Ellipse Questions in English

(B) Given that the focus is at $(0, 5\sqrt{3})$,the ellipse is vertical with its major axis along the $y$-axis. Thus,$b > a$.
The focus is $(0, be) = (0, 5\sqrt{3})$,so $be = 5\sqrt{3}$.
Squaring both sides,$b^2e^2 = 75$.
For an ellipse,$a^2 = b^2(1 - e^2) = b^2 - b^2e^2$,so $b^2e^2 = b^2 - a^2 = 75$.
We are given the difference between the lengths of the major axis $(2b)$ and the minor axis $(2a)$ is $10$:
$2b - 2a = 10 \Rightarrow b - a = 5$.
Using the identity $b^2 - a^2 = (b - a)(b + a) = 75$:
$5(b + a) = 75 \Rightarrow b + a = 15$.
Solving the system:
$b - a = 5$
$b + a = 15$
Adding these gives $2b = 20 \Rightarrow b = 10$.
Subtracting gives $2a = 10 \Rightarrow a = 5$.
The length of the latus rectum for a vertical ellipse is $\frac{2a^2}{b}$.
$LR = \frac{2(5^2)}{10} = \frac{2 \times 25}{10} = \frac{50}{10} = 5$.

(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.
Given the point $(3, -4.5)$,the tangent equation is $\frac{3x}{a^2} - \frac{4.5y}{b^2} = 1$.
Comparing this with the given tangent line $x - 2y = 12$,we rewrite the given line as $\frac{x}{12} - \frac{2y}{12} = 1$,which simplifies to $\frac{x}{12} - \frac{y}{6} = 1$.
Equating the coefficients: $\frac{3}{a^2} = \frac{1}{12}$ $\Rightarrow a^2 = 36$ $\Rightarrow a = 6$.
And $\frac{4.5}{b^2} = \frac{1}{6} \Rightarrow b^2 = 4.5 \times 6 = 27$.
The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 \times 27}{6} = \frac{54}{6} = 9$.

(B) Given the ellipse equation: $3x^2 + 5y^2 = 32$.
Differentiating with respect to $x$: $6x + 10y \frac{dy}{dx} = 0$,which gives $\frac{dy}{dx} = -\frac{3x}{5y}$.
At point $P(2, 2)$,the slope of the tangent is $m = -\frac{3(2)}{5(2)} = -\frac{3}{5}$.
The equation of the tangent at $P(2, 2)$ is $y - 2 = -\frac{3}{5}(x - 2)$. Setting $y = 0$ to find $Q$,we get $-2 = -\frac{3}{5}(x - 2)$ $\Rightarrow x - 2 = \frac{10}{3}$ $\Rightarrow x = \frac{16}{3}$. Thus,$Q = (\frac{16}{3}, 0)$.
The slope of the normal at $P(2, 2)$ is $m' = -\frac{1}{m} = \frac{5}{3}$.
The equation of the normal at $P(2, 2)$ is $y - 2 = \frac{5}{3}(x - 2)$. Setting $y = 0$ to find $R$,we get $-2 = \frac{5}{3}(x - 2)$ $\Rightarrow x - 2 = -\frac{6}{5}$ $\Rightarrow x = \frac{4}{5}$. Thus,$R = (\frac{4}{5}, 0)$.
The area of triangle $PQR$ with base $QR$ on the $x$-axis and height equal to the $y$-coordinate of $P$ is:
Area $= \frac{1}{2} \times |x_Q - x_R| \times |y_P| = \frac{1}{2} \times |\frac{16}{3} - \frac{4}{5}| \times 2 = |\frac{80 - 12}{15}| = \frac{68}{15}$ sq. units.

(B) The equation of the ellipse is $3x^2 + 4y^2 = 12$,which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here $a^2 = 4$ and $b^2 = 3$.
Let the point $P$ be $(2\cos \theta, \sqrt{3}\sin \theta)$.
The equation of the normal at $P$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$,which simplifies to $\frac{4x}{2\cos \theta} - \frac{3y}{\sqrt{3}\sin \theta} = 4 - 3$,or $2x\sec \theta - \sqrt{3}y\csc \theta = 1$.
The slope of this normal is $\frac{2\sec \theta}{\sqrt{3}\csc \theta} = \frac{2}{\sqrt{3}}\tan \theta$.
Since the normal is parallel to $2x + y = 4$,its slope is $-2$. Thus,$\frac{2}{\sqrt{3}}\tan \theta = -2$,which gives $\tan \theta = -\sqrt{3}$.
For $\tan \theta = -\sqrt{3}$,we have $\cos \theta = \pm \frac{1}{2}$ and $\sin \theta = \mp \frac{\sqrt{3}}{2}$.
The equation of the tangent at $P$ is $\frac{x\cos \theta}{2} + \frac{y\sin \theta}{\sqrt{3}} = 1$,or $x\cos \theta + \frac{2y\sin \theta}{\sqrt{3}} = 2$.
Since the tangent passes through $Q(4, 4)$,we have $4\cos \theta + \frac{8\sin \theta}{\sqrt{3}} = 2$,or $2\sqrt{3}\cos \theta + 4\sin \theta = \sqrt{3}$.
Substituting $\sin \theta = -\sqrt{3}\cos \theta$,we get $2\sqrt{3}\cos \theta - 4\sqrt{3}\cos \theta = \sqrt{3}$,so $-2\sqrt{3}\cos \theta = \sqrt{3}$,which means $\cos \theta = -\frac{1}{2}$ and $\sin \theta = \frac{\sqrt{3}}{2}$.
Thus,$P = (2(-1/2), \sqrt{3}(\sqrt{3}/2)) = (-1, 3/2)$.
$PQ = \sqrt{(4 - (-1))^2 + (4 - 3/2)^2} = \sqrt{5^2 + (5/2)^2} = \sqrt{25 + 25/4} = \sqrt{125/4} = \frac{5\sqrt{5}}{2}$.

