Mix Examples-Conic Sections Questions in English

(B) The parametric coordinates of a point on the rectangular hyperbola $xy = c^2$ are $(ct, c/t)$.
The slope of the tangent at point $t$ is given by $\frac{dy}{dx} = -\frac{1}{t^2}$.
Therefore,the slope of the normal at point $t$ is $m_N = t^2$.
The equation of the normal at point $t$ is $y - \frac{c}{t} = t^2(x - ct)$,which simplifies to $y - \frac{c}{t} = t^2x - c t^3$,or $y = t^2x - c t^3 + \frac{c}{t}$.
Since this normal meets the curve again at $t_1$,the point $(ct_1, c/t_1)$ must satisfy the equation of the normal:
$\frac{c}{t_1} = t^2(ct_1) - ct^3 + \frac{c}{t}$
Dividing by $c$ (assuming $c \neq 0$):
$\frac{1}{t_1} = t^2 t_1 - t^3 + \frac{1}{t}$
$\frac{1}{t_1} - \frac{1}{t} = t^2 t_1 - t^3$
$\frac{t - t_1}{t t_1} = -t^2(t - t_1)$
Since $t \neq t_1$ for the normal to meet the curve at a different point,we can divide by $(t - t_1)$:
$\frac{1}{t t_1} = -t^2$
$1 = -t^3 t_1$
$t^3 t_1 = -1$.

(B) The given hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$.
Here,$a^2 = 9$ and $b^2 = 16$,so $a = 3$ and $b = 4$.
The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{16}{9} = \frac{25}{9}$,which gives $e = \frac{5}{3}$.
The coordinates of the foci are $(\pm ae, 0) = (\pm 3 \times \frac{5}{3}, 0) = (\pm 5, 0)$.
Let the centre of the circle be the focus $(5, 0)$.
For the circle to be tangent to the hyperbola without any part of it lying outside,it must touch the vertex of the hyperbola.
The vertex of the hyperbola is $(a, 0) = (3, 0)$.
The radius $r$ of the circle is the distance between the focus $(5, 0)$ and the vertex $(3, 0)$,which is $r = |5 - 3| = 2$.
Thus,the radius of the circle is $2$.

(D) Let the coordinates of $A$ be $(x, l)$. Since $AB$ is a double ordinate,the coordinates of $B$ are $(x, -l)$.
Given that $\Delta AOB$ is an equilateral triangle,the side length $OA = OB = AB = 2l$.
In $\Delta OMA$ (where $M$ is the midpoint of $AB$ on the $x$-axis),$OM = x$ and $AM = l$.
Since $\angle AOM = 30^\circ$,we have $\tan(30^\circ) = \frac{l}{x} = \frac{1}{\sqrt{3}}$,so $x = l\sqrt{3}$.
Also,$OA^2 = x^2 + l^2 = (l\sqrt{3})^2 + l^2 = 3l^2 + l^2 = 4l^2$,which is consistent with $OA = 2l$.
Since $A(x, l)$ lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $\frac{3l^2}{a^2} - \frac{l^2}{b^2} = 1$.
$l^2 \left( \frac{3}{a^2} - \frac{1}{b^2} \right) = 1 \Rightarrow l^2 \left( \frac{3b^2 - a^2}{a^2b^2} \right) = 1$.
Since $l^2 > 0$,we must have $3b^2 - a^2 > 0$,which implies $3b^2 > a^2$,or $\frac{b^2}{a^2} > \frac{1}{3}$.
Using the relation $e^2 = 1 + \frac{b^2}{a^2}$,we get $e^2 > 1 + \frac{1}{3} = \frac{4}{3}$.
Therefore,$e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

(C) Let the coordinates of $R$ be $(at^2, 2at)$. Since $QR$ is a focal chord,the coordinates of $Q$ are $(\frac{a}{t^2}, -\frac{2a}{t})$.
The difference of the ordinates is $y_R - y_Q = 2at - (-\frac{2a}{t}) = 2a(t + \frac{1}{t})$.
The modulus of this difference is $d = 2a|t + \frac{1}{t}|$.
The area $A$ of triangle $PQR$ with vertices $P(0,0)$,$R(at^2, 2at)$,and $Q(\frac{a}{t^2}, -\frac{2a}{t})$ is given by:
$A = \frac{1}{2} |x_P(y_R - y_Q) + x_R(y_Q - y_P) + x_Q(y_P - y_R)|$
$A = \frac{1}{2} |0 + at^2(-\frac{2a}{t} - 0) + \frac{a}{t^2}(0 - 2at)|$
$A = \frac{1}{2} |-2a^2t - \frac{2a^2}{t}| = a^2|t + \frac{1}{t}|$.
From this,$|t + \frac{1}{t}| = \frac{A}{a^2}$.
Substituting this into the expression for $d$:
$d = 2a(\frac{A}{a^2}) = \frac{2A}{a}$.

(C) The equation of the circle is $x^2 + y^2 - 24y + 128 = 0$. Completing the square for $y$,we get $x^2 + (y - 12)^2 = 16$. The center of the circle is $C(0, 12)$ and its radius is $r = 4$.
Let a point on the parabola $y^2 = 4x$ be $P(t^2, 2t)$.
The distance between the center $C(0, 12)$ and point $P(t^2, 2t)$ is minimized when the normal to the parabola at $P$ passes through the center $C$.
The slope of the tangent at $P(t^2, 2t)$ is $\frac{dy}{dx} = \frac{2}{y} = \frac{2}{2t} = \frac{1}{t}$.
Thus,the slope of the normal at $P$ is $-t$.
The slope of the line segment $CP$ is $\frac{2t - 12}{t^2 - 0} = \frac{2t - 12}{t^2}$.
Equating the slopes: $-t = \frac{2t - 12}{t^2}$ $\Rightarrow -t^3 = 2t - 12$ $\Rightarrow t^3 + 2t - 12 = 0$.
Testing integer roots,for $t = 2$: $2^3 + 2(2) - 12 = 8 + 4 - 12 = 0$. So,$t = 2$ is a solution.
For $t = 2$,the point $P$ is $(t^2, 2t) = (2^2, 2(2)) = (4, 4)$.
Thus,the point on the parabola closest to the circle is $(4, 4)$.

(D) The vertex of the parabola $y^2 = 4ax$ is $(0,0)$. The length of the latus rectum is $4a$.
The diameter of the circle is $\frac{3}{4} \times 4a = 3a$. Thus,the radius $r = \frac{3a}{2}$.
The equation of the circle is $x^2 + y^2 = \left(\frac{3a}{2}\right)^2 = \frac{9a^2}{4}$,or $4(x^2 + y^2) = 9a^2$.
Substituting $y^2 = 4ax$ into the circle equation: $4(x^2 + 4ax) = 9a^2$,which simplifies to $4x^2 + 16ax - 9a^2 = 0$.
Solving for $x$: $x = \frac{-16a \pm \sqrt{(16a)^2 - 4(4)(-9a^2)}}{2(4)} = \frac{-16a \pm \sqrt{256a^2 + 144a^2}}{8} = \frac{-16a \pm 20a}{8}$.
Since $x > 0$,we take $x = \frac{4a}{8} = \frac{a}{2}$.
For $x = \frac{a}{2}$,$y^2 = 4a(\frac{a}{2}) = 2a^2$,so $y = \pm \sqrt{2}a$.
The points are $P(\frac{a}{2}, \sqrt{2}a)$ and $Q(\frac{a}{2}, -\sqrt{2}a)$. The length of the chord $PQ = 2\sqrt{2}a$.
The latus rectum $L_1L_2$ is at $x = a$,with $y^2 = 4a(a) = 4a^2$,so $L_1(a, 2a)$ and $L_2(a, -2a)$. The length $L_1L_2 = 4a$.
The area of the trapezium $PL_1L_2Q$ is $\frac{1}{2} \times (PQ + L_1L_2) \times \text{height}$.
The height is the distance between $x = a$ and $x = \frac{a}{2}$,which is $a - \frac{a}{2} = \frac{a}{2}$.
Area $= \frac{1}{2} \times (2\sqrt{2}a + 4a) \times \frac{a}{2} = \frac{1}{4} \times 2a(\sqrt{2} + 2) = \left(\frac{2+\sqrt{2}}{2}\right)a^2$.

