Ellipse Questions in English

(D) For the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$,$a^2=9$ and $b^2=5$. The eccentricity $e = \sqrt{1-\frac{5}{9}} = \frac{2}{3}$. The foci are $(\pm ae, 0) = (\pm 2, 0)$.
For the ellipse $\frac{x^2}{5}+\frac{y^2}{9}=1$,$a^2=9$ and $b^2=5$ (with major axis along the $y$-axis). The eccentricity $e = \sqrt{1-\frac{5}{9}} = \frac{2}{3}$. The foci are $(0, \pm ae) = (0, \pm 2)$.
The vertices of the quadrilateral are $(2, 0), (0, 2), (-2, 0),$ and $(0, -2)$.
This quadrilateral is a rhombus with diagonals of length $d_1 = 4$ and $d_2 = 4$.
The area of the quadrilateral is $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4 \times 4 = 8$ square units.

(D) Given the ellipse $E: 4x^2 + 9y^2 - 36 = 0$ and the circle $C: x^2 + y^2 - 9 = 0$.
For point $A(1, 2)$:
$E(1, 2) = 4(1)^2 + 9(2)^2 - 36 = 4 + 36 - 36 = 4 > 0$,so $A$ lies outside $E$.
$C(1, 2) = (1)^2 + (2)^2 - 9 = 1 + 4 - 9 = -4 < 0$,so $A$ lies inside $C$.
For point $B(2, 1)$:
$E(2, 1) = 4(2)^2 + 9(1)^2 - 36 = 16 + 9 - 36 = -11 < 0$,so $B$ lies inside $E$.
$C(2, 1) = (2)^2 + (1)^2 - 9 = 4 + 1 - 9 = -4 < 0$,so $B$ lies inside $C$.
Thus,$A$ lies inside $C$ but outside $E$ is the correct statement.

(B) The equation of the ellipse is $x^2+4y^2=64$,which can be written as $\frac{x^2}{64} + \frac{y^2}{16} = 1$.
This is in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2=64$ $(a=8)$ and $b^2=16$ $(b=4)$.
Let a vertex of the rectangle in the first quadrant be $(x, y) = (a \cos \theta, b \sin \theta) = (8 \cos \theta, 4 \sin \theta)$.
The sides of the rectangle are $2x$ and $2y$,so the area $A = (2x)(2y) = 4xy = 4(8 \cos \theta)(4 \sin \theta) = 128 \sin \theta \cos \theta = 64 \sin(2\theta)$.
The area is maximum when $\sin(2\theta) = 1$,i.e.,$2\theta = 90^\circ$ or $\theta = 45^\circ$.
Substituting $\theta = 45^\circ$,we get $x = 8 \cos 45^\circ = 8 \times \frac{1}{\sqrt{2}} = 4\sqrt{2}$ and $y = 4 \sin 45^\circ = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2}$.
The sides of the rectangle are $2x = 8\sqrt{2}$ and $2y = 4\sqrt{2}$.

(B) The foci are $(-2,0)$ and $(8,0)$.
The distance between the foci is $2ae = 8 - (-2) = 10$.
$\Rightarrow ae = 5$.
Given $e = \frac{1}{\sqrt{2}}$,we have $a \left(\frac{1}{\sqrt{2}}\right) = 5 \Rightarrow a = 5\sqrt{2}$.
Now,$b^2 = a^2(1 - e^2) = (5\sqrt{2})^2 \left(1 - \frac{1}{2}\right) = 50 \left(\frac{1}{2}\right) = 25$.
Thus,$b = 5$.
The centre of the ellipse is the mid-point of the line joining the two foci: $\left(\frac{-2+8}{2}, \frac{0+0}{2}\right) = (3,0)$.
The equation of the ellipse is $\frac{(x-3)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$,which is $\frac{(x-3)^2}{50} + \frac{y^2}{25} = 1$.
The parametric coordinates are $(x, y) = (h + a \cos \theta, k + b \sin \theta)$,where $(h, k)$ is the centre.
Substituting the values,we get $(3 + 5\sqrt{2} \cos \theta, 5 \sin \theta)$.

(B) Let the midpoint of the chord be $M(h, k)$.
The equation of the chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with midpoint $(h, k)$ is given by $T = S_1$,where $T = \frac{xh}{a^2} + \frac{yk}{b^2}$ and $S_1 = \frac{h^2}{a^2} + \frac{k^2}{b^2}$.
Here,$a^2 = 1$ and $b^2 = 4$. So,the equation is $xh + \frac{yk}{4} = h^2 + \frac{k^2}{4}$.
This chord is the same as the line $y = x + 1$,which can be written as $x - y + 1 = 0$,or $x - y = -1$.
Comparing the coefficients of $xh + \frac{yk}{4} = h^2 + \frac{k^2}{4}$ and $x - y = -1$:
$\frac{h}{1} = \frac{k/4}{-1} = \frac{h^2 + k^2/4}{-1}$.
From $\frac{h}{1} = \frac{k}{-4}$,we get $k = -4h$.
Substitute $k = -4h$ into $\frac{h}{1} = \frac{h^2 + k^2/4}{-1}$:
$-h = h^2 + \frac{(-4h)^2}{4} = h^2 + 4h^2 = 5h^2$.
$5h^2 + h = 0 \implies h(5h + 1) = 0$.
Since $h$ cannot be $0$ (as the line $y = x + 1$ does not pass through the origin),$h = -1/5$.
Then $k = -4(-1/5) = 4/5$.
The midpoint is $(-\frac{1}{5}, \frac{4}{5})$.

