) The given equation of the ellipse is $9x^2 + 5y^2 - 18x - 20y - 16 = 0$.
Rearranging the terms,we get $9(x^2 - 2x) + 5(y^2 - 4y) = 16$.
Completing the square,we have $9(x^2 - 2x + 1) + 5(y^2 - 4y + 4) = 16 + 9 + 20$.
This simplifies to $9(x - 1)^2 + 5(y - 2)^2 = 45$.
Dividing by $45$,we obtain the standard form $\frac{(x - 1)^2}{5} + \frac{(y - 2)^2}{9} = 1$.
Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the ellipse is vertical.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

(C) The given equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we get $a^2 = 16$ and $b^2 = 9$.
The formula for eccentricity $e$ is $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting the values,we get $e = \sqrt{1 - \frac{9}{16}}$.
$e = \sqrt{\frac{16 - 9}{16}} = \sqrt{\frac{7}{16}}$.
Therefore,$e = \frac{\sqrt{7}}{4}$.