(A) The given equations are $4x^2 + 9y^2 = 36$ (Ellipse) and $4x^2 - y^2 = 4$ (Hyperbola).
Dividing the ellipse equation by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Dividing the hyperbola equation by $4$,we get $\frac{x^2}{1} - \frac{y^2}{4} = 1$.
Let the points of intersection be $(x_1, y_1)$.
Adding the two equations: $(4x^2 + 9y^2) + k(4x^2 - y^2) = 36 + 4k$ represents a family of curves passing through the intersection points.
For this to be a circle,the coefficient of $x^2$ must equal the coefficient of $y^2$ and the $xy$ term must be zero.
$4 + 4k = 9 - k$ $\Rightarrow 5k = 5$ $\Rightarrow k = 1$.
Substituting $k = 1$ into the equation: $(4x^2 + 9y^2) + 1(4x^2 - y^2) = 36 + 4(1)$.
$8x^2 + 8y^2 = 40 \Rightarrow x^2 + y^2 = 5$.

(A) Let $(x_1, y_1)$ be the point of intersection.
Then $y_1^2 = 4ax_1$ and $x_1y_1 = c^2$.
For the parabola $y^2 = 4ax$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y}$.
Thus,$\tan \phi = \frac{2a}{y_1}$.
For the hyperbola $xy = c^2$,differentiating with respect to $x$ gives $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
Thus,$\tan \theta = -\frac{y_1}{x_1}$.
Now,$\frac{\tan \theta}{\tan \phi} = \frac{-y_1/x_1}{2a/y_1} = -\frac{y_1^2}{2ax_1}$.
Substituting $y_1^2 = 4ax_1$,we get $\frac{\tan \theta}{\tan \phi} = -\frac{4ax_1}{2ax_1} = -2$.
Therefore,$\tan \theta = -2 \tan \phi$,which implies $\theta = \tan^{-1}(-2 \tan \phi)$.

(C) The coordinates of $P$ are $(at^2, 2at)$. The tangent at $P$ is $ty = x + at^2$,so its slope $m_1 = \frac{1}{t}$.
The normal at $P$ is $y = -tx + 2at + at^3$. Setting $y=0$,we get $G(a(t^2+2), 0)$.
The tangent at $P$ meets the axis at $T(-at^2, 0)$.
The circle passes through $P(at^2, 2at)$,$T(-at^2, 0)$,and $G(a(t^2+2), 0)$.
The center of the circle is the midpoint of $TG$,which is $(a, 0)$,and the radius is $R = \sqrt{(at^2-a)^2 + (2at)^2} = a(t^2+1)$.
The slope of the radius $SP$ is $m_{SP} = \frac{2at-0}{at^2-a} = \frac{2t}{t^2-1}$.
The tangent to the circle at $P$ is perpendicular to the radius $SP$,so its slope $m_2 = -\frac{1}{m_{SP}} = \frac{1-t^2}{2t}$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{\frac{1}{t} - \frac{1-t^2}{2t}}{1 + \frac{1}{t} \cdot \frac{1-t^2}{2t}} \right| = \left| \frac{\frac{2-1+t^2}{2t}}{\frac{2t^2+1-t^2}{2t^2}} \right| = \left| \frac{1+t^2}{2t} \cdot \frac{2t^2}{1+t^2} \right| = t$.
Therefore,$\theta = \tan^{-1}t$.

(B) The given hyperbolas are conjugate to each other. The foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $(\pm ae_1, 0)$,where $e_1 = \sqrt{1 + \frac{b^2}{a^2}}$.
The foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ are $(0, \pm be_2)$,where $e_2 = \sqrt{1 + \frac{a^2}{b^2}}$.
The quadrilateral formed by these four foci is a rhombus with diagonals of lengths $2ae_1$ and $2be_2$.
The area of the rhombus is given by $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times (2ae_1) \times (2be_2) = 2abe_1e_2$.
We have $e_1 = \frac{\sqrt{a^2 + b^2}}{a}$ and $e_2 = \frac{\sqrt{a^2 + b^2}}{b}$.
Substituting these values,we get Area $= 2ab \times \frac{\sqrt{a^2 + b^2}}{a} \times \frac{\sqrt{a^2 + b^2}}{b} = 2(a^2 + b^2)$.

(D) Let the two fixed lines be the coordinate axes $OX$ and $OY$. Let the bar be $AB$ with length $L = 20$.
Let $P$ be a point on $AB$ such that $AP = 8$ and $PB = 20 - 8 = 12$.
Let $\angle OAB = \theta$. Then the coordinates of $P(x, y)$ are given by:
$x = PB \cos \theta = 12 \cos \theta$
$y = AP \sin \theta = 8 \sin \theta$
Thus,$\cos \theta = \frac{x}{12}$ and $\sin \theta = \frac{y}{8}$.
Using $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$\frac{x^2}{12^2} + \frac{y^2}{8^2} = 1$
This is the equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a = 12$ and $b = 8$.
Since $a > b$,the eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{8^2}{12^2} = 1 - \frac{64}{144} = 1 - \frac{4}{9} = \frac{5}{9}$.
Therefore,$e = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.

(D) Given the parabola $y = \frac{x^2}{4} - 2$,we have $4y = x^2 - 8$.
Differentiating with respect to $x$,we get $4 \frac{dy}{dx} = 2x$,so $\frac{dy}{dx} = \frac{x}{2}$.
Let the point of contact be $P(x_1, y_1)$. The slope of the tangent at $P$ is $m_t = \frac{x_1}{2}$.
The slope of the normal at $P$ is $m_n = -\frac{1}{m_t} = -\frac{2}{x_1}$.
The line passes through $A(10, -1)$ and $P(x_1, y_1)$,so its slope is $\frac{y_1 - (-1)}{x_1 - 10} = \frac{y_1 + 1}{x_1 - 10}$.
Since this line is the normal,$\frac{y_1 + 1}{x_1 - 10} = -\frac{2}{x_1}$.
$x_1(y_1 + 1) = -2(x_1 - 10)$ $\Rightarrow x_1y_1 + x_1 = -2x_1 + 20$ $\Rightarrow x_1y_1 + 3x_1 = 20$.
Substituting $y_1 = \frac{x_1^2}{4} - 2$,we get $x_1(\frac{x_1^2}{4} - 2) + 3x_1 = 20$.
$\frac{x_1^3}{4} - 2x_1 + 3x_1 = 20$ $\Rightarrow \frac{x_1^3}{4} + x_1 = 20$ $\Rightarrow x_1^3 + 4x_1 - 80 = 0$.
By inspection,$x_1 = 4$ is a root: $64 + 16 - 80 = 0$.
For $x_1 = 4$,$y_1 = \frac{4^2}{4} - 2 = 4 - 2 = 2$. So $P = (4, 2)$.
The slope of the normal is $m_n = -\frac{2}{4} = -\frac{1}{2}$.
The equation of the line passing through $(10, -1)$ with slope $-\frac{1}{2}$ is $y - (-1) = -\frac{1}{2}(x - 10)$.
$2(y + 1) = -(x - 10)$ $\Rightarrow 2y + 2 = -x + 10$ $\Rightarrow x + 2y = 8$.

(C) Let the quadratic polynomial be $f(x) = ax^2 + bx + c$. Given the leading coefficient $a = 1$ and constant term $c = 3$,we have $f(x) = x^2 + bx + 3$.
Since the curve is symmetric about the line $x = 1$,the $x$-coordinate of the vertex is $h = -b/(2a) = 1$. Thus,$-b/2 = 1$,which gives $b = -2$.
Therefore,the polynomial is $f(x) = x^2 - 2x + 3$.
The vertex $(h, k)$ has $h = 1$. The $y$-coordinate of the vertex is $k = f(1) = (1)^2 - 2(1) + 3 = 1 - 2 + 3 = 2$.
Thus,the vertex is $(1, 2)$.