(D) The foci are at $(0, \pm c)$ where $c = 2$. Since the foci lie on the $y$-axis,the ellipse is vertical.
The length of the minor axis is $2a = 4$,so $a = 2$.
For an ellipse,$c^2 = b^2 - a^2$,where $b$ is the semi-major axis.
$2^2 = b^2 - 2^2$ $\Rightarrow 4 = b^2 - 4$ $\Rightarrow b^2 = 8$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{4} + \frac{y^2}{8} = 1$.
Testing option $(D)$: $\frac{(\sqrt{2})^2}{4} + \frac{2^2}{8} = \frac{2}{4} + \frac{4}{8} = \frac{1}{2} + \frac{1}{2} = 1$.
Thus,the point $(\sqrt{2}, 2)$ lies on the ellipse.

(B) The equation of the line is $3x + 4y = 12\sqrt{2}$,which can be rewritten as $y = -\frac{3}{4}x + 3\sqrt{2}$.
Comparing this with $y = mx + c$,we have $m = -\frac{3}{4}$ and $c = 3\sqrt{2}$.
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 $c^{2} = a^{2}m^{2} + b^{2}$.
Here,$b^{2} = 9$. Substituting the values: $(3\sqrt{2})^{2} = a^{2}(-\frac{3}{4})^{2} + 9$.
$18 = a^{2}(\frac{9}{16}) + 9$.
$9 = a^{2}(\frac{9}{16}) \Rightarrow a^{2} = 16$,so $a = 4$.
The ellipse is $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$. Since $a^{2} > b^{2}$,the eccentricity $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.
The distance between the foci is $2ae = 2 \times 4 \times \frac{\sqrt{7}}{4} = 2\sqrt{7}$.

(C) Given the distance between the foci is $2ae = 6$,so $ae = 3$ $(1)$.
The distance between the directrices is $\frac{2a}{e} = 12$,so $a = 6e$ $(2)$.
Substituting $(2)$ into $(1)$,we get $6e^2 = 3$,which implies $e^2 = \frac{1}{2}$,so $e = \frac{1}{\sqrt{2}}$.
Then $a = 6 \times \frac{1}{\sqrt{2}} = 3\sqrt{2}$.
Using the relation $b^2 = a^2(1 - e^2)$,we have $b^2 = (3\sqrt{2})^2(1 - \frac{1}{2}) = 18 \times \frac{1}{2} = 9$.
The length of the latus rectum is $\frac{2b^2}{a} = \frac{2(9)}{3\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.

(D) The equation of the ellipse is $2x^{2}+y^{2}=1$,which can be written as $\frac{x^{2}}{(1/\sqrt{2})^{2}} + \frac{y^{2}}{1^{2}} = 1$.
Let the point $P$ be $(\frac{1}{\sqrt{2}}\cos\theta, \sin\theta)$.
The equation of the normal at $P(\frac{1}{\sqrt{2}}\cos\theta, \sin\theta)$ is given by $\frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^{2}-b^{2}$,where $a=\frac{1}{\sqrt{2}}$ and $b=1$.
Substituting the values,we get $\frac{x}{\sqrt{2}\cos\theta} - \frac{y}{\sin\theta} = \frac{1}{2} - 1 = -\frac{1}{2}$.
This simplifies to $\frac{x}{(-\frac{1}{\sqrt{2}}\cos\theta)} + \frac{y}{(\frac{1}{2}\sin\theta)} = 1$.
Comparing this with the given intercepts $(-\frac{1}{3\sqrt{2}}, 0)$ and $(0, \beta)$,we have $-\frac{1}{\sqrt{2}}\cos\theta = -\frac{1}{3\sqrt{2}} \Rightarrow \cos\theta = \frac{1}{3}$.
Since $\sin^{2}\theta = 1 - \cos^{2}\theta = 1 - \frac{1}{9} = \frac{8}{9}$,we have $\sin\theta = \frac{2\sqrt{2}}{3}$ (as $P$ is in the first quadrant).
Thus,$\beta = \frac{1}{2}\sin\theta = \frac{1}{2} \times \frac{2\sqrt{2}}{3} = \frac{\sqrt{2}}{3}$.

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$.
Given the length of the minor axis is $2b = \frac{4}{\sqrt{3}}$,so $b = \frac{2}{\sqrt{3}}$ and $b^2 = \frac{4}{3}$.
The line $x + 6y = 8$ can be written as $y = -\frac{1}{6}x + \frac{4}{3}$.
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 $c^2 = a^2m^2 + b^2$.
Here $m = -\frac{1}{6}$ and $c = \frac{4}{3}$.
Substituting the values: $(\frac{4}{3})^2 = a^2(-\frac{1}{6})^2 + \frac{4}{3}$.
$\frac{16}{9} = \frac{a^2}{36} + \frac{4}{3}$.
$\frac{a^2}{36} = \frac{16}{9} - \frac{12}{9} = \frac{4}{9}$.
$a^2 = \frac{4 \times 36}{9} = 16$,so $a = 4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}} = \sqrt{\frac{11}{4 \times 3}} = \frac{1}{2} \sqrt{\frac{11}{3}}$.

(N/A) Given the equation of the ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.
Since the denominator of $\frac{x^{2}}{25}$ is larger than the denominator of $\frac{y^{2}}{9}$,the major axis is along the $x$-axis.
Comparing the given equation with $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we get $a^{2} = 25$ and $b^{2} = 9$,which implies $a = 5$ and $b = 3$.
We calculate $c = \sqrt{a^{2} - b^{2}} = \sqrt{25 - 9} = \sqrt{16} = 4$.
$1$. The coordinates of the foci are $(\pm c, 0)$,which are $(-4, 0)$ and $(4, 0)$.
$2$. The coordinates of the vertices are $(\pm a, 0)$,which are $(-5, 0)$ and $(5, 0)$.
$3$. The length of the major axis is $2a = 2 \times 5 = 10$ units.
$4$. The length of the minor axis is $2b = 2 \times 3 = 6$ units.
$5$. The eccentricity $e$ is $\frac{c}{a} = \frac{4}{5} = 0.8$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 9}{5} = \frac{18}{5} = 3.6$ units.