(C) For a focal chord $PSQ$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passing through the focus $S(ae, 0)$,the relation between the eccentric angles $\theta_1$ and $\theta_2$ is given by $\tan \left(\frac{\theta_1}{2}\right) \tan \left(\frac{\theta_2}{2}\right) = \frac{e-1}{e+1}$.
Given $\tan \left(\frac{\theta_1}{2}\right) \tan \left(\frac{\theta_2}{2}\right) = -(2\sqrt{2}+3)$.
So,$\frac{e-1}{e+1} = -(2\sqrt{2}+3)$.
Let $k = 2\sqrt{2}+3$. Then $\frac{e-1}{e+1} = -k$.
$e-1 = -ke - k$ $\Rightarrow e(1+k) = 1-k$ $\Rightarrow e = \frac{1-k}{1+k}$.
$e = \frac{1-(2\sqrt{2}+3)}{1+(2\sqrt{2}+3)} = \frac{-2-2\sqrt{2}}{4+2\sqrt{2}} = \frac{-2(1+\sqrt{2})}{2(2+\sqrt{2})} = \frac{-(1+\sqrt{2})}{\sqrt{2}(\sqrt{2}+1)} = -\frac{1}{\sqrt{2}}$.
Since eccentricity $e$ must be positive,we take the magnitude,so $e = \frac{1}{\sqrt{2}}$.
The focus $S$ is $(ae, 0)$ or $(-ae, 0)$.
Substituting $e = \frac{1}{\sqrt{2}}$,we get $S = \left(\pm \frac{a}{\sqrt{2}}, 0\right)$.
Comparing with the options,option $C$ is $\left(\frac{a}{\sqrt{2}}, 0\right)$ and option $D$ is $\left(\frac{-a}{\sqrt{2}}, 0\right)$. Given the standard form,both are valid,but usually,the positive focus is considered. Based on the provided options,$C$ is the correct choice.

(D) Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b > a$ and eccentricity $e = \frac{1}{\sqrt{2}}$.
Since $e^2 = 1 - \frac{a^2}{b^2}$,we have $\frac{1}{2} = 1 - \frac{a^2}{b^2}$,which implies $\frac{a^2}{b^2} = \frac{1}{2}$,so $b^2 = 2a^2$.
The ellipse equation is $\frac{x^2}{a^2} + \frac{y^2}{2a^2} = 1$,or $2x^2 + y^2 = 2a^2$.
For the parabola $y^2 = 4ax$,substituting $y^2$ into the ellipse equation: $2x^2 + 4ax - 2a^2 = 0$,or $x^2 + 2ax - a^2 = 0$.
Solving for $x$: $x = \frac{-2a \pm \sqrt{4a^2 + 4a^2}}{2} = -a \pm a\sqrt{2}$. Since $x > 0$,$x = a(\sqrt{2} - 1)$.
Then $y^2 = 4a^2(\sqrt{2} - 1)$,so $y = 2a\sqrt{\sqrt{2} - 1}$.
The slopes $m_1$ (ellipse) and $m_2$ (parabola) at the intersection point $(x_0, y_0)$ are found by differentiation.
For $2x^2 + y^2 = 2a^2$,$4x + 2y y' = 0 \implies m_1 = -\frac{2x_0}{y_0}$.
For $y^2 = 4ax$,$2y y' = 4a \implies m_2 = \frac{2a}{y_0}$.
$\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{-2x_0 - 2a}{y_0(1 - \frac{4ax_0}{y_0^2})}| = |\frac{-2(x_0 + a)}{y_0(1 - \frac{4ax_0}{4ax_0})}|$ is undefined,meaning the curves intersect at $90^\circ$ or $\theta = \frac{\pi}{2}$.
Thus,$\theta = \frac{\pi}{2}$. The point on the ellipse is determined by the parameter $\phi$ where $x = a \cos \phi, y = b \sin \phi = a\sqrt{2} \sin \phi$. For $\theta = \frac{\pi}{2}$,the coordinates are $(\frac{a}{2}, \frac{\sqrt{3}a}{\sqrt{2}})$.

(B) The given line is $x \cos \alpha + y \sin \alpha = 2 \sqrt{3}$ and the ellipse is $\frac{x^2}{16} + \frac{y^2}{8} = 1$.
Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 8$.
The condition for the line $x \cos \alpha + y \sin \alpha = p$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$.
Here $p = 2 \sqrt{3}$,so $p^2 = (2 \sqrt{3})^2 = 12$.
Substituting the values: $12 = 16 \cos^2 \alpha + 8 \sin^2 \alpha$.
Using $\cos^2 \alpha = 1 - \sin^2 \alpha$,we get $12 = 16(1 - \sin^2 \alpha) + 8 \sin^2 \alpha$.
$12 = 16 - 16 \sin^2 \alpha + 8 \sin^2 \alpha$.
$8 \sin^2 \alpha = 4$.
$\sin^2 \alpha = \frac{4}{8} = \frac{1}{2}$.
Since $\alpha$ is an acute angle,$\sin \alpha = \frac{1}{\sqrt{2}}$.
Therefore,$\alpha = \frac{\pi}{4}$.

(C) The equation of the line is $y=mx+c$,where $m=4$.
The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,where $a^2=4$ and $b^2=1$.
$A$ line $y=mx+c$ touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ if $c^2=a^2m^2+b^2$.
Substituting the values $a^2=4$,$b^2=1$,and $m=4$:
$c^2 = 4(4)^2 + 1$
$c^2 = 4(16) + 1$
$c^2 = 64 + 1 = 65$
Therefore,$c = \pm \sqrt{65}$.

(A) Given the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,we have $a^2=16$ and $b^2=12$.
The eccentricity $e$ is given by $b^2=a^2(1-e^2)$,so $12=16(1-e^2)$,which gives $1-e^2=\frac{3}{4}$,so $e^2=\frac{1}{4}$ and $e=\frac{1}{2}$.
For a focal chord with eccentric angles $\alpha$ and $\beta$,the condition is $\tan(\frac{\alpha}{2}) \tan(\frac{\beta}{2}) = \frac{e-1}{e+1}$.
Here $\alpha = \frac{\pi}{3}$,so $\frac{\alpha}{2} = \frac{\pi}{6}$.
$\tan(\frac{\pi}{6}) \tan(\frac{\theta}{2}) = \frac{1/2 - 1}{1/2 + 1} = \frac{-1/2}{3/2} = -\frac{1}{3}$.
Since $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} \tan(\frac{\theta}{2}) = -\frac{1}{3}$,so $\tan(\frac{\theta}{2}) = -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}$.
Thus,$\frac{\theta}{2} = -\frac{\pi}{6}$,which means $\theta = -\frac{\pi}{3}$.
Therefore,$\tan \theta = \tan(-\frac{\pi}{3}) = -\sqrt{3}$.

