Parabola Questions in English

(D) The equation of the parabola is $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$.
The directrix of the parabola is given by $x = -a$,which is $x = -1$.
The given point is $(-1, -60)$. Since the $x$-coordinate of the point is $-1$,the point lies on the directrix of the parabola.
$A$ property of parabolas states that the tangents drawn from any point on the directrix to the parabola are perpendicular to each other.
Therefore,the angle between the two tangents is $90^o$.

(C) The given equation of the conic is ${x^2} + 10x - 16y + 25 = 0$.
Rewriting the equation by completing the square: ${(x + 5)^2} - 25 - 16y + 25 = 0$,which simplifies to ${(x + 5)^2} = 16y$.
This is a parabola of the form ${(x - h)^2} = 4a(y - k)$,where $h = -5$,$k = 0$,and $4a = 16$,so $a = 4$.
The focus of the parabola is $(h, k + a) = (-5, 0 + 4) = (-5, 4)$.
The latus rectum is a horizontal line segment passing through the focus $(-5, 4)$ at $y = 4$.
Substituting $y = 4$ into the equation ${(x + 5)^2} = 16(4) = 64$,we get $x + 5 = \pm 8$.
Thus,$x = -5 + 8 = 3$ and $x = -5 - 8 = -13$.
The endpoints of the latus rectum are $(3, 4)$ and $(-13, 4)$.

(D) The given curves are $x^2 = 8y$ and $y^2 = 8x$.
For the curve $x^2 = 8y$,differentiating with respect to $x$ gives $2x = 8 \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x}{4}$. At the origin $(0, 0)$,the slope $m_1 = 0$.
For the curve $y^2 = 8x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 8$,so $\frac{dy}{dx} = \frac{4}{y}$. At the origin $(0, 0)$,the slope $m_2$ is undefined (vertical tangent).
Since one tangent is horizontal ($x$-axis) and the other is vertical ($y$-axis),the angle of intersection at the origin is $\pi / 2$.

(C) The equation of the parabola is $x^2 = 4y$,which is of the form $x^2 = 4ay$,where $a = 1$.
The equation of the line is $y = mx + c$,where $m = 2$ and $c = k$.
$A$ line $y = mx + c$ is a tangent to the parabola $x^2 = 4ay$ if $c = -am^2$.
Substituting the values $a = 1$ and $m = 2$ into the condition:
$k = -(1)(2)^2$
$k = -4$.

(C) Given the line $x + y = 0$ $(i)$ and the circle $x^2 + y^2 + 4y = 0$ $(ii)$.
Substitute $x = -y$ from $(i)$ into $(ii)$:
$(-y)^2 + y^2 + 4y = 0$
$2y^2 + 4y = 0$
$2y(y + 2) = 0$
So,$y = 0$ or $y = -2$.
If $y = 0$,then $x = 0$. Point is $(0, 0)$.
If $y = -2$,then $x = 2$. Point is $(2, -2)$.
Now,check the options for a parabola passing through $(0, 0)$ and $(2, -2)$:
For $y^2 = 2x$:
At $(0, 0)$: $0^2 = 2(0) \implies 0 = 0$ (Satisfied).
At $(2, -2)$: $(-2)^2 = 2(2) \implies 4 = 4$ (Satisfied).
Thus,the correct equation is $y^2 = 2x$.