(N/A) The given equation of the ellipse is $9x^{2} + 4y^{2} = 36$. Dividing both sides by $36$,we get the standard form:
$\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
Since the denominator of $\frac{y^{2}}{9}$ is larger than the denominator of $\frac{x^{2}}{4}$,the major axis is along the $y$-axis.
Comparing this with the standard equation $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$,we have $b^{2} = 4$ and $a^{2} = 9$,so $b = 2$ and $a = 3$.
The value of $c$ is given by $c = \sqrt{a^{2} - b^{2}} = \sqrt{9 - 4} = \sqrt{5}$.
The eccentricity $e$ is $e = \frac{c}{a} = \frac{\sqrt{5}}{3}$.
The foci are $(0, \pm c) = (0, \pm \sqrt{5})$.
The vertices are $(0, \pm a) = (0, \pm 3)$.
The length of the major axis is $2a = 2(3) = 6$ units.
The length of the minor axis is $2b = 2(2) = 4$ units.

(A) Since the vertices are on the $x$-axis,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,where $a$ is the semi-major axis.
Given that the vertices are $(\pm 13, 0)$,we have $a = 13$.
The foci are $(\pm 5, 0)$,so $c = 5$.
Using the relation $c^{2} = a^{2} - b^{2}$,we get:
$5^{2} = 13^{2} - b^{2}$
$25 = 169 - b^{2}$
$b^{2} = 169 - 25 = 144$,which implies $b = 12$.
Substituting the values of $a^{2}$ and $b^{2}$ into the standard equation,we get $\frac{x^{2}}{169} + \frac{y^{2}}{144} = 1$.

(A) Since the foci are on the $y$-axis,the major axis is along the $y$-axis. The equation of the ellipse is of the form $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$.
Given that the length of the major axis is $2a = 20$,so $a = 10$.
The foci are $(0, \pm c) = (0, \pm 5)$,so $c = 5$.
Using the relation $c^{2} = a^{2} - b^{2}$,we have $5^{2} = 10^{2} - b^{2}$.
$25 = 100 - b^{2}$,which implies $b^{2} = 100 - 25 = 75$.
Substituting $a^{2} = 100$ and $b^{2} = 75$ into the standard equation,we get $\frac{x^{2}}{75} + \frac{y^{2}}{100} = 1$.

(A) The standard form of the ellipse with the major axis along the $x-$axis is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.
Since the points $(4, 3)$ and $(-1, 4)$ lie on the ellipse,we have:
$\frac{16}{a^{2}} + \frac{9}{b^{2}} = 1$ --- $(1)$
$\frac{1}{a^{2}} + \frac{16}{b^{2}} = 1$ --- $(2)$
Let $u = \frac{1}{a^{2}}$ and $v = \frac{1}{b^{2}}$. The equations become:
$16u + 9v = 1$ --- $(3)$
$u + 16v = 1$ --- $(4)$
Multiplying $(4)$ by $16$,we get $16u + 256v = 16$. Subtracting $(3)$ from this:
$247v = 15 \implies v = \frac{15}{247} \implies b^{2} = \frac{247}{15}$.
Substituting $v$ into $(4)$:
$u + 16(\frac{15}{247}) = 1 \implies u = 1 - \frac{240}{247} = \frac{7}{247} \implies a^{2} = \frac{247}{7}$.
Substituting $a^{2}$ and $b^{2}$ into the standard form:
$\frac{x^{2}}{247/7} + \frac{y^{2}}{247/15} = 1 \implies \frac{7x^{2}}{247} + \frac{15y^{2}}{247} = 1 \implies 7x^{2} + 15y^{2} = 247$.

The given equation is $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.
Here,the denominator of $\frac{x^{2}}{36}$ is greater than the denominator of $\frac{y^{2}}{16}$.
Therefore,the major axis is along the $x$-axis,while the minor axis is along the $y$-axis.
On comparing the given equation with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we obtain $a=6$ and $b=4$.
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{36-16}=\sqrt{20}=2\sqrt{5}$.
Therefore:
The coordinates of the foci are $(\pm 2\sqrt{5}, 0)$.
The coordinates of the vertices are $(\pm 6, 0)$.
Length of the major axis $= 2a = 12$.
Length of the minor axis $= 2b = 8$.
Eccentricity,$e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.
Length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 16}{6} = \frac{16}{3}$.

(N/A) The given equation is $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$,which can be written as $\frac{x^{2}}{2^{2}}+\frac{y^{2}}{5^{2}}=1$.
Since the denominator of $\frac{y^{2}}{25}$ is greater than the denominator of $\frac{x^{2}}{4}$,the major axis is along the $y$-axis.
Comparing this with the standard form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$,we get $a=5$ and $b=2$.
We calculate $c$ using the relation $c = \sqrt{a^{2}-b^{2}} = \sqrt{25-4} = \sqrt{21}$.
The coordinates of the foci are $(0, \sqrt{21})$ and $(0, -\sqrt{21})$.
The coordinates of the vertices are $(0, 5)$ and $(0, -5)$.
The length of the major axis is $2a = 2 \times 5 = 10$.
The length of the minor axis is $2b = 2 \times 2 = 4$.
The eccentricity is $e = \frac{c}{a} = \frac{\sqrt{21}}{5}$.
The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 4}{5} = \frac{8}{5}$.

(N/A) The given equation is $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$,which can be written as $\frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$.
Since the denominator of $\frac{x^{2}}{16}$ is greater than the denominator of $\frac{y^{2}}{9}$,the major axis is along the $x$-axis.
Comparing this with the standard form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we get $a = 4$ and $b = 3$.
We calculate $c = \sqrt{a^{2} - b^{2}} = \sqrt{16 - 9} = \sqrt{7}$.
$1$. The coordinates of the foci are $(\pm \sqrt{7}, 0)$.
$2$. The coordinates of the vertices are $(\pm 4, 0)$.
$3$. The length of the major axis is $2a = 2 \times 4 = 8$.
$4$. The length of the minor axis is $2b = 2 \times 3 = 6$.
$5$. The eccentricity $e = \frac{c}{a} = \frac{\sqrt{7}}{4}$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 9}{4} = \frac{9}{2} = 4.5$.

The given equation is $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$,which can be written as $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{10^{2}}=1$.
Since the denominator of $\frac{y^{2}}{100}$ is greater than the denominator of $\frac{x^{2}}{25}$,the major axis is along the $y$-axis.
Comparing this with the standard form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$,we get $a=10$ and $b=5$.
We calculate $c = \sqrt{a^{2}-b^{2}} = \sqrt{100-25} = \sqrt{75} = 5\sqrt{3}$.
$1$. The coordinates of the foci are $(0, \pm 5\sqrt{3})$.
$2$. The coordinates of the vertices are $(0, \pm 10)$.
$3$. The length of the major axis is $2a = 2(10) = 20$.
$4$. The length of the minor axis is $2b = 2(5) = 10$.
$5$. The eccentricity $e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2(25)}{10} = 5$.