(B) The quadratic polynomial is of the form $y = ax^2 + bx + c$. Given the absolute term $c = 3$ and the leading coefficient $a = 1$,we have $f(x) = x^2 + bx + 3$.
Since the curve is symmetric about the line $x = 1$,the vertex occurs at $x = 1$. Thus,the derivative $f'(x) = 2x + b$ must be zero at $x = 1$.
$2(1) + b = 0 \Rightarrow b = -2$.
So,the polynomial is $f(x) = x^2 - 2x + 3$.
For point $A$,$x_1 = 1$,so $y_1 = f(1) = 1^2 - 2(1) + 3 = 2$. Thus,$A = (1, 2)$.
For point $B$,$y_2 = 11$,so $11 = x^2 - 2x + 3 \Rightarrow x^2 - 2x - 8 = 0$.
$(x - 4)(x + 2) = 0$. Since $B$ is in the first quadrant,$x_2 = 4$. Thus,$B = (4, 11)$.
The vectors are $\vec{OA} = \hat{i} + 2\hat{j}$ and $\vec{OB} = 4\hat{i} + 11\hat{j}$.
The scalar product $\vec{OA} \cdot \vec{OB} = (1)(4) + (2)(11) = 4 + 22 = 26$.

(D) The equation of the hyperbola is $x^2 - (y - 1)^2 = 1$.
Let the tangent line passing through the origin $(0, 0)$ be $y = mx$.
Since the line is tangent to the hyperbola,the perpendicular distance from the center $(0, 1)$ to the line $mx - y = 0$ must be equal to the semi-transverse axis $a = 1$.
Using the formula for distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$,$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
$1 = \frac{|m(0) - 1(1)|}{\sqrt{m^2 + (-1)^2}}$ $\Rightarrow 1 = \frac{1}{\sqrt{m^2 + 1}}$ $\Rightarrow m^2 + 1 = 1$ $\Rightarrow m = 0$.
Wait,the slope must be positive. Let's re-evaluate. The tangent line is $y = mx$. The point of tangency $(a, b)$ satisfies $b = ma$ and $a^2 - (b - 1)^2 = 1$.
Differentiating $x^2 - (y - 1)^2 = 1$ with respect to $x$: $2x - 2(y - 1)\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y - 1}$.
At $(a, b)$,the slope is $m = \frac{a}{b - 1}$. Since $m = \frac{b}{a}$,we have $\frac{b}{a} = \frac{a}{b - 1} \Rightarrow a^2 = b^2 - b$.
Substituting $a^2 = b^2 - b$ into the hyperbola equation $a^2 - (b - 1)^2 = 1$:
$(b^2 - b) - (b^2 - 2b + 1) = 1$ $\Rightarrow b^2 - b - b^2 + 2b - 1 = 1$ $\Rightarrow b - 1 = 1$ $\Rightarrow b = 2$.
Then $a^2 = 2^2 - 2 = 2 \Rightarrow a = \sqrt{2}$ (since the point is in the first quadrant).
Thus,$\sin^{-1}\left(\frac{a}{b}\right) = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$.

(D) Statement $A$: The ellipse is $\frac{x^2}{9} + \frac{y^2}{5} = 1$. Here $a^2 = 9$ and $b^2 = 5$. The director circle is $x^2 + y^2 = a^2 + b^2 = 9 + 5 = 14$. This is true.
Statement $B$: The coordinates of $P$ and $Q$ are $(a \cos \theta, b \sin \theta)$ and $(a \cos(\theta + \alpha), b \sin(\theta + \alpha))$. The area of $\triangle OPQ$ is $\frac{1}{2} |x_P y_Q - x_Q y_P| = \frac{1}{2} |ab \cos \theta \sin(\theta + \alpha) - ab \cos(\theta + \alpha) \sin \theta| = \frac{1}{2} ab |\sin(\theta + \alpha - \theta)| = \frac{1}{2} ab |\sin \alpha|$. This is independent of $\theta$. This is true.
Statement $C$: This is a standard property of parabolas. The locus of the foot of the perpendicular from the focus to any tangent is the tangent at the vertex. This is true.
Since all statements are true,the correct option is $D$.

(D) Given hyperbola: $\frac{x^2}{9} - \frac{y^2}{3} = 1$. Here $a^2 = 9$ and $b^2 = 3$.
Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3}{9}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$. Thus,statement $(B)$ is incorrect.
Length of latus rectum $LLR = \frac{2b^2}{a} = \frac{2 \times 3}{3} = 2$. Thus,statement $(C)$ is correct.
The product of perpendicular distances from any point $(x_1, y_1)$ on the hyperbola to the asymptotes $bx \pm ay = 0$ is given by $p_1 p_2 = \frac{a^2 b^2}{a^2 + b^2} = \frac{9 \times 3}{9 + 3} = \frac{27}{12} = \frac{9}{4} = 2.25$.
Since $2.25 > 2$ $(LLR)$,the statement in $(A)$ is incorrect.
Since both $(A)$ and $(B)$ are incorrect,the correct option is $(D)$.

(D) The equation of the parabola is $y^2 = 4x$. The tangent to this parabola is $y = mx + \frac{1}{m}$.
The equation of the ellipse is $x^2 + 4y^2 = 8$,which can be written as $\frac{x^2}{8} + \frac{y^2}{2} = 1$. 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,$a^2 = 8$ and $b^2 = 2$. Substituting $c = \frac{1}{m}$,we get $\frac{1}{m^2} = 8m^2 + 2$.
Multiplying by $m^2$,we get $1 = 8m^4 + 2m^2$,or $8m^4 + 2m^2 - 1 = 0$.
Let $m^2 = t$. Then $8t^2 + 2t - 1 = 0$. Solving for $t$,we get $(4t - 1)(2t + 1) = 0$. Since $t = m^2 > 0$,we have $t = \frac{1}{4}$,so $m = \pm \frac{1}{2}$.
For $m = \frac{1}{2}$,$c = \frac{1}{1/2} = 2$. The tangent is $y = \frac{1}{2}x + 2$,which simplifies to $x - 2y + 4 = 0$.
For $m = -\frac{1}{2}$,$c = \frac{1}{-1/2} = -2$. The tangent is $y = -\frac{1}{2}x - 2$,which simplifies to $x + 2y + 4 = 0$.
Thus,both $(A)$ and $(B)$ are correct.

(D) Let the two hyperbolas be $H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $H_2: \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.
Any tangent to $H_1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.
For this line to be a tangent to $H_2$,we rewrite $H_2$ as $\frac{x^2}{(-b^2)} - \frac{y^2}{(-a^2)} = 1$. The condition for tangency $y = mx + c$ to $\frac{x^2}{A} - \frac{y^2}{B} = 1$ is $c^2 = Am^2 - B$.
Here $A = -b^2$ and $B = -a^2$,so $c^2 = (-b^2)m^2 - (-a^2) = a^2 - b^2m^2$.
Equating the two expressions for $c^2$: $a^2m^2 - b^2 = a^2 - b^2m^2$.
$(a^2 + b^2)m^2 = a^2 + b^2$,which gives $m^2 = 1$,so $m = \pm 1$.
Substituting $m^2 = 1$ into $c^2 = a^2m^2 - b^2$,we get $c^2 = a^2 - b^2$,so $c = \pm \sqrt{a^2 - b^2}$.
Thus,the common tangents are $y = \pm x \pm \sqrt{a^2 - b^2}$.
All given options are included in this set.

(A) For any conic section with focus $S$ and directrix $L$,the vertex $V$ lies on the axis of the conic section,which is the line passing through the focus $S(2, 3)$ and perpendicular to the directrix $x + y - 10 = 0$.
The slope of the directrix is $-1$,so the slope of the axis is $1$. The equation of the axis is $y - 3 = 1(x - 2)$,which simplifies to $y = x + 1$.
The intersection of the axis $y = x + 1$ and the directrix $x + y = 10$ is found by $x + (x + 1) = 10$,giving $2x = 9$,so $x = 4.5$ and $y = 5.5$. Let this point be $Z(4.5, 5.5)$.
The distance $SZ = \sqrt{(4.5 - 2)^2 + (5.5 - 3)^2} = \sqrt{2.5^2 + 2.5^2} = 2.5\sqrt{2}$.
For a conic with eccentricity $e$,the distance from the focus to the vertex is $SV = \frac{e}{1+e} SZ$ (for the vertex between focus and directrix).
For an ellipse,$0 < e < 1$,so $SV_E = \frac{e_E}{1+e_E} SZ$. As $e_E$ increases,$SV_E$ increases.
For a parabola,$e = 1$,so $SV_P = \frac{1}{2} SZ$.
For a hyperbola,$e > 1$,so $SV_H = \frac{e_H}{1+e_H} SZ$. As $e_H$ increases,$SV_H$ approaches $SZ$.
Since the vertex is between the focus and the directrix,the distance from the directrix to the vertex is $VZ = SZ - SV = SZ - \frac{e}{1+e} SZ = \frac{1}{1+e} SZ$.
For the ellipse $(e < 1)$,$VZ_E = \frac{1}{1+e_E} SZ > \frac{1}{2} SZ$.
For the parabola $(e = 1)$,$VZ_P = \frac{1}{2} SZ$.
For the hyperbola $(e > 1)$,$VZ_H = \frac{1}{1+e_H} SZ < \frac{1}{2} SZ$.
Thus,$VZ_E > VZ_P > VZ_H$. Since the vertices lie on the same line,the coordinates follow the order $\alpha > \beta > \gamma$.