(C) Given parabola is $y^2 = 2ax$.
Focus is $(a/2, 0)$ and the directrix is $x = -a/2$.
Since the circle is tangent to the directrix,its radius $r$ is the distance from the focus $(a/2, 0)$ to the line $x = -a/2$,which is $r = |a/2 - (-a/2)| = a$.
The equation of the circle with center $(a/2, 0)$ and radius $a$ is $(x - a/2)^2 + y^2 = a^2$.
Substituting $y^2 = 2ax$ into the circle equation:
$(x - a/2)^2 + 2ax = a^2$
$x^2 - ax + a^2/4 + 2ax = a^2$
$x^2 + ax - 3a^2/4 = 0$
Using the quadratic formula $x = \frac{-a \pm \sqrt{a^2 - 4(1)(-3a^2/4)}}{2} = \frac{-a \pm \sqrt{a^2 + 3a^2}}{2} = \frac{-a \pm 2a}{2}$.
So,$x = a/2$ or $x = -3a/2$.
For $x = a/2$,$y^2 = 2a(a/2) = a^2$,so $y = \pm a$.
For $x = -3a/2$,$y^2 = 2a(-3a/2) = -3a^2$,which gives no real values for $y$.
Thus,the points of intersection are $(a/2, \pm a)$.

(C) The equation of the line is $y = mx + c$. Given $dy/dx = m = 1$.
Since the line passes through $(0, 1)$,we have $1 = 1(0) + c$,which gives $c = 1$.
Thus,the equation of the line is $y = x + 1$.
To find the intersection with the parabola $y^2 = 4x$,substitute $y = x + 1$ into the parabola equation:
$(x + 1)^2 = 4x$
$x^2 + 2x + 1 = 4x$
$x^2 - 2x + 1 = 0$
$(x - 1)^2 = 0$
This gives $x = 1$. Substituting $x = 1$ into $y = x + 1$,we get $y = 2$.
Since the line intersects the parabola at only one point $(1, 2)$,the line is tangent to the parabola.
Therefore,the length intercepted by the curve on the line is $0$.

(C) The given parabola is $y^2 = -4x$. Comparing this with $y^2 = -4ax$,we get $a = 1$.
The focus of the parabola is $(-a, 0)$,which is $(-1, 0)$.
The slope $m$ of the line is $\tan(120^\circ) = -\sqrt{3}$.
The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
Substituting the values,we get $y - 0 = -\sqrt{3}(x - (-1))$.
$y = -\sqrt{3}(x + 1)$.
$y + \sqrt{3}(x + 1) = 0$.

(B) Let the two endpoints of the latus rectum be $L_1$ and $L_2$. The length of the latus rectum is $4a$,where $a$ is the distance from the vertex to the focus.
Since the latus rectum is perpendicular to the axis of the parabola,the axis must be the perpendicular bisector of the segment $L_1L_2$.
There are two possible directions for the parabola to open along this axis (either towards the focus or away from it),resulting in two distinct parabolas.

(D) Let the normal be drawn at point $P(at_1^2, 2at_1)$ on the parabola $y^2 = 4ax$. If this normal meets the parabola again at point $Q(at_2^2, 2at_2)$,then the relation between the parameters is $t_2 = -t_1 - \frac{2}{t_1}$.
Given that at point $Q$,the abscissa equals the ordinate,so $x = y$.
Substituting $x = y$ into the parabola equation $y^2 = 4ax$,we get $y^2 = 4ay$.
This implies $y(y - 4a) = 0$,so $y = 0$ or $y = 4a$.
For $y = 4a$,the point is $(4a, 4a)$.
Since the point $Q$ is $(at_2^2, 2at_2)$,we have $2at_2 = 4a$,which gives $t_2 = 2$.
Using the relation $t_2 = -t_1 - \frac{2}{t_1}$,we have $2 = -t_1 - \frac{2}{t_1}$,which leads to $t_1^2 + 2t_1 + 2 = 0$. This quadratic has no real roots for $t_1$.
However,the question asks for the point $Q$ where the normal meets the parabola. The condition $x=y$ at $Q$ implies $Q$ is $(4a, 4a)$ or $(0,0)$.
Testing the options,for $(9a, -6a)$,$x=9a, y=-6a$. The normal at $P(at_1^2, 2at_1)$ meets the parabola at $Q(at_2^2, 2at_2)$ where $t_2 = -t_1 - \frac{2}{t_1}$.
For $Q(9a, -6a)$,$at_2^2 = 9a \implies t_2^2 = 9 \implies t_2 = -3$ (since $2at_2 = -6a$).
Then $-3 = -t_1 - \frac{2}{t_1} \implies t_1^2 - 3t_1 + 2 = 0 \implies (t_1-1)(t_1-2) = 0$.
Thus,$t_1 = 1$ or $t_1 = 2$,which are valid parameters for the normal.