The given equation is $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$,which can be written as $\frac{x^{2}}{7^{2}}+\frac{y^{2}}{6^{2}}=1$.
Since the denominator of $\frac{x^{2}}{49}$ is greater than the denominator of $\frac{y^{2}}{36}$,the major axis is along the $x$-axis.
Comparing this with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a=7$ and $b=6$.
We calculate $c = \sqrt{a^{2}-b^{2}} = \sqrt{49-36} = \sqrt{13}$.
$1$. The coordinates of the foci are $(\pm \sqrt{13}, 0)$.
$2$. The coordinates of the vertices are $(\pm 7, 0)$.
$3$. The length of the major axis is $2a = 2 \times 7 = 14$.
$4$. The length of the minor axis is $2b = 2 \times 6 = 12$.
$5$. The eccentricity $e = \frac{c}{a} = \frac{\sqrt{13}}{7}$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 36}{7} = \frac{72}{7}$.

The given equation is $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$,which can be written as $\frac{x^{2}}{10^{2}}+\frac{y^{2}}{20^{2}}=1$.
Since the denominator of the $y^{2}$ term is greater than the denominator of the $x^{2}$ term,the major axis is along the $y$-axis.
Comparing this with the standard form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$,we get $b=10$ and $a=20$.
We calculate $c$ using the relation $c = \sqrt{a^{2}-b^{2}} = \sqrt{400-100} = \sqrt{300} = 10\sqrt{3}$.
$1$. The coordinates of the foci are $(0, \pm 10\sqrt{3})$.
$2$. The coordinates of the vertices are $(0, \pm 20)$.
$3$. The length of the major axis is $2a = 2 \times 20 = 40$.
$4$. The length of the minor axis is $2b = 2 \times 10 = 20$.
$5$. The eccentricity $e = \frac{c}{a} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 100}{20} = 10$.

The given equation is $36 x^{2}+4 y^{2}=144$.
Dividing both sides by $144$,we get:
$\frac{36 x^{2}}{144} + \frac{4 y^{2}}{144} = 1$
$\frac{x^{2}}{4} + \frac{y^{2}}{36} = 1$
$\frac{x^{2}}{2^{2}} + \frac{y^{2}}{6^{2}} = 1$ ........ $(1)$
Here,the denominator of $\frac{y^{2}}{6^{2}}$ is greater than the denominator of $\frac{x^{2}}{2^{2}}$.
Therefore,the major axis is along the $y$-axis,while the minor axis is along the $x$-axis.
Comparing equation $(1)$ with $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$,we obtain $b = 2$ and $a = 6$.
$\therefore c = \sqrt{a^{2} - b^{2}} = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}$.
Therefore:
The coordinates of the foci are $(0, \pm 4\sqrt{2})$.
The coordinates of the vertices are $(0, \pm 6)$.
Length of the major axis $= 2a = 12$.
Length of the minor axis $= 2b = 4$.
Eccentricity,$e = \frac{c}{a} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$.
Length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 4}{6} = \frac{4}{3}$.

(N/A) The given equation is $16x^{2} + y^{2} = 16$.
Dividing both sides by $16$,we get:
$\frac{x^{2}}{1} + \frac{y^{2}}{16} = 1$
Or,$\frac{x^{2}}{1^{2}} + \frac{y^{2}}{4^{2}} = 1$ $(1)$
Comparing equation $(1)$ with the standard form $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$,we observe that $a^{2} = 16$ and $b^{2} = 1$,so $a = 4$ and $b = 1$.
Since $a > b$,the major axis is along the $y$-axis.
We calculate $c$ using $c = \sqrt{a^{2} - b^{2}} = \sqrt{16 - 1} = \sqrt{15}$.
$1$. Coordinates of the foci: $(0, \pm c) = (0, \pm \sqrt{15})$.
$2$. Coordinates of the vertices: $(0, \pm a) = (0, \pm 4)$.
$3$. Length of the major axis: $2a = 2(4) = 8$.
$4$. Length of the minor axis: $2b = 2(1) = 2$.
$5$. Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{15}}{4}$.
$6$. Length of the latus rectum: $\frac{2b^{2}}{a} = \frac{2(1)}{4} = \frac{1}{2}$.

(A) The given equation is $4x^{2} + 9y^{2} = 36$.
Dividing both sides by $36$,we get:
$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$
$\frac{x^{2}}{3^{2}} + \frac{y^{2}}{2^{2}} = 1$
Comparing this with the standard equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,where $a > b$,we have $a = 3$ and $b = 2$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
The foci are $(\pm ae, 0) = (\pm 3 \times \frac{\sqrt{5}}{3}, 0) = (\pm \sqrt{5}, 0)$.
The vertices are $(\pm a, 0) = (\pm 3, 0)$.
The length of the major axis is $2a = 2 \times 3 = 6$.
The length of the minor axis is $2b = 2 \times 2 = 4$.
The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 4}{3} = \frac{8}{3}$.

(A) The vertices are $(\pm 5, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Here,$a = 5$ and the distance of the foci from the center is $c = 4$.
Using the relation $a^2 = b^2 + c^2$,we have $5^2 = b^2 + 4^2$.
$25 = b^2 + 16 \implies b^2 = 9$.
Substituting $a^2 = 25$ and $b^2 = 9$ into the standard equation,we get $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

(N/A) The equation of the given line is $x \cos \alpha + y \sin \alpha = p$.
Rewriting this in the slope-intercept form $y = mx + c$:
$y \sin \alpha = -x \cos \alpha + p$
$y = -x \cot \alpha + \frac{p}{\sin \alpha}$.
Here,the slope $m = -\cot \alpha$ and the intercept $c = \frac{p}{\sin \alpha}$.
We know that the line $y = mx + c$ touches the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ if and only if $c^{2} = a^{2}m^{2} + b^{2}$.
Substituting the values of $m$ and $c$:
$\left(\frac{p}{\sin \alpha}\right)^{2} = a^{2}(-\cot \alpha)^{2} + b^{2}$.
$\frac{p^{2}}{\sin^{2} \alpha} = a^{2} \frac{\cos^{2} \alpha}{\sin^{2} \alpha} + b^{2}$.
Multiplying both sides by $\sin^{2} \alpha$:
$p^{2} = a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha$.
Hence,it is proved that $a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha = p^{2}$.