(D) Let the midpoint of the chord be $(h, k).$
The equation of the chord of the hyperbola $x^2 - y^2 = a^2$ with midpoint $(h, k)$ is given by $T = S_1$,which is $hx - ky = h^2 - k^2.$
Rearranging this into the slope-intercept form $y = mx + c$,we get $ky = hx - (h^2 - k^2)$,or $y = \frac{h}{k}x - \frac{h^2 - k^2}{k}.$
This line is a tangent to the parabola $x^2 = 4by$,which can be written as $y = \frac{1}{4b}x^2.$ The condition for the line $y = mx + c$ to be a tangent to the parabola $x^2 = 4by$ is $c = -bm^2.$
Comparing the line $y = \frac{h}{k}x - \frac{h^2 - k^2}{k}$ with $y = mx + c$,we have $m = \frac{h}{k}$ and $c = -\frac{h^2 - k^2}{k}.$
Substituting these into the condition $c = -bm^2$,we get $-\frac{h^2 - k^2}{k} = -b\left(\frac{h}{k}\right)^2.$
Simplifying,we get $h^2 - k^2 = b\frac{h^2}{k},$ or $k(h^2 - k^2) = bh^2.$
Replacing $(h, k)$ with $(x, y)$,the locus is $y(x^2 - y^2) = bx^2.$
Clearly,this locus depends on $b$ but is independent of $a.$

(D) The equation of a tangent to the parabola $y^2 = 8x$ with slope $m$ is $y = mx + \frac{2}{m}$.
The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{32} + \frac{y^2}{8} = 1$ is $c^2 = a^2m^2 + b^2$,where $a^2 = 32$ and $b^2 = 8$.
Substituting $c = \frac{2}{m}$ into the condition: $(\frac{2}{m})^2 = 32m^2 + 8$.
$\frac{4}{m^2} = 32m^2 + 8$.
Dividing by $4$: $\frac{1}{m^2} = 8m^2 + 2$.
$8m^4 + 2m^2 - 1 = 0$.
Let $t = m^2$. Then $8t^2 + 2t - 1 = 0$.
$(4t - 1)(2t + 1) = 0$.
Since $m^2$ must be positive,$m^2 = \frac{1}{4}$.
Thus,$m = \frac{1}{2}$ or $m = -\frac{1}{2}$.
The product of the slopes is $(\frac{1}{2}) \times (-\frac{1}{2}) = -\frac{1}{4}$.

(A) The equation of the ellipse is $\frac{(x - 2)^2}{3^2} + \frac{(y + 2)^2}{2^2} = 1$.
The equation of the circle is $x^2 - 4x + 4 + y^2 + 2y + 1 = -4 + 4 + 1$,which simplifies to $(x - 2)^2 + (y + 1)^2 = 1$.
Let $X = x - 2$ and $Y = y + 2$. Then the ellipse is $\frac{X^2}{9} + \frac{Y^2}{4} = 1$.
The circle is $(X)^2 + (Y - 1)^2 = 1$,or $X^2 + Y^2 - 2Y = 0$.
Substituting $X^2 = 9(1 - \frac{Y^2}{4})$ into the circle equation: $9 - \frac{9Y^2}{4} + Y^2 - 2Y = 0$.
$9 - \frac{5Y^2}{4} - 2Y = 0 \implies 5Y^2 + 8Y - 36 = 0$.
Solving for $Y$: $Y = \frac{-8 \pm \sqrt{64 - 4(5)(-36)}}{10} = \frac{-8 \pm \sqrt{64 + 720}}{10} = \frac{-8 \pm \sqrt{784}}{10} = \frac{-8 \pm 28}{10}$.
$Y = 2$ or $Y = -3.6$. Since the ellipse is $\frac{Y^2}{4} \le 1$,$Y$ must be in $[-2, 2]$.
Thus $Y = 2$ is the only solution,which gives $X = 0$.
At $X = 0, Y = 2$,we have $x = 2$ and $y = 0$. The curves touch at $(2, 0)$.
Since they touch at a single point,the length of the common chord is $0$.

(B) The conic $y = x^2 - x + 3$ is a parabola,so its eccentricity is $e_1 = 1$.
The conic $\frac{x^2}{a^2} + \frac{y^2}{3a^4} = 1$ is an ellipse,so its eccentricity $e_2$ satisfies $0 < e_2 < 1$.
The conic $a^2x^2 - 3a^4y^2 = 1$ is a hyperbola,so its eccentricity $e_3$ satisfies $e_3 > 1$.
Comparing these values,we have $e_2 < 1 < e_3$,which implies $e_2 < e_1 < e_3$.

(C) Let $X = x - 1$ and $Y = y - 2$. The equations become $X^2 = 4Y$ and $X^2 + \frac{Y^2}{2} = 1$.
The equation of a tangent to the parabola $X^2 = 4Y$ with slope $m$ is $Y = mX - am^2$,where $a = 1$. Thus,$Y = mX - m^2$.
The equation of a tangent to the ellipse $X^2 + \frac{Y^2}{2} = 1$ with slope $m$ is $Y = mX \pm \sqrt{a^2m^2 + b^2}$,where $a^2 = 1$ and $b^2 = 2$. Thus,$Y = mX \pm \sqrt{m^2 + 2}$.
For the tangents to be common,the constant terms must be equal: $-m^2 = \pm \sqrt{m^2 + 2}$.
Squaring both sides,we get $m^4 = m^2 + 2$,which implies $m^4 - m^2 - 2 = 0$.
Let $t = m^2$. Then $t^2 - t - 2 = 0$,which factors as $(t - 2)(t + 1) = 0$.
Since $m^2$ must be non-negative,$m^2 = 2$. The slopes $m_1$ and $m_2$ are $\sqrt{2}$ and $-\sqrt{2}$.
Therefore,$m_1^2 + m_2^2 = 2 + 2 = 4$.

(D) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,differentiating with respect to $x$ gives $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{xb^2}{ya^2}$.
For the hyperbola $\frac{x^2}{l^2} - \frac{y^2}{m^2} = 1$,differentiating with respect to $x$ gives $\frac{2x}{l^2} - \frac{2y}{m^2} \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = \frac{xm^2}{yl^2}$.
Since the curves cut orthogonally,the product of their slopes is $-1$:
$(-\frac{xb^2}{ya^2}) \times (\frac{xm^2}{yl^2}) = -1 \Rightarrow \frac{x^2}{y^2} = \frac{a^2 l^2}{b^2 m^2}$.
Subtracting the two curve equations: $\frac{x^2}{a^2} + \frac{y^2}{b^2} - (\frac{x^2}{l^2} - \frac{y^2}{m^2}) = 1 - 1 = 0$.
$x^2(\frac{1}{a^2} - \frac{1}{l^2}) + y^2(\frac{1}{b^2} + \frac{1}{m^2}) = 0 \Rightarrow \frac{x^2}{y^2} = -\frac{(\frac{1}{b^2} + \frac{1}{m^2})}{(\frac{1}{a^2} - \frac{1}{l^2})} = \frac{(m^2 + b^2)a^2 l^2}{b^2 m^2 (l^2 - a^2)}$.
Equating the two expressions for $\frac{x^2}{y^2}$:
$\frac{a^2 l^2}{b^2 m^2} = \frac{(m^2 + b^2)a^2 l^2}{b^2 m^2 (l^2 - a^2)} \Rightarrow 1 = \frac{m^2 + b^2}{l^2 - a^2}$.
$l^2 - a^2 = m^2 + b^2 \Rightarrow a^2 - b^2 = l^2 - m^2$ is incorrect; re-evaluating: $l^2 - a^2 = m^2 + b^2 \Rightarrow a^2 - b^2 = l^2 - m^2$ is not the standard form. Let's re-check: $l^2 - a^2 = m^2 + b^2 \Rightarrow a^2 + b^2 = l^2 - m^2$.