(B) The equation of the ellipse is $x^2+3y^2=3$,which can be written as $\frac{x^2}{3}+y^2=1$. Here $a^2=3$ and $b^2=1$.
The condition for $lx+my+n=0$ to be a normal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2)^2}{n^2}$.
Substituting $l=1, m=1, a^2=3, b^2=1$,we get $\frac{3}{1^2}+\frac{1}{1^2}=\frac{(3-1)^2}{n^2} \Rightarrow 4=\frac{4}{n^2} \Rightarrow n^2=1$.
Since $n>0$,$n=1$. The normal is $x+y+1=0$.
The equation of the ellipse is $x^2+5y^2=5$,or $\frac{x^2}{5}+y^2=1$.
The condition for $lx+my+n=0$ to be a tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $n^2=a^2l^2+b^2m^2$.
Substituting $l=1, n=3, a^2=5, b^2=1$,we get $3^2=5(1)^2+1(m)^2 \Rightarrow 9=5+m^2 \Rightarrow m^2=4$.
Since $m < 0$,$m=-2$. The tangent is $x-2y+3=0$.
Solving $x+y+1=0$ and $x-2y+3=0$: Subtracting the equations gives $3y-2=0 \Rightarrow y=\frac{2}{3}$.
Then $x=-1-y=-1-\frac{2}{3}=-\frac{5}{3}$.
Checking the point $(-\frac{5}{3}, \frac{2}{3})$ in the options: For $x-5y+5=0$,we have $-\frac{5}{3}-5(\frac{2}{3})+5 = \frac{-5-10+15}{3} = 0$.
Thus,the point satisfies $x-5y+5=0$.

(D) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{8} = 1$.
Let the point of contact be $(a, b)$. The equation of the tangent at $(a, b)$ to the ellipse $\frac{x^2}{18} + \frac{y^2}{8} = 1$ is $\frac{ax}{18} + \frac{by}{8} = 1$.
Given the tangent line is $8x + 3\sqrt{2}y = 36$,which can be written as $\frac{8x}{36} + \frac{3\sqrt{2}y}{36} = 1$,or $\frac{2x}{9} + \frac{\sqrt{2}y}{12} = 1$.
Comparing the two equations of the tangent:
$\frac{a}{18} = \frac{2}{9} \implies a = \frac{18 \times 2}{9} = 4$.
$\frac{b}{8} = \frac{\sqrt{2}}{12} \implies b = \frac{8\sqrt{2}}{12} = \frac{2\sqrt{2}}{3}$.
Now,calculate $a + \sqrt{2}b = 4 + \sqrt{2} \times \frac{2\sqrt{2}}{3} = 4 + \frac{4}{3} = \frac{12 + 4}{3} = \frac{16}{3}$.

(A) For a line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the condition is $c^2=a^2m^2+b^2$.
Given $4x+y+p=0$,we have $y=-4x-p$.
For the ellipse $x^2+3y^2=3$,i.e.,$\frac{x^2}{3}+\frac{y^2}{1}=1$,we have $a^2=3, b^2=1, m=-4, c=-p$.
Substituting these values: $(-p)^2 = 3(-4)^2 + 1 \Rightarrow p^2 = 48+1 = 49$.
Since $p>0$,$p=7$.
For a line $y=mx+c$ to be a normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the condition is $c^2=\frac{m^2(a^2-b^2)^2}{a^2+b^2m^2}$.
Given $16x+qy+14=0$,we have $y=-\frac{16}{q}x-\frac{14}{q}$.
For the ellipse $x^2+8y^2=33$,i.e.,$\frac{x^2}{33}+\frac{y^2}{33/8}=1$,we have $a^2=33, b^2=\frac{33}{8}, m=-\frac{16}{q}, c=-\frac{14}{q}$.
Substituting these values: $\frac{196}{q^2} = \frac{(-16/q)^2(33-33/8)^2}{33+(33/8)(-16/q)^2} = \frac{(256/q^2)(33 \times 7/8)^2}{33+(33/8)(256/q^2)} = \frac{(256/q^2)(33^2 \times 49 / 64)}{33(1+32/q^2)} = \frac{4 \times 33^2 \times 49 / q^2}{33(q^2+32)/q^2} = \frac{4 \times 33 \times 49}{q^2+32}$.
Simplifying: $\frac{196}{q^2} = \frac{6468}{q^2+32}$ $\Rightarrow \frac{49}{q^2} = \frac{1617}{q^2+32}$ $\Rightarrow q^2+32 = 33q^2$ $\Rightarrow 32q^2=32$ $\Rightarrow q^2=1$.
Since $q>0$,$q=1$.
Therefore,$p+q = 7+1 = 8$.

(C) The given equation of the ellipse is $9x^2 + 4y^2 - 18x - 8y - 23 = 0$.
Rearranging the terms,we get $9(x^2 - 2x) + 4(y^2 - 2y) = 23$.
Completing the square,$9(x^2 - 2x + 1) + 4(y^2 - 2y + 1) = 23 + 9 + 4$.
$9(x - 1)^2 + 4(y - 1)^2 = 36$.
Dividing by $36$,we get $\frac{(x - 1)^2}{4} + \frac{(y - 1)^2}{9} = 1$.
Here,$a^2 = 4$ and $b^2 = 9$,so $a < b$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
The equations of the latus rectum for a vertical ellipse are $y - k = \pm be$,where $(h, k) = (1, 1)$.
$y - 1 = \pm 3 \times \frac{\sqrt{5}}{3} = \pm \sqrt{5}$.
Therefore,$y = 1 \pm \sqrt{5}$.

(B) The equation of the ellipse is $\frac{x^2}{100} + \frac{y^2}{25} = 1$.
The equation of the tangent at point $(x_1, y_1) = (-8, 3)$ is given by $\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$.
Substituting the values: $\frac{x(-8)}{100} + \frac{y(3)}{25} = 1$.
Simplifying the equation: $-\frac{2x}{25} + \frac{3y}{25} = 1$,which gives $-2x + 3y = 25$.
To find the point where the particle crosses the $Y$-axis,we set $x = 0$.
Substituting $x = 0$ into the tangent equation: $3y = 25$,which gives $y = \frac{25}{3}$.
Thus,the point is $\left(0, \frac{25}{3}\right)$.