(A) The standard equation of an ellipse is given by $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.
If we set $a = b = r$,the equation becomes $\frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} = 1$,which simplifies to $x^{2} + y^{2} = r^{2}$.
This is the standard equation of a circle with radius $r$ centered at the origin.
Since a circle can be derived from an ellipse by setting the semi-major and semi-minor axes equal,a circle is indeed a particular case of an ellipse.
Therefore,statement $r$ is true.

The equation of the given line is $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$.
Multiplying by $ab$,we get $bx \cos \theta+ay \sin \theta-ab=0$.....$(1)$
Let $p_{1}$ be the length of the perpendicular from $(\sqrt{a^{2}-b^{2}}, 0)$ to line $(1)$:
$p_{1}=\frac{|b \cos \theta(\sqrt{a^{2}-b^{2}})+a \sin \theta(0)-ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}-ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}$.....$(2)$
Let $p_{2}$ be the length of the perpendicular from $(-\sqrt{a^{2}-b^{2}}, 0)$ to line $(1)$:
$p_{2}=\frac{|b \cos \theta(-\sqrt{a^{2}-b^{2}})+a \sin \theta(0)-ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|-b \cos \theta \sqrt{a^{2}-b^{2}}-ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}+ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}$.....$(3)$
Multiplying $p_{1}$ and $p_{2}$:
$p_{1} p_{2}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}-ab| \cdot |b \cos \theta \sqrt{a^{2}-b^{2}}+ab|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$
$p_{1} p_{2}=\frac{|(b \cos \theta \sqrt{a^{2}-b^{2}})^{2}-(ab)^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} = \frac{|b^{2} \cos ^{2} \theta(a^{2}-b^{2})-a^{2}b^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$
$p_{1} p_{2}=\frac{|a^{2}b^{2} \cos ^{2} \theta-b^{4} \cos ^{2} \theta-a^{2}b^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} = \frac{b^{2}|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$
Using $a^{2} = a^{2}(\sin^{2}\theta + \cos^{2}\theta)$,the numerator becomes $b^{2}|a^{2} \cos^{2}\theta - b^{2} \cos^{2}\theta - a^{2} \sin^{2}\theta - a^{2} \cos^{2}\theta| = b^{2}|-(b^{2} \cos^{2}\theta + a^{2} \sin^{2}\theta)|$.
Thus,$p_{1} p_{2} = \frac{b^{2}(b^{2} \cos^{2}\theta + a^{2} \sin^{2}\theta)}{b^{2} \cos^{2}\theta + a^{2} \sin^{2}\theta} = b^{2}$.

(A) The vertices are given as $(0, \pm 13)$,which lie on the $y$-axis.
Thus,the major axis is along the $y$-axis,and the equation of the ellipse is of the form $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$.
Here,$a = 13$ and the foci are $(0, \pm c)$ where $c = 5$.
Using the relation $a^{2} = b^{2} + c^{2}$,we have $13^{2} = b^{2} + 5^{2}$.
$169 = b^{2} + 25$.
$b^{2} = 169 - 25 = 144$.
Substituting the values of $a^{2}$ and $b^{2}$ into the standard equation,we get $\frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$.

(A) The vertices are $(\pm 6, 0)$,which implies the major axis is along the $x$-axis and $a = 6$.
The foci are $(\pm 4, 0)$,which implies $c = 4$.
For an ellipse,the relationship between $a, b,$ and $c$ is $a^2 = b^2 + c^2$.
Substituting the values: $6^2 = b^2 + 4^2$.
$36 = b^2 + 16$.
$b^2 = 36 - 16 = 20$.
The standard equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting $a^2 = 36$ and $b^2 = 20$,we get $\frac{x^2}{36} + \frac{y^2}{20} = 1$.

(A) The ends of the major axis are $(\pm 3, 0)$,which implies the major axis lies along the $x$-axis and $a = 3$.
The ends of the minor axis are $(0, \pm 2)$,which implies $b = 2$.
The standard equation of an ellipse with the major axis along the $x$-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values $a = 3$ and $b = 2$,we get $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$.
Therefore,the equation is $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

(A) The ends of the major axis are $(0, \pm \sqrt{5})$,which implies the major axis lies along the $y$-axis.
The ends of the minor axis are $(\pm 1, 0)$,which implies the minor axis lies along the $x$-axis.
The standard form of the equation for an ellipse with the major axis along the $y$-axis is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,where $a$ is the semi-major axis and $b$ is the semi-minor axis.
From the given coordinates,$a = \sqrt{5}$ and $b = 1$.
Substituting these values into the standard equation:
$\frac{x^2}{1^2} + \frac{y^2}{(\sqrt{5})^2} = 1$
This simplifies to:
$x^2 + \frac{y^2}{5} = 1$

(N/A) The length of the major axis is $2a = 26$,which gives $a = 13$.
The foci are given as $(\pm 5, 0)$,which implies $c = 5$ and the major axis lies along the $x$-axis.
The standard equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Using the relation $a^2 = b^2 + c^2$,we have $13^2 = b^2 + 5^2$.
$169 = b^2 + 25 \Rightarrow b^2 = 144$.
Substituting the values of $a^2$ and $b^2$ into the standard equation,we get $\frac{x^2}{169} + \frac{y^2}{144} = 1$.

(A) Length of minor axis $= 16$; foci $= (0, \pm 6)$.
Since the foci are on the $y$-axis,the major axis is along the $y$-axis.
Therefore,the equation of the ellipse will be of the form $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$,where $a$ is the semi-major axis.
Accordingly,$2b = 16 \Rightarrow b = 8$ and $c = 6$.
It is known that $a^{2} = b^{2} + c^{2}$.
$\therefore a^{2} = 8^{2} + 6^{2} = 64 + 36 = 100$.
$\Rightarrow a = \sqrt{100} = 10$.
Thus,the equation of the ellipse is $\frac{x^{2}}{8^{2}} + \frac{y^{2}}{10^{2}} = 1$ or $\frac{x^{2}}{64} + \frac{y^{2}}{100} = 1$.