(C) Any point on the line $y = \sqrt{3}x$ can be represented in polar form as $(r \cos \theta, r \sin \theta)$,where $\theta = 60^\circ = \frac{\pi}{3}$.
Thus,the coordinates are $\left(r \cos 60^\circ, r \sin 60^\circ\right) = \left(\frac{r}{2}, \frac{r\sqrt{3}}{2}\right)$.
Substituting these into the curve equation $x^4 + ax^2y + bxy + cx + dy + 6 = 0$:
$\left(\frac{r}{2}\right)^4 + a\left(\frac{r}{2}\right)^2\left(\frac{r\sqrt{3}}{2}\right) + b\left(\frac{r}{2}\right)\left(\frac{r\sqrt{3}}{2}\right) + c\left(\frac{r}{2}\right) + d\left(\frac{r\sqrt{3}}{2}\right) + 6 = 0$
$\frac{r^4}{16} + \frac{a\sqrt{3}r^3}{8} + \frac{b\sqrt{3}r^2}{4} + \frac{r(c + d\sqrt{3})}{2} + 6 = 0$
Multiplying by $16$:
$r^4 + (2a\sqrt{3})r^3 + (4b\sqrt{3})r^2 + 8(c + d\sqrt{3})r + 96 = 0$
This is a quartic equation in $r$ where the roots $r_1, r_2, r_3, r_4$ represent the distances $OA, OB, OC, OD$ from the origin.
By Vieta's formulas,the product of the roots is the constant term divided by the leading coefficient:
$OA \cdot OB \cdot OC \cdot OD = r_1 r_2 r_3 r_4 = \frac{96}{1} = 96$.

(D) The equation of a tangent to the circle $x^2 + y^2 = 4$ with slope $m$ is $y = mx \pm 2\sqrt{1 + m^2}$.
Since this line is also a tangent to the parabola $x^2 = 4y$,we substitute $y = mx + c$ into the parabola equation: $x^2 = 4(mx + c) \Rightarrow x^2 - 4mx - 4c = 0$.
For tangency,the discriminant $D = 0$,so $(-4m)^2 - 4(1)(-4c) = 0$ $\Rightarrow 16m^2 + 16c = 0$ $\Rightarrow c = -m^2$.
Comparing this with the circle's tangent form $c = \pm 2\sqrt{1 + m^2}$,we get $-m^2 = \pm 2\sqrt{1 + m^2}$.
Squaring both sides: $m^4 = 4(1 + m^2) \Rightarrow m^4 - 4m^2 - 4 = 0$.
Using the quadratic formula for $m^2$: $m^2 = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}$.
Since $m^2$ must be positive,$m^2 = 2 + 2\sqrt{2}$.

(D) Given equations are:
$x^2 = 8y \quad (i)$
$\frac{x^2}{3} + y^2 = 1 \quad (ii)$
Substituting $(i)$ into $(ii)$:
$\frac{8y}{3} + y^2 = 1$
$3y^2 + 8y - 3 = 0$
$3y^2 + 9y - y - 3 = 0$
$3y(y + 3) - 1(y + 3) = 0$
$(3y - 1)(y + 3) = 0$
So,$y = \frac{1}{3}$ or $y = -3$.
Since $x^2 = 8y$,$y$ must be non-negative,so $y = -3$ is rejected.
For $y = \frac{1}{3}$,$x^2 = 8(\frac{1}{3}) = \frac{8}{3}$,so $x = \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{2}}{\sqrt{3}} = \pm \frac{2\sqrt{6}}{3}$.
The points of intersection are $(\frac{2\sqrt{6}}{3}, \frac{1}{3})$ and $(-\frac{2\sqrt{6}}{3}, \frac{1}{3})$.
The line passing through these points is a horizontal line $y = \frac{1}{3}$,which simplifies to $3y - 1 = 0$.

(B) Given curves are $C_1: \frac{x^2}{\alpha} + \frac{y^2}{4} = 1$ and $C_2: y^3 = 16x$.
Differentiating $C_1$ with respect to $x$: $\frac{2x}{\alpha} + \frac{2y}{4} \cdot \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{4x}{\alpha y} = m_1$.
Differentiating $C_2$ with respect to $x$: $3y^2 \cdot \frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{16}{3y^2} = m_2$.
Since the curves intersect at right angles,$m_1 \cdot m_2 = -1$.
Substituting the slopes: $\left( -\frac{4x}{\alpha y} \right) \cdot \left( \frac{16}{3y^2} \right) = -1$.
This simplifies to $\frac{64x}{3\alpha y^3} = 1 \Rightarrow 3\alpha y^3 = 64x$.
Substitute $y^3 = 16x$ into the equation: $3\alpha (16x) = 64x$.
Assuming $x \neq 0$,we get $48\alpha = 64 \Rightarrow \alpha = \frac{64}{48} = \frac{4}{3}$.

(A) Given conics are $x^2 = 6y$ $(i)$ and $2x^2 - 4y^2 = 9$ $(ii)$.
Consider the line $x - y = \frac{3}{2}$ $(iii)$,which implies $x = y + \frac{3}{2}$.
Substitute $x = y + \frac{3}{2}$ into $(i)$: $(y + \frac{3}{2})^2 = 6y \implies y^2 + 3y + \frac{9}{4} = 6y \implies y^2 - 3y + \frac{9}{4} = 0 \implies (y - \frac{3}{2})^2 = 0$. Thus,$y = \frac{3}{2}$ and $x = 3$.
Since there is only one point of intersection $(3, \frac{3}{2})$,the line is tangent to the parabola.
Substitute $x = y + \frac{3}{2}$ into $(ii)$: $2(y + \frac{3}{2})^2 - 4y^2 = 9 \implies 2(y^2 + 3y + \frac{9}{4}) - 4y^2 = 9 \implies 2y^2 + 6y + \frac{9}{2} - 4y^2 = 9 \implies -2y^2 + 6y - \frac{9}{2} = 0 \implies 4y^2 - 12y + 9 = 0 \implies (2y - 3)^2 = 0$. Thus,$y = \frac{3}{2}$ and $x = 3$.
Since there is only one point of intersection $(3, \frac{3}{2})$,the line is tangent to the hyperbola.
Therefore,$x - y = \frac{3}{2}$ is the common tangent.

(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$.
Here,$a^2 = 16$,so $a = 4$.
The foci of the ellipse are $(\pm ae, 0)$,where $e = \sqrt{1 - \frac{b^2}{16}}$.
Thus,the foci are $(\pm 4 \sqrt{1 - \frac{b^2}{16}}) = (\pm \sqrt{16 - b^2}, 0)$.
The equation of the hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,which can be written as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(9/5)^2} = 1$.
Here,$a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$.
The eccentricity $e_h$ of the hyperbola is $\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{81/25}{144/25}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$.
The foci of the hyperbola are $(\pm a_h e_h, 0) = (\pm \frac{12}{5} \times \frac{5}{4}, 0) = (\pm 3, 0)$.
Since the foci coincide,$\sqrt{16 - b^2} = 3$.
Squaring both sides,$16 - b^2 = 9$,which gives $b^2 = 7$.

(A) The given ellipse is $\frac{x^2}{16} + \frac{y^2}{3} = 1$.
Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 3$.
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 $(x_1, y_1) = \left( 2, \frac{3}{2} \right)$,$a^2 = 16$,and $b^2 = 3$:
$\frac{16x}{2} - \frac{3y}{3/2} = 16 - 3$
$8x - 2y = 13$
$2y = 8x - 13 \Rightarrow y = 4x - \frac{13}{2}$.
$A$ line $y = mx + c$ touches the parabola $y^2 = 4Ax$ if $c = \frac{A}{m}$.
Here $m = 4$ and $c = -\frac{13}{2}$.
$-\frac{13}{2} = \frac{A}{4} \Rightarrow A = -26$.
Thus,the equation of the parabola is $y^2 = 4(-26)x = -104x$.