(A) For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the ends of the latus rectum are $(\pm ae, \pm \frac{b^2}{a})$.
Given $\frac{x^2}{9}+\frac{y^2}{5}=1$,we have $a^2=9, b^2=5$.
$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{5}{9} = \frac{4}{9} \Rightarrow e = \frac{2}{3}$.
Ends of latus rectum are $(\pm 2, \pm \frac{5}{3})$.
The equation of the tangent at $P(2, \frac{5}{3})$ is $\frac{2x}{9} + \frac{5y/3}{5} = 1$,which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$.
This line intersects the axes at $A(\frac{9}{2}, 0)$ and $C(0, 3)$.
The area of the triangle formed in the first quadrant is $\frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$.
Since there are four such symmetric triangles formed by the tangents at the four ends of the latus rectum,the total area is $4 \times \frac{27}{4} = 27$ sq. units.

(D) The product of the lengths of the perpendiculars drawn from the two foci of an ellipse to any tangent is equal to the square of the semi-minor axis.
For the given ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$,we have $a^2 = 9$ and $b^2 = 25$. Since $b^2 > a^2$,the semi-minor axis is $b = \sqrt{9} = 3$.
Therefore,the product of the lengths of the perpendiculars is $b^2 = 3^2 = 9$.

(A) The given equation of the 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 $b^2=1$,so $a=\sqrt{2}$ and $b=1$.
The equation of the tangent to the ellipse at point $P(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
Substituting the values,we get $\frac{x \cos \theta}{\sqrt{2}} + y \sin \theta = 1$.
For the $x$-intercept $(A)$,set $y=0$: $\frac{x \cos \theta}{\sqrt{2}} = 1 \Rightarrow x = \sqrt{2} \sec \theta$. So,$A = (\sqrt{2} \sec \theta, 0)$.
For the $y$-intercept $(B)$,set $x=0$: $y \sin \theta = 1 \Rightarrow y = \operatorname{cosec} \theta$. So,$B = (0, \operatorname{cosec} \theta)$.
Let $M(h, k)$ be the midpoint of $AB$. Then:
$h = \frac{\sqrt{2} \sec \theta + 0}{2} = \frac{\sec \theta}{\sqrt{2}} \Rightarrow \sec \theta = \sqrt{2}h$
$k = \frac{0 + \operatorname{cosec} \theta}{2} = \frac{\operatorname{cosec} \theta}{2} \Rightarrow \operatorname{cosec} \theta = 2k$
We know that $\cos^2 \theta + \sin^2 \theta = 1$,which implies $\frac{1}{\sec^2 \theta} + \frac{1}{\operatorname{cosec}^2 \theta} = 1$.
Substituting the values of $\sec \theta$ and $\operatorname{cosec} \theta$:
$\frac{1}{(\sqrt{2}h)^2} + \frac{1}{(2k)^2} = 1$
$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(A) Given circle is $(x-1)^2+y^2=r^2 \dots (i)$
Here,the centre of the circle is $(1,0)$ and the radius is $r$.
Given ellipse is $x^2+4y^2=16$,which can be written as $\frac{x^2}{16}+\frac{y^2}{4}=1$.
Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we get $a=4$ and $b=2$.
Since the circle touches the ellipse internally,the normal to the ellipse at the point of contact $P$ must pass through the centre of the circle $(1,0)$.
Let the point of contact be $P(4\cos\theta, 2\sin\theta)$.
The equation of the normal to the ellipse at $P$ is given by $ax\sec\theta - by\operatorname{cosec}\theta = a^2-b^2$.
Substituting $a=4, b=2$ and $a^2-b^2 = 16-4=12$,we get $4x\sec\theta - 2y\operatorname{cosec}\theta = 12$.
Since this normal passes through $(1,0)$,we have $4(1)\sec\theta - 2(0)\operatorname{cosec}\theta = 12$,which implies $4\sec\theta = 12$,so $\sec\theta = 3$.
Thus,$\cos\theta = \frac{1}{3}$ and $\sin\theta = \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.
The point $P$ is $\left(4\left(\frac{1}{3}\right), 2\left(\frac{2\sqrt{2}}{3}\right)\right) = \left(\frac{4}{3}, \frac{4\sqrt{2}}{3}\right)$.
The radius $r$ is the distance between the centre $(1,0)$ and the point $P\left(\frac{4}{3}, \frac{4\sqrt{2}}{3}\right)$.
$r^2 = \left(\frac{4}{3}-1\right)^2 + \left(\frac{4\sqrt{2}}{3}-0\right)^2 = \left(\frac{1}{3}\right)^2 + \frac{32}{9} = \frac{1}{9} + \frac{32}{9} = \frac{33}{9} = \frac{11}{3}$.
Therefore,$r = \sqrt{\frac{11}{3}}$.

(C) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Let $P(5 \cos \theta, 4 \sin \theta)$ be a point on the ellipse.
The equation of the tangent at $P$ is $\frac{x(5 \cos \theta)}{25} + \frac{y(4 \sin \theta)}{16} = 1$,which simplifies to $\frac{x \cos \theta}{5} + \frac{y \sin \theta}{4} = 1$.
This tangent intersects the $X$-axis at point $A$ by setting $y=0$,giving $A = (\frac{5}{\cos \theta}, 0) = (5 \sec \theta, 0)$.
The image of $A(5 \sec \theta, 0)$ with respect to the line $y=x$ is $A^{\prime} = (0, 5 \sec \theta)$.
The equation of the circle with diameter $AA^{\prime}$ is $(x - 5 \sec \theta)(x - 0) + (y - 0)(y - 5 \sec \theta) = 0$.
This simplifies to $x^2 + y^2 - 5 \sec \theta (x + y) = 0$.
For this circle to pass through a fixed point independent of $\theta$,we observe that for any $\theta$,the point $(0, 0)$ satisfies the equation $0^2 + 0^2 - 5 \sec \theta (0 + 0) = 0$.

(D) We know that the product of the lengths of the perpendiculars drawn from the foci to any tangent of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is equal to $b^2$.
Given that the product is $9$,we have $b^2 = 9$.
The equation of the tangent is $y = \frac{-3}{4}x + 3\sqrt{2}$.
Comparing this with the standard tangent equation $y = mx \pm \sqrt{a^2m^2 + b^2}$,where $m = \frac{-3}{4}$,we have $3\sqrt{2} = \sqrt{a^2(\frac{-3}{4})^2 + b^2}$.
Squaring both sides,we get $18 = a^2(\frac{9}{16}) + 9$.
Subtracting $9$ from both sides,$9 = \frac{9a^2}{16}$,which gives $a^2 = 16$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.