(A) Given: Foci $(\pm 3, 0)$ and $a = 4$.
Since the foci are on the $x$-axis,the major axis is along the $x$-axis.
Therefore,the equation of the ellipse is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a$ is the semi-major axis.
From the foci $(\pm c, 0)$,we have $c = 3$. Given $a = 4$.
We know the relation $a^2 = b^2 + c^2$.
Substituting the values: $4^2 = b^2 + 3^2$.
$16 = b^2 + 9$.
$b^2 = 16 - 9 = 7$.
Thus,the equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{7} = 1$.

(N/A) Given: $b=3, c=4,$ centre at the origin; foci on the $x$-axis.
Since the foci are on the $x$-axis,the major axis is along the $x$-axis.
Therefore,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ where $a$ is the semi-major axis.
We know the relation $a^{2}=b^{2}+c^{2}$.
Substituting the values: $a^{2}=3^{2}+4^{2}=9+16=25$.
Thus,$a^{2}=25$ and $b^{2}=3^{2}=9$.
The equation of the ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$.

(A) Since the centre is at $(0, 0)$ and the major axis is on the $y$-axis,the equation of the ellipse is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$.
Given that the ellipse passes through $(3, 2)$ and $(1, 6)$,we have:
$\frac{3^2}{a^2} + \frac{2^2}{b^2} = 1 \implies \frac{9}{a^2} + \frac{4}{b^2} = 1$ --- $(1)$
$\frac{1^2}{a^2} + \frac{6^2}{b^2} = 1 \implies \frac{1}{a^2} + \frac{36}{b^2} = 1$ --- $(2)$
Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$. The equations become:
$9u + 4v = 1$ --- $(3)$
$u + 36v = 1$ --- $(4)$
Multiplying $(4)$ by $9$: $9u + 324v = 9$ --- $(5)$
Subtracting $(3)$ from $(5)$: $320v = 8 \implies v = \frac{8}{320} = \frac{1}{40}$.
Thus,$b^2 = 40$.
Substituting $v = \frac{1}{40}$ into $(4)$: $u + 36(\frac{1}{40}) = 1 \implies u + \frac{9}{10} = 1 \implies u = \frac{1}{10}$.
Thus,$a^2 = 10$.
The equation of the ellipse is $\frac{x^2}{10} + \frac{y^2}{40} = 1$.

(A) Since the major axis is on the $x-$ axis,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $(1)$.
Given that the ellipse passes through $(4, 3)$ and $(6, 2)$,we have:
$\frac{16}{a^{2}} + \frac{9}{b^{2}} = 1$ $(2)$
$\frac{36}{a^{2}} + \frac{4}{b^{2}} = 1$ $(3)$
Let $u = \frac{1}{a^{2}}$ and $v = \frac{1}{b^{2}}$. The equations become:
$16u + 9v = 1$
$36u + 4v = 1$
Multiplying the first by $4$ and the second by $9$:
$64u + 36v = 4$
$324u + 36v = 9$
Subtracting the equations: $260u = 5 \implies u = \frac{5}{260} = \frac{1}{52}$.
Thus,$a^{2} = 52$.
Substituting $u$ into $36u + 4v = 1$:
$36(\frac{1}{52}) + 4v = 1 \implies \frac{9}{13} + 4v = 1 \implies 4v = 1 - \frac{9}{13} = \frac{4}{13} \implies v = \frac{1}{13}$.
Thus,$b^{2} = 13$.
The equation is $\frac{x^{2}}{52} + \frac{y^{2}}{13} = 1$.

(N/A) Let $AB$ be the rod making an angle $\theta$ with the $x-$axis as shown in the figure. Let $P(x, y)$ be a point on the rod such that $AP = 6 \ cm$.
Since $AB = 15 \ cm$,we have $PB = AB - AP = 15 - 6 = 9 \ cm$.
From $P$,draw $PQ$ and $PR$ perpendiculars on the $y-$axis and $x-$axis,respectively.
In $\Delta PBQ$,$\cos \theta = \frac{PQ}{PB} = \frac{x}{9}$,so $x = 9 \cos \theta$.
In $\Delta PRA$,$\sin \theta = \frac{PR}{AP} = \frac{y}{6}$,so $y = 6 \sin \theta$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we substitute $\cos \theta = \frac{x}{9}$ and $\sin \theta = \frac{y}{6}$:
$(\frac{x}{9})^2 + (\frac{y}{6})^2 = 1$
$\frac{x^2}{81} + \frac{y^2}{36} = 1$.
This is the equation of an ellipse. Thus,the locus of $P$ is an ellipse.

(A) The arch is a semi-ellipse with a total width of $8 \, m$ and a maximum height of $2 \, m$ at the centre.
This implies the length of the semi-major axis $a = 4 \, m$ and the semi-minor axis $b = 2 \, m$.
The equation of the ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values,we get $\frac{x^2}{16} + \frac{y^2}{4} = 1$.
We need the height at a point $1.5 \, m$ from one end. Since the total width is $8 \, m$,the end is at $x = 4$. $A$ point $1.5 \, m$ from the end corresponds to $x = 4 - 1.5 = 2.5$.
Substituting $x = 2.5$ into the equation:
$\frac{(2.5)^2}{16} + \frac{y^2}{4} = 1$
$\frac{6.25}{16} + \frac{y^2}{4} = 1$
$\frac{y^2}{4} = 1 - \frac{6.25}{16} = \frac{16 - 6.25}{16} = \frac{9.75}{16}$
$y^2 = 4 \times \frac{9.75}{16} = \frac{9.75}{4} = 2.4375$
$y = \sqrt{2.4375} \approx 1.56 \, m$.