(C) The equation of a tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$ ....$(i)$
Substitute this into the equation of the hyperbola $xy = 2$:
$x(mx + \frac{1}{m}) = 2$
$mx^2 + \frac{x}{m} - 2 = 0$
For the line to be a tangent,the discriminant $D$ must be $0$:
$D = b^2 - 4ac = (\frac{1}{m})^2 - 4(m)(-2) = 0$
$\frac{1}{m^2} + 8m = 0$
$1 + 8m^3 = 0$
$m^3 = -\frac{1}{8}$
$m = -\frac{1}{2}$
Substitute $m = -\frac{1}{2}$ into equation $(i)$:
$y = -\frac{1}{2}x + \frac{1}{-1/2}$
$y = -\frac{1}{2}x - 2$
$2y = -x - 4$
$x + 2y + 4 = 0$

(C) The equation of the tangent to the parabola $y^2 = x$ at point $(\alpha, \beta)$ is given by $y\beta = \frac{x + \alpha}{2}$.
Since $(\alpha, \beta)$ lies on the parabola,$\beta^2 = \alpha$,so the equation becomes $y\beta = \frac{x + \beta^2}{2}$,which simplifies to $y = \frac{1}{2\beta}x + \frac{\beta}{2}$.
Here,the slope $m = \frac{1}{2\beta}$ and the intercept $c = \frac{\beta}{2}$.
This line is also a tangent to the ellipse $x^2 + 2y^2 = 1$,which can be written as $\frac{x^2}{1} + \frac{y^2}{1/2} = 1$.
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 $a^2 = 1$,$b^2 = 1/2$,$m = \frac{1}{2\beta}$,and $c = \frac{\beta}{2}$:
$(\frac{\beta}{2})^2 = 1(\frac{1}{2\beta})^2 + \frac{1}{2}$.
$\frac{\beta^2}{4} = \frac{1}{4\beta^2} + \frac{1}{2}$.
Multiplying by $4\beta^2$: $\beta^4 = 1 + 2\beta^2$.
$\beta^4 - 2\beta^2 - 1 = 0$.
Using the quadratic formula for $\beta^2$: $\beta^2 = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.
Since $\beta^2 > 0$,we have $\beta^2 = 1 + \sqrt{2}$.
Since $\alpha = \beta^2$,we get $\alpha = \sqrt{2} + 1$.

(C) The equation of the parabola is $y^2 = 12x$,so $a = 3$. The tangent is $y = mx + \frac{3}{m}$.
The equation of the hyperbola is $8x^2 - y^2 = 8$,which simplifies to $x^2 - \frac{y^2}{8} = 1$. Here $a^2 = 1$ and $b^2 = 8$.
The tangent is $y = mx \pm \sqrt{a^2m^2 - b^2} = mx \pm \sqrt{m^2 - 8}$.
For common tangents,$\frac{3}{m} = \pm \sqrt{m^2 - 8}$.
Squaring both sides,$\frac{9}{m^2} = m^2 - 8 \Rightarrow m^4 - 8m^2 - 9 = 0$.
$(m^2 - 9)(m^2 + 1) = 0$. Since $m$ is real,$m^2 = 9$,so $m = \pm 3$.
The common tangents are $y = 3x + 1$ and $y = -3x - 1$.
Solving these,the intersection point $P$ is $(-1/3, 0)$.
For the hyperbola $x^2 - \frac{y^2}{8} = 1$,$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 8} = 3$.
The foci are $S(ae, 0) = (3, 0)$ and $S'(-ae, 0) = (-3, 0)$.
Let $P$ divide $SS'$ in the ratio $k : 1$. Then $P = \left( \frac{k(-3) + 1(3)}{k+1}, 0 \right) = \left( \frac{3-3k}{k+1}, 0 \right)$.
Equating to $P(-1/3, 0)$,we get $\frac{3-3k}{k+1} = -\frac{1}{3}$.
$9 - 9k = -k - 1$ $\Rightarrow 8k = 10$ $\Rightarrow k = \frac{10}{8} = \frac{5}{4}$.
Thus,the ratio is $5 : 4$.

(D) The equation of a tangent to the parabola $y^2 = 16x$ is of the form $y = mx + \frac{a}{m}$,where $a = 4$. So,$y = mx + \frac{4}{m} \dots (i)$.
If this line is also a tangent to the hyperbola $xy = -4$,then substituting $y$ from $(i)$ into the hyperbola equation gives $x(mx + \frac{4}{m}) = -4$.
This simplifies to $mx^2 + \frac{4}{m}x + 4 = 0$,or $m^2x^2 + 4x + 4m = 0$.
For the line to be a tangent,the discriminant $D$ must be zero: $D = (4)^2 - 4(m^2)(4m) = 0$.
$16 - 16m^3 = 0$ $\Rightarrow m^3 = 1$ $\Rightarrow m = 1$.
Substituting $m = 1$ into equation $(i)$,we get $y = x + 4$,which can be rewritten as $x - y + 4 = 0$.
Thus,the correct option is $(D)$.

(D) For the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$,we have $a^{2}=18$ and $b^{2}=4$. The eccentricity $e_{1} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{4}{18}} = \sqrt{\frac{14}{18}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$.
For the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$,we have $a^{2}=9$ and $b^{2}=4$. The eccentricity $e_{2} = \sqrt{1+\frac{b^{2}}{a^{2}}} = \sqrt{1+\frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.
Since the point $(e_{1}, e_{2})$ lies on the ellipse $15x^{2}+3y^{2}=k$,we substitute the values of $e_{1}$ and $e_{2}$ into the equation:
$15(\frac{\sqrt{7}}{3})^{2} + 3(\frac{\sqrt{13}}{3})^{2} = k$
$15(\frac{7}{9}) + 3(\frac{13}{9}) = k$
$\frac{105}{9} + \frac{39}{9} = k$
$\frac{144}{9} = k$
$k = 16$.

(A) Given the curves $y^{2}=4ax \dots (i)$ and $x^{2}=4by \dots (ii)$.
Solving $(i)$ and $(ii)$,we substitute $y = \frac{x^{2}}{4b}$ into $(i)$:
$\left(\frac{x^{2}}{4b}\right)^{2} = 4ax \Rightarrow \frac{x^{4}}{16b^{2}} = 4ax \Rightarrow x^{4} = 64ab^{2}x$.
This gives $x(x^{3} - 64ab^{2}) = 0$,so $x = 0$ or $x = 4a^{1/3}b^{2/3}$.
The points of intersection are $(0,0)$ and $(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$.
For $(i)$,$\frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}$. For $(ii)$,$\frac{dy}{dx} = \frac{2x}{4b} = \frac{x}{2b}$.
At $(0,0)$,the tangent to $(i)$ is vertical $(x=0)$ and to $(ii)$ is horizontal $(y=0)$,so the angle is $\frac{\pi}{2}$.
At $(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$,slopes are $m_{1} = \frac{2a}{4a^{2/3}b^{1/3}} = \frac{1}{2}(\frac{a}{b})^{1/3}$ and $m_{2} = \frac{4a^{1/3}b^{2/3}}{2b} = 2(\frac{a}{b})^{1/3}$.
The angle $\theta$ is given by $\tan \theta = |\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}| = |\frac{2(a/b)^{1/3} - 0.5(a/b)^{1/3}}{1 + 2(a/b)^{1/3} \cdot 0.5(a/b)^{1/3}}| = \frac{1.5(a/b)^{1/3}}{1 + (a/b)^{2/3}} = \frac{3a^{1/3}b^{1/3}}{2(a^{2/3}+b^{2/3})}$.
Thus,$\theta = \tan^{-1}\left(\frac{3a^{1/3}b^{1/3}}{2(a^{2/3}+b^{2/3})}\right)$.