(A) Given the ellipse equation: $2x^2+y^2=2$,which can be written as $x^2+\frac{y^2}{2}=1$.
Here,$a^2=1$ and $b^2=2$.
The equation of the tangent is $x+2y+k=0$,which implies $y=-\frac{1}{2}x-\frac{k}{2}$.
Comparing this with $y=mx+c$,we get $m=-\frac{1}{2}$ and $c=-\frac{k}{2}$.
The condition for a line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2m^2+b^2$.
Substituting the values: $(-\frac{k}{2})^2 = (1)(-\frac{1}{2})^2 + 2$.
$\frac{k^2}{4} = \frac{1}{4} + 2 = \frac{9}{4}$.
$k^2=9$,so $k=\pm 3$. Since $k>0$,we have $k=3$.
The point of contact is $\left(\frac{1}{\sqrt{2}}, \frac{3}{3}\right) = \left(\frac{1}{\sqrt{2}}, 1\right)$.
The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(x_1, y_1)$ is $\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$.
Substituting $a^2=1, b^2=2, x_1=\frac{1}{\sqrt{2}}, y_1=1$:
$\frac{1 \cdot x}{1/\sqrt{2}} - \frac{2 \cdot y}{1} = 1-2$.
$\sqrt{2}x - 2y = -1$,which simplifies to $\sqrt{2}x-2y+1=0$.

(D) Given ellipse is $\frac{x^2}{16}+\frac{y^2}{12}=1$. Here $a^2=16$ and $b^2=12$.
Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{12}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2}$.
The end points of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 4 \times \frac{1}{2}, \pm \frac{12}{4}) = (\pm 2, \pm 3)$.
The equation of the tangent at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$.
For the point $(2, 3)$,the tangent is $\frac{2x}{16}+\frac{3y}{12}=1$,which simplifies to $\frac{x}{8}+\frac{y}{4}=1$.
This line intersects the $x$-axis at $(8, 0)$ and the $y$-axis at $(0, 4)$.
Due to symmetry,the quadrilateral formed by the four tangents at $(\pm 2, \pm 3)$ is a rhombus with vertices at $(\pm 8, 0)$ and $(0, \pm 4)$.
The area of this rhombus is $4 \times (\text{Area of triangle with vertices } (0,0), (8,0), (0,4)) = 4 \times (\frac{1}{2} \times 8 \times 4) = 64$ sq. units.

(C) The equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$.
Any point $P$ on the ellipse is $(\sqrt{2}\cos\theta, \sin\theta)$.
The equation of the tangent at $P$ is $\frac{x(\sqrt{2}\cos\theta)}{2} + \frac{y(\sin\theta)}{1} = 1$,which simplifies to $\frac{x\cos\theta}{\sqrt{2}} + y\sin\theta = 1$.
The intercepts made by the tangent on the coordinate axes are $A\left(\frac{\sqrt{2}}{\cos\theta}, 0\right)$ and $B\left(0, \frac{1}{\sin\theta}\right)$.
Let $(h, k)$ be the mid-point of $AB$. Then $h = \frac{\sqrt{2}}{2\cos\theta}$ and $k = \frac{1}{2\sin\theta}$.
This implies $\cos\theta = \frac{1}{\sqrt{2}h}$ and $\sin\theta = \frac{1}{2k}$.
Using the identity $\cos^2\theta + \sin^2\theta = 1$,we get $\left(\frac{1}{\sqrt{2}h}\right)^2 + \left(\frac{1}{2k}\right)^2 = 1$.
Thus,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(A) The equation of the tangent at $(3 \sqrt{3} \cos \theta, \sin \theta)$ to the ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$ is given by $\frac{x(3 \sqrt{3} \cos \theta)}{27} + \frac{y \sin \theta}{1} = 1$,which simplifies to $\frac{x \cos \theta}{3 \sqrt{3}} + y \sin \theta = 1$.
The $x$-intercept is $a = 3 \sqrt{3} \sec \theta$ and the $y$-intercept is $b = \operatorname{cosec} \theta$.
The sum of the intercepts is $L(\theta) = 3 \sqrt{3} \sec \theta + \operatorname{cosec} \theta$.
To find the minimum,we differentiate with respect to $\theta$: $\frac{dL}{d\theta} = 3 \sqrt{3} \sec \theta \tan \theta - \operatorname{cosec} \theta \cot \theta$.
Setting $\frac{dL}{d\theta} = 0$,we get $3 \sqrt{3} \frac{\sin \theta}{\cos^2 \theta} = \frac{\cos \theta}{\sin^2 \theta}$,which implies $\tan^3 \theta = \frac{1}{3 \sqrt{3}} = \left(\frac{1}{\sqrt{3}}\right)^3$.
Thus,$\tan \theta = \frac{1}{\sqrt{3}}$,which gives $\theta = \frac{\pi}{6}$.

(A) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$,so $a^2 = 16$ and $b^2 = 9$.
The tangent at $P(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$. It meets the major axis $(y=0)$ at $Q(\frac{a}{\cos \theta}, 0)$.
The normal at $P$ is $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2 - b^2$. It meets the major axis at $R(\frac{(a^2 - b^2) \cos \theta}{a}, 0)$.
The distance $QR = \left| \frac{a}{\cos \theta} - \frac{(a^2 - b^2) \cos \theta}{a} \right| = \left| \frac{a^2 - (a^2 - b^2) \cos^2 \theta}{a \cos \theta} \right| = \left| \frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{a \cos \theta} \right|$.
Thus,Reason $(R)$ is true.
For $a=4, b=3, \theta = \frac{\pi}{3}$,$QR = \left| \frac{16 \cdot \frac{3}{4} + 9 \cdot \frac{1}{4}}{4 \cdot \frac{1}{2}} \right| = \left| \frac{12 + 2.25}{2} \right| = \frac{14.25}{2} = \frac{57}{8}$.
Thus,Assertion $(A)$ is also true and $(R)$ is the correct explanation.