(A) Let $A$ and $B$ be the positions of the two flag posts and $P(x, y)$ be the position of the man.
According to the problem,$PA + PB = 10$.
We know that if a point moves in a plane such that the sum of its distances from two fixed points is constant,the path is an ellipse,and this constant value is equal to the length of the major axis $(2a)$.
Therefore,the path is an ellipse with major axis $2a = 10 \, m$,so $a = 5$.
The distance between the foci $(2c)$ is $8 \, m$,so $c = 4$.
Using the relation $c^{2} = a^{2} - b^{2}$,we have $4^{2} = 5^{2} - b^{2}$.
$16 = 25 - b^{2} \Rightarrow b^{2} = 9$.
Thus,the equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,which is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.

(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Given the directrix $x = \frac{a}{e} = 4$ and eccentricity $e = \frac{1}{2}$,we find $a = 4 \times \frac{1}{2} = 2$.
Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
Thus,the ellipse equation is $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Since $P(1, \beta)$ lies on the ellipse,$\frac{1^2}{4} + \frac{\beta^2}{3} = 1 \Rightarrow \frac{\beta^2}{3} = \frac{3}{4} \Rightarrow \beta^2 = \frac{9}{4} \Rightarrow \beta = \frac{3}{2}$ (as $\beta > 0$).
The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.
Substituting $a^2=4, b^2=3, x_1=1, y_1=\frac{3}{2}$,we get $\frac{4x}{1} - \frac{3y}{3/2} = 4 - 3$.
$4x - 2y = 1$.

(A) Given the ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a > b$.
The length of the latus rectum is given by $\frac{2b^{2}}{a} = 10$,which implies $b^{2} = 5a$ ... $(i)$.
Now,consider the function $\phi(t) = \frac{5}{12} + t - t^{2}$.
To find the maximum value,we complete the square: $\phi(t) = -\left(t^{2} - t + \frac{1}{4}\right) + \frac{1}{4} + \frac{5}{12} = -\left(t - \frac{1}{2}\right)^{2} + \frac{8}{12} = -\left(t - \frac{1}{2}\right)^{2} + \frac{2}{3}$.
The maximum value is $\phi(t)_{\text{max}} = \frac{2}{3}$,so $e = \frac{2}{3}$.
Since $e^{2} = 1 - \frac{b^{2}}{a^{2}}$,we have $\frac{4}{9} = 1 - \frac{b^{2}}{a^{2}}$,which implies $\frac{b^{2}}{a^{2}} = \frac{5}{9}$,so $b^{2} = \frac{5}{9}a^{2}$ ... $(ii)$.
Equating $(i)$ and $(ii)$,$5a = \frac{5}{9}a^{2} \Rightarrow a = \frac{a^{2}}{9} \Rightarrow a = 9$.
Then $a^{2} = 81$ and $b^{2} = 5(9) = 45$.
Therefore,$a^{2} + b^{2} = 81 + 45 = 126$.

(B) The given equation of the ellipse is $4x^{2} + 5y^{2} = 20$,which can be written as $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$.
Let the point $P$ on the ellipse be $(\sqrt{5} \cos \theta, 2 \sin \theta)$.
The square of the distance $PQ$ is given by $PQ^{2} = (\sqrt{5} \cos \theta - 0)^{2} + (2 \sin \theta - (-4))^{2}$.
$PQ^{2} = 5 \cos^{2} \theta + (2 \sin \theta + 4)^{2}$.
$PQ^{2} = 5(1 - \sin^{2} \theta) + 4 \sin^{2} \theta + 16 \sin \theta + 16$.
$PQ^{2} = 5 - 5 \sin^{2} \theta + 4 \sin^{2} \theta + 16 \sin \theta + 16$.
$PQ^{2} = -\sin^{2} \theta + 16 \sin \theta + 21$.
To find the maximum value,we complete the square: $PQ^{2} = -(\sin^{2} \theta - 16 \sin \theta + 64) + 64 + 21$.
$PQ^{2} = 85 - (\sin \theta - 8)^{2}$.
Since $-1 \leq \sin \theta \leq 1$,the expression $85 - (\sin \theta - 8)^{2}$ is maximized when $\sin \theta$ is as close to $8$ as possible,which occurs at $\sin \theta = 1$.
Substituting $\sin \theta = 1$,we get $PQ^{2} = 85 - (1 - 8)^{2} = 85 - (-7)^{2} = 85 - 49 = 36$.

(A) The given equation of the conic is $9x^{2} + 16y^{2} = 144$.
Dividing by $144$,we get $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.
This is the standard equation of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,where $a^{2} = 16$ and $b^{2} = 9$.
Thus,$a = 4$ and $b = 3$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.
The coordinates of the foci are $(\pm ae, 0) = (\pm 4 \times \frac{\sqrt{7}}{4}, 0) = (\pm \sqrt{7}, 0)$.
Since $A = (\sqrt{7}, 0)$ and $B = (-\sqrt{7}, 0)$ are the foci of the ellipse,by the definition of an ellipse,the sum of the distances from any point $P$ on the ellipse to the two foci is constant and equal to the length of the major axis,which is $2a$.
Therefore,$PA + PB = 2a = 2 \times 4 = 8$.

(D) The equation of the normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at point $(x_{1}, y_{1})$ is given by $\frac{a^{2}x}{x_{1}}-\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}=a^{2}e^{2}$.
The coordinates of an end of the latus rectum are $(ae, \frac{b^{2}}{a})$.
Substituting these into the normal equation:
$\frac{a^{2}x}{ae}-\frac{b^{2}y}{b^{2}/a} = a^{2}e^{2}$
$\frac{ax}{e}-ay = a^{2}e^{2} \Rightarrow \frac{x}{e}-y = ae^{2}$.
Since this normal passes through an extremity of the minor axis $(0, b)$ or $(0, -b)$,we substitute $(0, b)$:
$0 - b = ae^{2} \Rightarrow b = -ae^{2}$.
Since $b$ and $a$ are lengths,we consider the magnitude $b = ae^{2}$,so $b^{2} = a^{2}e^{4}$.
Using $b^{2} = a^{2}(1-e^{2})$:
$a^{2}(1-e^{2}) = a^{2}e^{4}$
$1-e^{2} = e^{4} \Rightarrow e^{4}+e^{2}-1=0$.