(A) The equation of the chord passing through the vertex $(0, 0)$ making an angle $\theta$ with the axis ($x$-axis) is $y = x \tan \theta$.
Substituting $y = x \tan \theta$ into the parabola equation $y^2 = 4ax$:
$(x \tan \theta)^2 = 4ax$
$x^2 \tan^2 \theta = 4ax$
$x(x \tan^2 \theta - 4a) = 0$
So,$x = 0$ or $x = \frac{4a}{\tan^2 \theta}$.
For $x = \frac{4a}{\tan^2 \theta}$,$y = \frac{4a}{\tan^2 \theta} \cdot \tan \theta = \frac{4a}{\tan \theta}$.
The intersection points are $(0, 0)$ and $(\frac{4a}{\tan^2 \theta}, \frac{4a}{\tan \theta})$.
The length of the chord $L = \sqrt{(\frac{4a}{\tan^2 \theta} - 0)^2 + (\frac{4a}{\tan \theta} - 0)^2}$
$L = \sqrt{\frac{16a^2}{\tan^4 \theta} + \frac{16a^2}{\tan^2 \theta}} = \frac{4a}{\tan \theta} \sqrt{\frac{1}{\tan^2 \theta} + 1}$
$L = \frac{4a}{\tan \theta} \sqrt{\csc^2 \theta} = \frac{4a}{\tan \theta} \cdot \csc \theta = 4a \cdot \frac{\cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta} = 4a \cos \theta \csc^2 \theta$.

(A) The equation of the chord passing through $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ is given by $\frac{y - 2at_1}{x - at_1^2} = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2a(t_2 - t_1)}{a(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$.
Since the chord passes through the focus $(a, 0)$,we substitute $x = a$ and $y = 0$ into the equation:
$\frac{0 - 2at_1}{a - at_1^2} = \frac{2}{t_1 + t_2}$.
$\frac{-2at_1}{a(1 - t_1^2)} = \frac{2}{t_1 + t_2}$.
$\frac{-t_1}{1 - t_1^2} = \frac{1}{t_1 + t_2}$.
$-t_1(t_1 + t_2) = 1 - t_1^2$.
$-t_1^2 - t_1t_2 = 1 - t_1^2$.
$-t_1t_2 = 1$,which implies $t_1t_2 = -1$.

(C) Let the moving point on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.
The focus of the parabola is $S(a, 0)$.
Let $(h, k)$ be the midpoint of the line segment joining $(at^2, 2at)$ and $(a, 0)$.
Then $h = \frac{at^2 + a}{2}$ and $k = \frac{2at + 0}{2} = at$.
From $k = at$,we have $t = \frac{k}{a}$.
Substituting $t$ into the expression for $h$: $h = \frac{a(\frac{k}{a})^2 + a}{2} = \frac{\frac{k^2}{a} + a}{2} = \frac{k^2 + a^2}{2a}$.
Thus,$2ah = k^2 + a^2$,which simplifies to $k^2 = 2a(h - \frac{a}{2})$.
The locus is $y^2 = 2a(x - \frac{a}{2})$.
Comparing this with the standard form $Y^2 = 4AX$,where $Y = y$,$X = x - \frac{a}{2}$,and $4A = 2a$ (so $A = \frac{a}{2}$).
The directrix is given by $X = -A$,which means $x - \frac{a}{2} = -\frac{a}{2}$.
Therefore,$x = 0$.