(B) Given the hyperbola equation: $x^{2} - y^{2} \sec^{2} \theta = 10 \Rightarrow \frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} \theta} = 1.$
For a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ the eccentricity $e_{H} = \sqrt{1 + \frac{b^{2}}{a^{2}}}.$
Thus,$e_{H} = \sqrt{1 + \frac{10 \cos^{2} \theta}{10}} = \sqrt{1 + \cos^{2} \theta}.$
Given the ellipse equation: $x^{2} \sec^{2} \theta + y^{2} = 5 \Rightarrow \frac{x^{2}}{5 \cos^{2} \theta} + \frac{y^{2}}{5} = 1.$
For an ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$ (where $a > b$),$e_{E} = \sqrt{1 - \frac{b^{2}}{a^{2}}}.$
Here $a^{2} = 5$ and $b^{2} = 5 \cos^{2} \theta,$ so $e_{E} = \sqrt{1 - \frac{5 \cos^{2} \theta}{5}} = \sqrt{1 - \cos^{2} \theta} = \sin \theta.$
Given $e_{H} = \sqrt{5} e_{E},$ we have $\sqrt{1 + \cos^{2} \theta} = \sqrt{5} \sin \theta.$
Squaring both sides: $1 + \cos^{2} \theta = 5 \sin^{2} \theta = 5(1 - \cos^{2} \theta).$
$1 + \cos^{2} \theta = 5 - 5 \cos^{2} \theta$ $\Rightarrow 6 \cos^{2} \theta = 4$ $\Rightarrow \cos^{2} \theta = \frac{2}{3}.$
Then $\sin^{2} \theta = 1 - \frac{2}{3} = \frac{1}{3}.$
The length of the latus rectum of the ellipse is $\frac{2b^{2}}{a} = \frac{2(5 \cos^{2} \theta)}{\sqrt{5}} = \frac{10 \times (2/3)}{\sqrt{5}} = \frac{20}{3 \sqrt{5}} = \frac{4 \sqrt{5}}{3}.$

(A) For the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1$ $(b < 5)$,the eccentricity $e_{1}$ satisfies $b^{2}=25(1-e_{1}^{2})$.
For the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$,the eccentricity $e_{2}$ satisfies $b^{2}=16(e_{2}^{2}-1)$.
Equating the two expressions for $b^{2}$,we get $25(1-e_{1}^{2})=16(e_{2}^{2}-1)$.
Given $e_{1}e_{2}=1$,we substitute $e_{2}=\frac{1}{e_{1}}$:
$25(1-e_{1}^{2})=16(\frac{1}{e_{1}^{2}}-1) = 16(\frac{1-e_{1}^{2}}{e_{1}^{2}})$.
Since $b  < 5$,$e_{1} \neq 1$,so we divide by $(1-e_{1}^{2})$ to get $25 = \frac{16}{e_{1}^{2}}$,which implies $e_{1}^{2}=\frac{16}{25}$,so $e_{1}=\frac{4}{5}$.
Then $e_{2}=\frac{1}{e_{1}}=\frac{5}{4}$.
The distance between the foci of the ellipse is $2ae_{1} = 2(5)(\frac{4}{5}) = 8 = \alpha$.
The distance between the foci of the hyperbola is $2ae_{2} = 2(4)(\frac{5}{4}) = 10 = \beta$.
Thus,the ordered pair $(\alpha, \beta) = (8, 10)$.

(B) The equation of the parabola is $y^{2} = 4(x-5)$.
The equation of the tangent to the parabola $y^{2} = 4a(x-h)$ at $(x_{1}, y_{1})$ is $yy_{1} = 2a(x+x_{1}) - 4ah$.
Here,$a=1$,$h=5$,$x_{1}=6$,and $y_{1}=2$. Substituting these values:
$2y = 2(x+6) - 20$
$2y = 2x + 12 - 20$
$2y = 2x - 8$
$y = x - 4$,which can be written as $x - y - 4 = 0$.
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,$m = 1$,$c = -4$,$A^{2} = 2$,and $B^{2} = b$.
Substituting these into the condition:
$(-4)^{2} = 2(1)^{2} + b$
$16 = 2 + b$
$b = 14$.

(D) To find the angle subtended by the chord $PQ$ at the origin,we homogenize the equation of the curve $x^{2} + 2y^{2} = 2$ using the line $x + y = 1$.
The equation of the line can be written as $x + y = 1$,so $1 = x + y$.
Substituting this into the curve equation:
$x^{2} + 2y^{2} = 2(1)^{2}$
$x^{2} + 2y^{2} = 2(x + y)^{2}$
$x^{2} + 2y^{2} = 2(x^{2} + 2xy + y^{2})$
$x^{2} + 2y^{2} = 2x^{2} + 4xy + 2y^{2}$
$x^{2} + 4xy = 0$
This represents a pair of lines passing through the origin. Let these lines be $y = m_{1}x$ and $y = m_{2}x$.
From $x(x + 4y) = 0$,we get $x = 0$ (slope $m_{1} = \infty$) and $y = -\frac{1}{4}x$ (slope $m_{2} = -\frac{1}{4}$).
The angle $\theta$ between these two lines is given by:
$\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$
Since one line is vertical $(x=0)$,the angle $\theta$ is the angle between the vertical line and the line with slope $-\frac{1}{4}$.
The angle of the line $y = -\frac{1}{4}x$ with the $x$-axis is $\alpha = \tan^{-1}(-\frac{1}{4}) = -\tan^{-1}(\frac{1}{4})$.
The angle between the vertical line $(90^{\circ})$ and this line is $\frac{\pi}{2} - (-\tan^{-1}(\frac{1}{4})) = \frac{\pi}{2} + \tan^{-1}(\frac{1}{4})$.

(C) Let the midpoint of the chord be $(h, k)$. The equation of the chord of the hyperbola $x^{2}-y^{2}=4$ with midpoint $(h, k)$ is given by $T=S_{1}$,which is $xh-yk=h^{2}-k^{2}$.
Rewriting this in the slope-intercept form $y=mx+c$,we get $y=\frac{h}{k}x - \frac{h^{2}-k^{2}}{k}$.
This line touches the parabola $y^{2}=8x$ (where $a=2$). The condition for the line $y=mx+c$ to touch the parabola $y^{2}=4ax$ is $c=\frac{a}{m}$.
Substituting $m=\frac{h}{k}$ and $c=-\frac{h^{2}-k^{2}}{k}$,we get $-\frac{h^{2}-k^{2}}{k} = \frac{2}{h/k} = \frac{2k}{h}$.
Simplifying,$-(h^{2}-k^{2})h = 2k^{2}$,which gives $h^{2}h - k^{2}h = -2k^{2}$,or $h^{3} = k^{2}(h-2)$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^{2}(x-2)=x^{3}$.

(C) The equation of the line is $y = -2x + k$. Since it is a tangent to the hyperbola $x^2 - y^2 = 3$ (where $a^2 = 3, b^2 = 3$),the condition for tangency $y = mx \pm \sqrt{a^2m^2 - b^2}$ gives:
$k = \sqrt{3(-2)^2 - 3} = \sqrt{3(4) - 3} = \sqrt{9} = 3$ (since $k > 0$).
So,the line is $y = -2x + 3$.
For this line to be a tangent to the parabola $y^2 = \alpha x$,it must satisfy the condition $c = \frac{a}{m}$,where $y = mx + c$ and $a$ in the parabola equation $y^2 = 4Ax$ is $\frac{\alpha}{4}$.
Here,$m = -2$ and $c = 3$. The condition for tangency to $y^2 = \alpha x$ is $c = \frac{\alpha}{4m}$.
Substituting the values: $3 = \frac{\alpha}{4(-2)} = \frac{\alpha}{-8}$.
Therefore,$\alpha = 3 \times (-8) = -24$.