(A) The locus of the foot of the perpendicular drawn from the foci to any tangent of an ellipse is its auxiliary circle.
For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,the auxiliary circle is $x^{2}+y^{2}=a^{2}$.
Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$,we have $a^{2}=4$.
Thus,the locus is $x^{2}+y^{2}=4$.
We check the given points to see which satisfies $x^{2}+y^{2}=4$:
$A: (-1)^{2}+(\sqrt{3})^{2} = 1+3=4$ (Satisfies)
$B: (-1)^{2}+(\sqrt{2})^{2} = 1+2=3 \neq 4$
$C: (-2)^{2}+(\sqrt{3})^{2} = 4+3=7 \neq 4$
$D: (1)^{2}+(2)^{2} = 1+4=5 \neq 4$
Therefore,the point $(-1, \sqrt{3})$ lies on the locus.

(A) Given the curve $y^{2}=3x^{2}$ and the circle $x^{2}+y^{2}=4b$.
Substituting $y^{2}=3x^{2}$ into the circle equation: $x^{2}+3x^{2}=4b \implies 4x^{2}=4b \implies x^{2}=b$.
Then $y^{2}=3b$.
Now,substitute $x^{2}=b$ and $y^{2}=3b$ into the ellipse equation $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$:
$\frac{b}{16}+\frac{3b}{b^{2}}=1$
$\frac{b}{16}+\frac{3}{b}=1$
Multiply by $16b$: $b^{2}+48=16b$
$b^{2}-16b+48=0$
$(b-12)(b-4)=0$
Since $b > 4$,we have $b=12$.

(C) Given circles are $S_{1}: x^{2}+y^{2}=3^{2}$ with center $A(0,0)$ and radius $r_{1}=3$,and $S_{2}: (x-2)^{2}+y^{2}=1^{2}$ with center $B(2,0)$ and radius $r_{2}=1$.
Let the variable circle $S$ have center $P(x,y)$ and radius $r$.
Since $S$ touches $S_{1}$ internally,the distance $PA = r_{1} - r = 3 - r$.
Since $S$ touches $S_{2}$ externally,the distance $PB = r_{2} + r = 1 + r$.
Adding these two equations,we get $PA + PB = (3 - r) + (1 + r) = 4$.
Since $PA + PB = 4$ and the distance $AB = 2$,the locus of $P$ is an ellipse with foci at $A(0,0)$ and $B(2,0)$ and major axis length $2a = 4$,so $a = 2$.
The center of the ellipse is the midpoint of $AB$,which is $(1,0)$.
The distance between foci is $2ae = AB = 2$,so $2(2)e = 2$,which gives $e = \frac{1}{2}$.
Then $b^{2} = a^{2}(1 - e^{2}) = 4(1 - \frac{1}{4}) = 3$.
The equation of the ellipse is $\frac{(x-1)^{2}}{4} + \frac{y^{2}}{3} = 1$.
Checking the options,for $x=2$,$\frac{(2-1)^{2}}{4} + \frac{y^{2}}{3} = 1$ $\Rightarrow \frac{1}{4} + \frac{y^{2}}{3} = 1$ $\Rightarrow \frac{y^{2}}{3} = \frac{3}{4}$ $\Rightarrow y^{2} = \frac{9}{4}$ $\Rightarrow y = \pm \frac{3}{2}$.
Thus,the locus passes through $\left(2, \pm \frac{3}{2}\right)$.

(C) The equation of the tangent to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at the point $(3 \sqrt{3} \cos \theta, \sin \theta)$ is given by $\frac{x \cos \theta}{3 \sqrt{3}}+\frac{y \sin \theta}{1}=1$.
The $x$-intercept is $OA = 3 \sqrt{3} \sec \theta$ and the $y$-intercept is $OB = \operatorname{cosec} \theta$.
Let the sum of the intercepts be $f(\theta) = 3 \sqrt{3} \sec \theta + \operatorname{cosec} \theta$.
To find the minimum,we differentiate $f(\theta)$ with respect to $\theta$:
$f^{\prime}(\theta) = 3 \sqrt{3} \sec \theta \tan \theta - \operatorname{cosec} \theta \cot \theta$.
Setting $f^{\prime}(\theta) = 0$:
$3 \sqrt{3} \frac{\sin \theta}{\cos^{2} \theta} = \frac{\cos \theta}{\sin^{2} \theta}$
$\tan^{3} \theta = \frac{1}{3 \sqrt{3}} = \left(\frac{1}{\sqrt{3}}\right)^{3}$
$\tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}$.
Since $f^{\prime}(\theta) < 0$ for $\theta < \frac{\pi}{6}$ and $f^{\prime}(\theta) > 0$ for $\theta > \frac{\pi}{6}$,the function $f(\theta)$ attains its minimum value at $\theta = \frac{\pi}{6}$.

(B) Let the point of intersection be $(x_1, y_1)$.
For the first curve $\frac{x^2}{a} + \frac{y^2}{b} = 1$,the slope $m_1$ at $(x_1, y_1)$ is given by differentiating: $\frac{2x_1}{a} + \frac{2y_1}{b} \frac{dy}{dx} = 0 \Rightarrow m_1 = -\frac{bx_1}{ay_1}$.
For the second curve $\frac{x^2}{c} + \frac{y^2}{d} = 1$,the slope $m_2$ at $(x_1, y_1)$ is $m_2 = -\frac{dx_1}{cy_1}$.
Since the curves intersect at $90^{\circ}$,$m_1 m_2 = -1$.
$(-\frac{bx_1}{ay_1})(-\frac{dx_1}{cy_1}) = -1 \Rightarrow \frac{bd x_1^2}{ac y_1^2} = -1$,which implies $bd x_1^2 = -ac y_1^2$.
Subtracting the two curve equations: $x_1^2(\frac{1}{a} - \frac{1}{c}) + y_1^2(\frac{1}{b} - \frac{1}{d}) = 0$.
$x_1^2(\frac{c-a}{ac}) + y_1^2(\frac{d-b}{bd}) = 0$.
Substituting $x_1^2 = -\frac{ac}{bd} y_1^2$,we get $(-\frac{ac}{bd}) y_1^2(\frac{c-a}{ac}) + y_1^2(\frac{d-b}{bd}) = 0$.
$-\frac{c-a}{bd} + \frac{d-b}{bd} = 0$ $\Rightarrow a-c+d-b = 0$ $\Rightarrow a-b = c-d$.