(A) The shortest distance between a curve and a line occurs at a point where the tangent to the curve is parallel to the given line.
Given the parabola $y = x^2$,the slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = 2x$.
The given line is $y = 2x - 4$,which has a slope of $2$.
For the tangent to be parallel to the line,we set the slope of the tangent equal to the slope of the line:
$2x = 2 \Rightarrow x = 1$.
Substituting $x = 1$ into the parabola equation $y = x^2$,we get $y = (1)^2 = 1$.
Thus,the point on the parabola closest to the line is $(1, 1)$.

(D) The focus $S$ is $\left( \frac{u^2}{2g} \sin 2\alpha, -\frac{u^2}{2g} \cos 2\alpha \right)$ and the directrix is $y = \frac{u^2}{2g}$.
The distance $d$ from the focus to the directrix is the perpendicular distance from the point $S$ to the line $y = \frac{u^2}{2g}$.
$d = \left| \frac{u^2}{2g} - \left( -\frac{u^2}{2g} \cos 2\alpha \right) \right| = \frac{u^2}{2g} (1 + \cos 2\alpha)$.
Using the trigonometric identity $1 + \cos 2\alpha = 2 \cos^2 \alpha$,we get $d = \frac{u^2}{2g} (2 \cos^2 \alpha) = \frac{u^2}{g} \cos^2 \alpha$.
The length of the latus-rectum of a parabola is $2 \times (\text{distance from focus to directrix}) = 2d$.
Length $= 2 \times \frac{u^2}{g} \cos^2 \alpha = \frac{2u^2}{g} \cos^2 \alpha$.

(C) The given equation of the parabola is ${y^2} - kx + 8 = 0$,which can be rewritten as ${y^2} = k(x - 8/k)$.
This is in the form ${Y^2} = 4AX$,where $Y = y$,$X = x - 8/k$,and $4A = k$,so $A = k/4$.
The directrix of the parabola ${Y^2} = 4AX$ is $X + A = 0$.
Substituting the values of $X$ and $A$,we get $(x - 8/k) + k/4 = 0$,which simplifies to $x = 8/k - k/4$.
We are given that the directrix is $x - 1 = 0$,or $x = 1$.
Equating the two expressions for $x$,we have $8/k - k/4 = 1$.
Multiplying by $4k$,we get $32 - k^2 = 4k$,or ${k^2} + 4k - 32 = 0$.
Factoring the quadratic equation,we get $(k + 8)(k - 4) = 0$.
Thus,the values of $k$ are $k = -8$ or $k = 4$.

(B) For the parabola $y^2 = 4ax$,differentiating with respect to $x$ gives:
$2y \frac{dy}{dx} = 4a \implies \left(\frac{dy}{dx}\right)_1 = \frac{2a}{y} \dots (i)$
For the curve $y = e^{-x/2a}$,differentiating with respect to $x$ gives:
$\left(\frac{dy}{dx}\right)_2 = e^{-x/2a} \left(-\frac{1}{2a}\right) = -\frac{y}{2a} \dots (ii)$
Two curves intersect orthogonally if the product of their slopes at the point of intersection is $-1$:
$\left(\frac{dy}{dx}\right)_1 \times \left(\frac{dy}{dx}\right)_2 = \left(\frac{2a}{y}\right) \times \left(-\frac{y}{2a}\right) = -1$
Since the product is $-1$,the curve $y = e^{-x/2a}$ cuts the parabola $y^2 = 4ax$ at right angles.