(A) The equation of the given ellipse is $\frac{x^2}{64}+\frac{y^2}{4}=1$.
Let a parametric point on the ellipse be $P(8 \cos \theta, 2 \sin \theta)$.
The equation of the tangent to the ellipse at point $P$ is $\frac{x \cos \theta}{8}+\frac{y \sin \theta}{2}=1$.
This can be rewritten as $\frac{x}{8/\cos \theta} + \frac{y}{2/\sin \theta} = 1$.
The length of the intercept $l$ between the coordinate axes is $l = \sqrt{(\frac{8}{\cos \theta})^2 + (\frac{2}{\sin \theta})^2} = \sqrt{64 \sec^2 \theta + 4 \operatorname{cosec}^2 \theta}$.
Using $\sec^2 \theta = 1 + \tan^2 \theta$ and $\operatorname{cosec}^2 \theta = 1 + \cot^2 \theta$,we get $l = \sqrt{64(1 + \tan^2 \theta) + 4(1 + \cot^2 \theta)} = \sqrt{68 + 64 \tan^2 \theta + 4 \cot^2 \theta}$.
By the $AM-GM$ inequality,$64 \tan^2 \theta + 4 \cot^2 \theta \geq 2 \sqrt{64 \tan^2 \theta \cdot 4 \cot^2 \theta} = 2 \sqrt{256} = 32$.
Thus,the minimum value of $l$ is $\sqrt{68 + 32} = \sqrt{100} = 10$.
Therefore,option $A$ is correct.

(B) Given the ellipse equation $\frac{x^2}{25}+\frac{y^2}{16}=1$,we have $a^2=25$ and $b^2=16$,so $a=5$ and $b=4$.
Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The ends of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 3, \pm \frac{16}{5})$.
The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.
For the point $(3, \frac{16}{5})$,the tangent is $\frac{3x}{25} + \frac{(16/5)y}{16} = 1$,which simplifies to $\frac{3x}{25} + \frac{y}{5} = 1$.
This line intersects the $x$-axis at $P(\frac{25}{3}, 0)$ and the $y$-axis at $Q(0, 5)$.
The area of the triangle $OPQ$ in the first quadrant is $\frac{1}{2} \times \frac{25}{3} \times 5 = \frac{125}{6}$.
By symmetry,the total area of the quadrilateral formed by the four tangents is $4 \times \frac{125}{6} = \frac{250}{3}$ sq units.

(A) The given equation of the curve is $9x^2 + 16y^2 = 144$. Dividing by $144$,we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$. This is an ellipse with $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$. The centre of the ellipse is $(0, 0)$. The equation of the normal at any point $(x_1, y_1)$ on the ellipse is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Substituting $a^2 = 16$ and $b^2 = 9$,we get $\frac{16x}{x_1} - \frac{9y}{y_1} = 7$. The distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$. Here,the line is $\frac{16}{x_1}x - \frac{9}{y_1}y - 7 = 0$. Thus,$d = \frac{|-7|}{\sqrt{(\frac{16}{x_1})^2 + (\frac{9}{y_1})^2}} = \frac{7}{\sqrt{\frac{256}{x_1^2} + \frac{81}{y_1^2}}}$. Since $x_1 = 4\cos\theta$ and $y_1 = 3\sin\theta$,we have $d = \frac{7}{\sqrt{\frac{256}{16\cos^2\theta} + \frac{81}{9\sin^2\theta}}} = \frac{7}{\sqrt{16\sec^2\theta + 9\csc^2\theta}}$. To maximize $d$,we must minimize $f(\theta) = 16\sec^2\theta + 9\csc^2\theta = 16(1 + \tan^2\theta) + 9(1 + \cot^2\theta) = 25 + 16\tan^2\theta + 9\cot^2\theta$. By $AM$-$GM$ inequality,$16\tan^2\theta + 9\cot^2\theta \ge 2\sqrt{16 \times 9} = 2 \times 12 = 24$. The minimum value of $f(\theta)$ is $25 + 24 = 49$. Thus,the maximum distance is $d_{max} = \frac{7}{\sqrt{49}} = \frac{7}{7} = 1$.

(C) Given the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$,we have $a=5$ and $b=4$.
The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The end of the latus rectum in the first quadrant is $(ae, \frac{b^2}{a}) = (5 \times \frac{3}{5}, \frac{16}{5}) = (3, 3.2)$.
The equation of the normal at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.
Substituting $x_1 = 3, y_1 = 3.2, a^2 = 25, b^2 = 16$:
$\frac{25x}{3} - \frac{16y}{3.2} = 25 - 16 = 9$.
$\frac{25x}{3} - 5y = 9$.
Multiplying by $3$,we get $25x - 15y = 27$.
Comparing this with $lx + my = 27$,we get $l = 25$ and $m = -15$.
Thus,$l + m = 25 - 15 = 10$.
Since $e = 0.6$,we have $\frac{6}{e} = \frac{6}{0.6} = 10$.
Therefore,$l + m = \frac{6}{e}$.

(D) The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $y = mx \pm \frac{m(a^2 - b^2)}{\sqrt{a^2 + b^2m^2}}$.
Given the normal equation $6x - 5y - 20 = 0$,we can rewrite it as $y = \frac{6}{5}x - 4$. Thus,$m = \frac{6}{5}$ and the constant term is $-4$.
For the ellipse $x^2 + 3y^2 = k$,we have $\frac{x^2}{k} + \frac{y^2}{k/3} = 1$,so $a^2 = k$ and $b^2 = \frac{k}{3}$.
Substituting these into the normal equation constant term: $\frac{m(a^2 - b^2)}{\sqrt{a^2 + b^2m^2}} = 4$.
$\frac{\frac{6}{5}(k - \frac{k}{3})}{\sqrt{k + \frac{k}{3} \cdot (\frac{6}{5})^2}} = 4$.
$\frac{\frac{6}{5} \cdot \frac{2k}{3}}{\sqrt{k + \frac{k}{3} \cdot \frac{36}{25}}} = 4$ $\Rightarrow \frac{\frac{4k}{5}}{\sqrt{k(1 + \frac{12}{25})}} = 4$.
$\frac{4k}{5\sqrt{k} \cdot \sqrt{\frac{37}{25}}} = 4 \Rightarrow \frac{4\sqrt{k}}{5 \cdot \frac{\sqrt{37}}{5}} = 4$.
$\frac{\sqrt{k}}{\sqrt{37}} = 1$ $\Rightarrow \sqrt{k} = \sqrt{37}$ $\Rightarrow k = 37$.