(A) For the hyperbola $H : \frac{x^2}{a^2} - y^2 = 1$,the length of the latus rectum is $LR_H = \frac{2b^2}{a} = \frac{2(1)^2}{a} = \frac{2}{a}$.
For the ellipse $E : 3x^2 + 4y^2 = 12$,we rewrite it as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here $a^2 = 4$ and $b^2 = 3$. The length of the latus rectum is $LR_E = \frac{2b^2}{a} = \frac{2(3)}{2} = 3$.
Given $LR_H = LR_E$,we have $\frac{2}{a} = 3$,which implies $a = \frac{2}{3}$.
The eccentricity of the hyperbola is $e_H = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{(2/3)^2}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$. Thus,$e_H^2 = \frac{13}{4}$.
The eccentricity of the ellipse is $e_E = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$. Thus,$e_E^2 = \frac{1}{4}$.
Finally,$12(e_H^2 + e_E^2) = 12(\frac{13}{4} + \frac{1}{4}) = 12(\frac{14}{4}) = 3 \times 14 = 42$.

(C) The point $(\alpha, \beta)$ lies on the ellipse $25x^{2} + 4y^{2} = 1$,so we can write $\alpha = \frac{1}{5} \cos \theta$ and $\beta = \frac{1}{2} \sin \theta$.
The equation of a tangent to the parabola $y^{2} = 4x$ with slope $m$ is $y = mx + \frac{1}{m}$.
Since it passes through $(\alpha, \beta)$,we have $\beta = m\alpha + \frac{1}{m}$,which simplifies to $m^{2}\alpha - m\beta + 1 = 0$.
Let the slopes be $m_{1}$ and $m_{2}$ such that $m_{1} = 4m_{2}$.
From the quadratic equation $m^{2}\alpha - m\beta + 1 = 0$,we have $m_{1} + m_{2} = \frac{\beta}{\alpha}$ and $m_{1}m_{2} = \frac{1}{\alpha}$.
Substituting $m_{1} = 4m_{2}$,we get $5m_{2} = \frac{\beta}{\alpha}$ and $4m_{2}^{2} = \frac{1}{\alpha}$.
Thus,$m_{2} = \frac{\beta}{5\alpha}$,so $4(\frac{\beta}{5\alpha})^{2} = \frac{1}{\alpha}$,which gives $4\beta^{2} = 25\alpha$.
Substituting $\alpha = \frac{1}{5} \cos \theta$ and $\beta^{2} = \frac{1}{4} \sin^{2} \theta$,we get $4(\frac{1}{4} \sin^{2} \theta) = 25(\frac{1}{5} \cos \theta)$,so $\sin^{2} \theta = 5 \cos \theta$.
$1 - \cos^{2} \theta = 5 \cos \theta \Rightarrow \cos^{2} \theta + 5 \cos \theta - 1 = 0$.
Solving for $\cos \theta$,we get $\cos \theta = \frac{-5 \pm \sqrt{25 + 4}}{2} = \frac{-5 \pm \sqrt{29}}{2}$.
Then $10\alpha = 2 \cos \theta = -5 \pm \sqrt{29}$,so $(10\alpha + 5)^{2} = 29$.
Also,$16\beta^{2} = 16(\frac{1}{4} \sin^{2} \theta) = 4(1 - \cos^{2} \theta) = 4(1 - (1 - 5 \cos \theta)) = 20 \cos \theta = 20(\frac{-5 \pm \sqrt{29}}{2}) = -50 \pm 10\sqrt{29}$.
Thus,$(16\beta^{2} + 50)^{2} = (\pm 10\sqrt{29})^{2} = 100 \times 29 = 2900$.
Finally,$(10\alpha + 5)^{2} + (16\beta^{2} + 50)^{2} = 29 + 2900 = 2929$.

(B) The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. The equation of a tangent to the ellipse with slope $m$ is given by $y=mx \pm \sqrt{a^{2}m^{2}+b^{2}}$.
Substituting $a^{2}=16$ and $b^{2}=9$,we get $y=mx \pm \sqrt{16m^{2}+9}$ $(i)$.
The equation of the circle is $x^{2}+y^{2}=12$. The equation of a tangent to the circle with slope $m$ is given by $y=mx \pm r\sqrt{1+m^{2}}$.
Substituting $r^{2}=12$,we get $y=mx \pm \sqrt{12(1+m^{2})}$ $(ii)$.
For the common tangent,the constant terms in $(i)$ and $(ii)$ must be equal:
$16m^{2}+9 = 12(1+m^{2})$
$16m^{2}+9 = 12+12m^{2}$
$16m^{2}-12m^{2} = 12-9$
$4m^{2} = 3$
Multiplying both sides by $3$,we get $12m^{2} = 9$.

(C) The equation $|z - (4 + 3i)| = 2$ represents a circle with center $(4, 3)$ and radius $r = 2$.
In Cartesian coordinates,this is $(x - 4)^2 + (y - 3)^2 = 4$.
The equation $|z| + |z - 4| = 6$ represents an ellipse with foci at $(0, 0)$ and $(4, 0)$.
The sum of distances from the foci is $2a = 6$,so $a = 3$. The center is $(2, 0)$.
The distance between foci is $2ae = 4$,so $ae = 2$. Since $a = 3$,$e = 2/3$.
$b^2 = a^2(1 - e^2) = 9(1 - 4/9) = 5$.
The equation of the ellipse is $\frac{(x - 2)^2}{9} + \frac{y^2}{5} = 1$.
Check the position of the circle relative to the ellipse.
The center of the circle is $(4, 3)$. The highest point of the ellipse is $(2, \sqrt{5}) \approx (2, 2.236)$.
The circle's lowest point is $(4, 3 - 2) = (4, 1)$.
Substitute $(4, 1)$ into the ellipse equation: $\frac{(4 - 2)^2}{9} + \frac{1^2}{5} = \frac{4}{9} + \frac{1}{5} = \frac{20 + 9}{45} = \frac{29}{45} < 1$.
Since the point $(4, 1)$ lies inside the ellipse and the center $(4, 3)$ lies outside the ellipse,the circle must intersect the ellipse at two points.

(B) Let the coordinates of $P$ be $(t^{2}, 2t)$ and $Q$ be $(\frac{1}{t^{2}}, -\frac{2}{t})$.
Since $PQ$ subtends an angle of $\frac{\pi}{2}$ at $R(3, 0)$,the product of the slopes of $PR$ and $QR$ is $-1$.
Slope of $PR = \frac{2t-0}{t^{2}-3} = \frac{2t}{t^{2}-3}$.
Slope of $QR = \frac{-2/t-0}{1/t^{2}-3} = \frac{-2/t}{(1-3t^{2})/t^{2}} = \frac{-2t}{1-3t^{2}}$.
Since the product is $-1$,we have $\frac{2t}{t^{2}-3} \times \frac{-2t}{1-3t^{2}} = -1$.
$\frac{-4t^{2}}{(t^{2}-3)(1-3t^{2})} = -1 \Rightarrow 4t^{2} = (t^{2}-3)(1-3t^{2}) = t^{2} - 3t^{4} - 3 + 9t^{2} = -3t^{4} + 10t^{2} - 3$.
$3t^{4} - 6t^{2} + 3 = 0 \Rightarrow 3(t^{2}-1)^{2} = 0 \Rightarrow t^{2} = 1$.
Thus,$P$ is $(1, 2)$ and $Q$ is $(1, -2)$.
The length of the chord $PQ$ is $4$,which is the length of the latus rectum of the ellipse $E$. Thus,$\frac{2b^{2}}{a} = 4 \Rightarrow b^{2} = 2a$.
The focus of the ellipse is $(ae, 0)$. Since $PQ$ is a focal chord,the line $x=1$ must pass through the focus $(ae, 0)$,so $ae = 1$.
Using $b^{2} = a^{2}(1-e^{2})$,we substitute $b^{2} = 2a$ and $e^{2} = \frac{1}{a^{2}}$:
$2a = a^{2}(1 - \frac{1}{a^{2}}) = a^{2} - 1$.
$a^{2} - 2a - 1 = 0$. Solving for $a$,$a = \frac{2 \pm \sqrt{4+4}}{2} = 1 \pm \sqrt{2}$. Since $a>0$,$a = 1+\sqrt{2}$.
Then $e^{2} = \frac{1}{a^{2}} = \frac{1}{(1+\sqrt{2})^{2}} = \frac{1}{1+2+2\sqrt{2}} = \frac{1}{3+2\sqrt{2}} = 3-2\sqrt{2}$.
Therefore,$\frac{1}{e^{2}} = \frac{1}{3-2\sqrt{2}} = 3+2\sqrt{2}$.