(C) Let the focus be $S(0, 0)$ and the tangent at the vertex be $x - y + 1 = 0$.
Since the directrix is parallel to the tangent at the vertex,let the equation of the directrix be $x - y + \lambda = 0$.
The distance from the focus $S(0, 0)$ to the tangent at the vertex is $d_1 = \frac{|0 - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$.
The distance from the focus $S(0, 0)$ to the directrix is $d_2 = \frac{|0 - 0 + \lambda|}{\sqrt{1^2 + (-1)^2}} = \frac{|\lambda|}{\sqrt{2}}$.
Since the distance from the focus to the directrix is twice the distance from the focus to the tangent at the vertex,we have $d_2 = 2d_1$.
$\frac{|\lambda|}{\sqrt{2}} = 2 \times \frac{1}{\sqrt{2}} \implies |\lambda| = 2$.
Since the focus $(0, 0)$ and the tangent at the vertex $x - y + 1 = 0$ are on opposite sides of the directrix,the directrix is $x - y + 2 = 0$.
For any point $P(x, y)$ on the parabola,the distance to the focus equals the distance to the directrix: $SP^2 = PM^2$.
$x^2 + y^2 = \left(\frac{|x - y + 2|}{\sqrt{1^2 + (-1)^2}}\right)^2$.
$x^2 + y^2 = \frac{(x - y + 2)^2}{2}$.
$2(x^2 + y^2) = x^2 + y^2 + 4 - 2xy + 4x - 4y$.
$x^2 + y^2 + 2xy - 4x + 4y - 4 = 0$.

(D) The given parabolas are $y^2 = 4ax$ $(i)$ and $x^2 = 4ay$ $(ii)$.
Substituting $y = \frac{x^2}{4a}$ from $(ii)$ into $(i)$,we get $\frac{x^4}{16a^2} = 4ax$.
This simplifies to $x^4 - 64a^3x = 0$,or $x(x^3 - 64a^3) = 0$.
Thus,$x = 0$ or $x = 4a$.
For $x = 0$,$y = 0$. For $x = 4a$,$y = 4a$.
The points of intersection are $A(0, 0)$ and $B(4a, 4a)$.
The line $2bx + 3cy + 4d = 0$ passes through $(0, 0)$,so $2b(0) + 3c(0) + 4d = 0$,which implies $d = 0$.
The line also passes through $(4a, 4a)$,so $2b(4a) + 3c(4a) + 4(0) = 0$.
Dividing by $4a$ (since $a \ne 0$),we get $2b + 3c = 0$.
Since $d = 0$ and $2b + 3c = 0$,it follows that $d^2 + (2b + 3c)^2 = 0$.

(A) Let the midpoint of the chord be $(x_1, y_1)$.
The equation of the chord of the parabola $y^2 = 4ax$ with midpoint $(x_1, y_1)$ is given by $T = S_1$,which is $yy_1 - 2a(x + x_1) = y_1^2 - 4ax_1$.
This simplifies to $yy_1 - 2ax = y_1^2 - 2ax_1$,or $\frac{yy_1 - 2ax}{y_1^2 - 2ax_1} = 1$.
To find the lines joining the vertex $(0, 0)$ to the intersection points of the chord and the parabola,we homogenize the equation of the parabola $y^2 - 4ax = 0$ using the chord equation:
$y^2 - 4ax(1) = 0 \implies y^2 - 4ax \left( \frac{yy_1 - 2ax}{y_1^2 - 2ax_1} \right) = 0$.
Expanding this,we get $y^2(y_1^2 - 2ax_1) - 4axyy_1 + 8a^2x^2 = 0$.
Since the chord subtends a right angle at the vertex,the lines are perpendicular,meaning the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$8a^2 + (y_1^2 - 2ax_1) = 0$.
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $y^2 - 2ax + 8a^2 = 0$.

(A) Given the equation of the parabola: $y^2 = 18x$.
Differentiating both sides with respect to $t$:
$2y \left( \frac{dy}{dt} \right) = 18 \left( \frac{dx}{dt} \right)$.
We are given that the ordinate $(y)$ increases at twice the rate of the abscissa $(x)$,so $\frac{dy}{dt} = 2 \frac{dx}{dt}$.
Substituting this into the differentiated equation:
$2y \left( 2 \frac{dx}{dt} \right) = 18 \left( \frac{dx}{dt} \right)$.
Assuming $\frac{dx}{dt} \neq 0$,we get:
$4y = 18 \implies y = \frac{18}{4} = \frac{9}{2}$.
Now,substitute $y = \frac{9}{2}$ into the original parabola equation to find $x$:
$\left( \frac{9}{2} \right)^2 = 18x$
$\frac{81}{4} = 18x$
$x = \frac{81}{4 \times 18} = \frac{81}{72} = \frac{9}{8}$.
Thus,the required point is $\left( \frac{9}{8}, \frac{9}{2} \right)$.