(B) The equation of the ellipse is $\frac{x^2}{8} + \frac{y^2}{2} = 1$.
The equation of the normal at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$.
Substituting $(x_1, y_1) = (2, -1)$,$a^2 = 8$,and $b^2 = 2$:
$\frac{8x}{2} - \frac{2y}{-1} = 8 - 2
$ $\Rightarrow 4x + 2y = 6
$ $\Rightarrow y = 3 - 2x$.
Substituting $y = 3 - 2x$ into the ellipse equation $x^2 + 4y^2 = 8$:
$x^2 + 4(3 - 2x)^2 = 8
$ $\Rightarrow x^2 + 4(9 - 12x + 4x^2) = 8
$ $\Rightarrow 17x^2 - 48x + 28 = 0$.
Since $(2, -1)$ is a point on the ellipse,$x = 2$ is a root of this quadratic equation.
Let the other root be $a$. Using the product of roots formula $x_1 x_2 = \frac{c}{A}$:
$2 \times a = \frac{28}{17}
$ $\Rightarrow a = \frac{14}{17}
$ $\Rightarrow 17a = 14$.

(D) The equation of the line is $4x + 2y + n = 0$,which can be rewritten as $y = -2x - \frac{n}{2}$.
Comparing this with $y = mx + c$,we get $m = -2$ and $c = -\frac{n}{2}$.
The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$,so $a^2 = 36$ and $b^2 = 16$.
The condition for the line $y = mx + c$ to be a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = \frac{(a^2 - b^2)^2 m^2}{a^2 + b^2 m^2}$.
Substituting the values: $(-\frac{n}{2})^2 = \frac{(36 - 16)^2 (-2)^2}{36 + 16(-2)^2}$.
$\frac{n^2}{4} = \frac{(20)^2 \times 4}{36 + 16(4)} = \frac{400 \times 4}{36 + 64} = \frac{1600}{100} = 16$.
$n^2 = 16 \times 4 = 64$.
Therefore,$n = \pm 8$.

(C) The given equations are $x = 1 + 2 \cos \theta$ and $y = 2 + \sin \theta$.
This represents an ellipse centered at $(h, k) = (1, 2)$ with $a = 2$ and $b = 1$.
The equation of the ellipse is $\frac{(x-1)^2}{4} + \frac{(y-2)^2}{1} = 1$.
The parametric point $P$ at $\theta = \pi/4$ is $(1 + 2 \cos(\pi/4), 2 + \sin(\pi/4)) = (1 + \sqrt{2}, 2 + 1/\sqrt{2})$.
The equation of the normal at point $(a \cos \theta, b \sin \theta)$ for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $ax \sec \theta - by \csc \theta = a^2 - b^2$.
Shifting the origin to $(1, 2)$,the normal equation is $2(x-1) \sec(\pi/4) - 1(y-2) \csc(\pi/4) = 2^2 - 1^2$.
$2(x-1) \sqrt{2} - (y-2) \sqrt{2} = 3$.
$2\sqrt{2}(x-1) - \sqrt{2}(y-2) = 3$.
The major axis of this ellipse is the line $y = 2$.
Substituting $y = 2$ into the normal equation:
$2\sqrt{2}(x-1) - \sqrt{2}(2-2) = 3$.
$2\sqrt{2}(x-1) = 3$.
$x-1 = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4}$.
$x = 1 + \frac{3\sqrt{2}}{4} = \frac{4+3\sqrt{2}}{4}$.
Note: Re-evaluating the normal equation $ax \sec \theta - by \csc \theta = a^2 - b^2$ with $a=2, b=1, \theta=\pi/4$:
$2(x-1)\sqrt{2} - 1(y-2)\sqrt{2} = 3$.
At $y=2$,$2\sqrt{2}(x-1) = 3 \implies x = 1 + \frac{3}{2\sqrt{2}} = 1 + \frac{3\sqrt{2}}{4}$.
Given the options provided,there appears to be a discrepancy in the standard form. Based on the calculation,the intersection point is $(\frac{4+3\sqrt{2}}{4}, 2)$.

(D) Given the tangent line $x+\sqrt{3} y=3$ ... $(i)$ and the ellipse $2 x^2+3 y^2=k$.
The equation of the tangent at point $P(x_1, y_1)$ is $2 x x_1+3 y y_1=k$ ... (ii).
Comparing $(i)$ and (ii),we have $\frac{2 x_1}{1} = \frac{3 y_1}{\sqrt{3}} = \frac{k}{3}$.
Thus,$x_1 = \frac{k}{6}$ and $y_1 = \frac{k}{3 \sqrt{3}}$.
Since $P(x_1, y_1)$ lies on the ellipse,$2(\frac{k}{6})^2 + 3(\frac{k}{3 \sqrt{3}})^2 = k$.
$\frac{k^2}{18} + \frac{k^2}{9} = k$ $\Rightarrow \frac{k^2}{6} = k$ $\Rightarrow k = 6$.
So,$x_1 = 1$ and $y_1 = \frac{6}{3 \sqrt{3}} = \frac{2}{\sqrt{3}}$.
The slope of the tangent is $m_t = -\frac{1}{\sqrt{3}}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = \sqrt{3}$.
The equation of the normal at $P(1, \frac{2}{\sqrt{3}})$ is $y - \frac{2}{\sqrt{3}} = \sqrt{3}(x - 1)$.
$\sqrt{3} y - 2 = 3x - 3 \Rightarrow 3x - \sqrt{3} y = 1$.

(C) Given the ellipse equation: $9x^2+4y^2-18x+16y-11=0$.
Since $P(1, k)$ lies on the ellipse,substitute $x=1$:
$9(1)^2+4k^2-18(1)+16k-11=0$
$9+4k^2-18+16k-11=0$
$4k^2+16k-20=0$
$k^2+4k-5=0$
$(k+5)(k-1)=0$.
Since $k>0$,we have $k=1$. Thus,$P$ is $(1, 1)$.
Rewrite the ellipse equation in standard form:
$9(x-1)^2+4(y+2)^2 = 11+9+16 = 36$
$\frac{(x-1)^2}{4} + \frac{(y+2)^2}{9} = 1$.
Here $a^2=4$ and $b^2=9$,so $b>a$.
Eccentricity $e = \sqrt{1-\frac{a^2}{b^2}} = \sqrt{1-\frac{4}{9}} = \frac{\sqrt{5}}{3}$.
The directrices are $y+2 = \pm \frac{b}{e} = \pm \frac{3}{\sqrt{5}/3} = \pm \frac{9}{\sqrt{5}}$.
So,$y = -2 \pm \frac{9}{\sqrt{5}}$.
The distances from $P(1, 1)$ to the directrices $y = -2 + \frac{9}{\sqrt{5}}$ and $y = -2 - \frac{9}{\sqrt{5}}$ are:
$d_1 = |1 - (-2 + \frac{9}{\sqrt{5}})| = |3 - \frac{9}{\sqrt{5}}| = \frac{9}{\sqrt{5}} - 3$ (since $\frac{9}{\sqrt{5}} \approx 4.02 > 3$).
$d_2 = |1 - (-2 - \frac{9}{\sqrt{5}})| = |3 + \frac{9}{\sqrt{5}}| = 3 + \frac{9}{\sqrt{5}}$.
The shortest distance is $\frac{9}{\sqrt{5}} - 3$.