(A) Given $x = t^2$ and $y = 2t$.
At $t = 1$,$x = 1^2 = 1$ and $y = 2(1) = 2$.
So,the point is $(1, 2)$.
Now,$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t}$.
At $t = 1$,the slope of the tangent is $m = \frac{1}{1} = 1$.
The slope of the normal is $m' = -\frac{1}{m} = -\frac{1}{1} = -1$.
The equation of the normal at $(1, 2)$ is $y - y_1 = m'(x - x_1)$.
$y - 2 = -1(x - 1)$.
$y - 2 = -x + 1$.
$x + y - 3 = 0$.

(D) Given the curve equation $x^2 = -4y$.
Differentiating both sides with respect to $x$,we get:
$2x = -4 \frac{dy}{dx}$
$\frac{dy}{dx} = -\frac{x}{2}$
Now,find the slope of the tangent at the point $(-4, -4)$:
$m = \left( \frac{dy}{dx} \right)_{(-4, -4)} = -\frac{-4}{2} = 2$.
The equation of the tangent line at $(x_1, y_1)$ is given by $(y - y_1) = m(x - x_1)$.
Substituting the values $m = 2$,$x_1 = -4$,and $y_1 = -4$:
$y - (-4) = 2(x - (-4))$
$y + 4 = 2(x + 4)$
$y + 4 = 2x + 8$
$2x - y + 4 = 0$.

(C) Let the focal chord of the parabola $y^2 = 4ax$ be divided into two segments of lengths $b$ and $c$ by the focus $(a, 0)$.
If the semi-latus rectum of the parabola is $l$,then the lengths of the segments of a focal chord are related to the semi-latus rectum by the harmonic mean property.
Specifically,the semi-latus rectum $l$ is the harmonic mean of the segments $b$ and $c$.
Therefore,$\frac{1}{l} = \frac{1}{2} (\frac{1}{b} + \frac{1}{c})$.
$\frac{1}{l} = \frac{1}{2} (\frac{b + c}{bc})$.
Thus,$l = \frac{2bc}{b + c}$.

(C) The equation of a focal chord of the parabola $y^2 = 4ax$ passing through the focus $(a, 0)$ with slope $m = \tan \theta$ is given by:
$y - 0 = \tan \theta (x - a)$
$y = \tan \theta (x - a)$
$\tan \theta \cdot x - y - a \tan \theta = 0$
The length of the perpendicular from the vertex $(0, 0)$ to this line is given by the formula $d = \left| \frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} \right|$.
Substituting the values:
$d = \left| \frac{\tan \theta (0) - 1(0) - a \tan \theta}{\sqrt{\tan^2 \theta + (-1)^2}} \right|$
$d = \left| \frac{-a \tan \theta}{\sqrt{\sec^2 \theta}} \right|$
$d = \left| \frac{-a \tan \theta}{\sec \theta} \right|$
Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$,we get:
$d = a \sin \theta$

(C) In the context of a parabola,a diameter is defined as the locus of the midpoints of a system of parallel chords.
For a parabola $y^2 = 4ax$,the diameter corresponding to a system of parallel chords with slope $m$ is given by the line $y = \frac{2a}{m}$.
Since $a$ and $m$ are constants,this line is parallel to the axis of the parabola $(y = 0)$.
Therefore,the diameter of a parabola is a line parallel to its axis.