(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{7} = 1$,where $a^2 = 9$ and $b^2 = 7$.
The point given is $(x_1, y_1) = \left(3 \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right) = \left(\frac{3}{\sqrt{2}}, \sqrt{\frac{7}{2}}\right)$.
The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$.
Substituting the values: $\frac{9x}{3/\sqrt{2}} - \frac{7y}{\sqrt{7/2}} = 9 - 7$.
This simplifies to $3\sqrt{2}x - \sqrt{14}y = 2$.
To find the intersection with the major axis ($x$-axis),set $y = 0$:
$3\sqrt{2}x = 2 \implies x = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$.
Thus,the point of intersection is $\left(\sqrt{\frac{2}{9}}, 0\right)$.

(C) The equation of the ellipse is $b^2 x^2 + a^2 y^2 = a^2 b^2$,which can be written as $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
One end of the latus rectum is $(ae, \frac{b^2}{a})$.
The equation of the normal at $(x_1, y_1)$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$.
Substituting $(ae, \frac{b^2}{a})$ into the normal equation:
$\frac{a^2 x}{ae} - \frac{b^2 y}{b^2/a} = a^2 - b^2 \Rightarrow \frac{ax}{e} - ay = a^2 - b^2$.
Since the normal passes through one end of the minor axis $(0, -b)$,we substitute $x=0$ and $y=-b$:
$0 - a(-b) = a^2 - b^2 \Rightarrow ab = a^2 - b^2$.
Using $b^2 = a^2(1 - e^2)$,we have $b^2 = a^2 - a^2 e^2$,so $a^2 - b^2 = a^2 e^2$.
Thus,$ab = a^2 e^2 \Rightarrow b = ae^2$.
Squaring both sides,$b^2 = a^2 e^4$.
Since $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2 e^4$.
Dividing by $a^2$,$1 - e^2 = e^4 \Rightarrow e^4 + e^2 = 1$.

(C) Given,the curve is $2x^2+y^2=2x$.
Rearranging the terms,we get $2x^2-2x+y^2=0$.
Completing the square for $x$: $2(x^2-x)+y^2=0 \Rightarrow 2(x-\frac{1}{2})^2+y^2=\frac{1}{2}$.
Dividing by $\frac{1}{2}$,we get $\frac{(x-\frac{1}{2})^2}{1/4} + \frac{y^2}{1/2} = 1$,which represents an ellipse.
Let a point $P$ on the ellipse be $P(\frac{1}{2}+\frac{1}{2}\cos\theta, \frac{1}{\sqrt{2}}\sin\theta)$.
The distance $PQ$ from $Q(a, 0)$ is given by $PQ^2 = (\frac{1}{2}+\frac{1}{2}\cos\theta-a)^2 + (\frac{1}{\sqrt{2}}\sin\theta)^2$.
$PQ^2 = ((\frac{1}{2}-a)+\frac{1}{2}\cos\theta)^2 + \frac{1}{2}\sin^2\theta$.
$PQ^2 = (\frac{1}{2}-a)^2 + \frac{1}{4}\cos^2\theta + (\frac{1}{2}-a)\cos\theta + \frac{1}{2}(1-\cos^2\theta)$.
$PQ^2 = \frac{1}{4}-a+a^2 + \frac{1}{4}\cos^2\theta + (\frac{1}{2}-a)\cos\theta + \frac{1}{2} - \frac{1}{2}\cos^2\theta$.
$PQ^2 = a^2-a+\frac{3}{4} - \frac{1}{4}\cos^2\theta + (\frac{1}{2}-a)\cos\theta$.
To find the maximum distance,we differentiate with respect to $\cos\theta$ and set to $0$,or analyze the quadratic in $\cos\theta$.
After simplification,the maximum distance is $\sqrt{1-2a+2a^2}$.

(D) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$,where $a^2 = 16$ and $b^2 = 9$.
The director circle of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $x^2 + y^2 = a^2 + b^2$.
Here,$a^2 + b^2 = 16 + 9 = 25$.
Thus,the given circle $x^2 + y^2 = 25$ is the director circle of the ellipse.
By definition,the locus of the point of intersection of perpendicular tangents to an ellipse is its director circle.
Therefore,the angle between the tangents drawn from any point on the director circle to the ellipse is $\frac{\pi}{2}$.

(B) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{25} = 1$.
Given the line $2x - 3y + 4 = 0$,we can write $y = \frac{2x + 4}{3}$.
Substituting this into the ellipse equation: $25x^2 + 9(\frac{2x + 4}{3})^2 = 225$.
$25x^2 + (2x + 4)^2 = 225$ $\Rightarrow 25x^2 + 4x^2 + 16x + 16 = 225$ $\Rightarrow 29x^2 + 16x - 209 = 0$.
The midpoint $\alpha$ of the $x$-coordinates is $\frac{x_1 + x_2}{2} = \frac{-16/29}{2} = -\frac{8}{29}$.
Similarly,for $y$,$x = \frac{3y - 4}{2}$. Substituting into the ellipse equation: $25(\frac{3y - 4}{2})^2 + 9y^2 = 225$.
$\frac{25}{4}(9y^2 - 24y + 16) + 9y^2 = 225$ $\Rightarrow 225y^2 - 600y + 400 + 36y^2 = 900$ $\Rightarrow 261y^2 - 600y - 500 = 0$.
The midpoint $\beta$ of the $y$-coordinates is $\frac{y_1 + y_2}{2} = \frac{600/261}{2} = \frac{300}{261} = \frac{100}{87}$.
Now,$3\beta - 2\alpha = 3(\frac{100}{87}) - 2(-\frac{8}{29}) = \frac{100}{29} + \frac{16}{29} = \frac{116}{29} = 4$.