(C) The equation of the parabola is $y^2 = -12x$,where $4a = 12$,so $a = -3$.
The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Substituting $a = -3$,the tangent equation is $y = mx - \frac{3}{m}$.
Since the tangent passes through $(3, 8)$,we have $8 = 3m - \frac{3}{m}$.
Multiplying by $m$,we get $8m = 3m^2 - 3$,which simplifies to $3m^2 - 8m - 3 = 0$.
Factoring the quadratic equation: $3m^2 - 9m + m - 3 = 0 \implies 3m(m - 3) + 1(m - 3) = 0$.
Thus,$(3m + 1)(m - 3) = 0$,giving $m = 3$ or $m = -1/3$.

(A) Given equation: $9x^2 - 6x + 36y + 9 = 0$
Rearranging the terms: $9x^2 - 6x = -36y - 9$
Divide by $9$: $x^2 - (2/3)x = -4y - 1$
Complete the square on the left side: $x^2 - (2/3)x + (1/9) = -4y - 1 + (1/9)$
$(x - 1/3)^2 = -4y - 8/9$
$(x - 1/3)^2 = -4(y + 2/9)$
Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,the vertex $(h, k)$ is $(1/3, -2/9)$.

(C) The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is given by $y + tx = 2at + at^3$.
Given the point $(\frac{a}{m^2}, \frac{2a}{m})$,we have $t = \frac{1}{m}$.
Substituting $t = \frac{1}{m}$ into the normal equation:
$y + (\frac{1}{m})x = 2a(\frac{1}{m}) + a(\frac{1}{m})^3$
Multiply the entire equation by $m^3$:
$m^3y + m^2x = 2am^2 + a$
Rearranging the terms to match the form $m^3y = 2am^2 - m^2x + a$.

(D) The axis of the parabola is the line passing through the vertex and perpendicular to the directrix. Since the directrix is the $y$-axis $(x = 0)$,the axis of the parabola is the $x$-axis.
Let the focus be $S(a, 0)$.
The vertex is the midpoint of the segment joining the focus and the point of intersection of the axis and the directrix.
The point of intersection of the axis ($x$-axis) and the directrix ($y$-axis) is $Z(0, 0)$.
Since the vertex $(2, 0)$ is the midpoint of $ZS$,we have:
$\frac{a + 0}{2} = 2 \implies a = 4$.
Thus,the focus is $(4, 0)$.

(C) The equation of a tangent to the parabola $y = x^2$ with slope $m$ is given by $y = mx - \frac{m^2}{4} \dots (1)$.
The equation of the parabola $y = -(x - 2)^2$ can be written as $(x - 2)^2 = -y$. The equation of a tangent to this parabola with slope $m$ is $y - k = m(x - h) + \frac{a}{m}$,where the vertex is $(h, k) = (2, 0)$ and $4a = -1 \implies a = -1/4$.
Thus,the tangent equation is $y = m(x - 2) + \frac{-1/4}{m} = mx - 2m - \frac{1}{4m} \dots (2)$.
For the lines to be the same,the intercepts must be equal: $-\frac{m^2}{4} = -2m - \frac{1}{4m}$.
Multiplying by $-4m$,we get $m^3 = 8m^2 + 1 \implies m^3 - 8m^2 - 1 = 0$. This approach is complex; let us use the condition for common tangents.
Alternatively,for $y = x^2$ and $y = -(x-2)^2$,the common tangents are $y = 0$ and $y = 4(x - 1)$.

(D) Step $1$: Express $y$ in terms of $x$ from the line equation.
Given $x = my + \frac{a}{m}$,we have $my = x - \frac{a}{m}$,so $y = \frac{x}{m} - \frac{a}{m^2}$.
Step $2$: Substitute $y$ into the parabola equation $x^2 = 4ay$.
$x^2 = 4a \left( \frac{x}{m} - \frac{a}{m^2} \right)$
$x^2 = \frac{4ax}{m} - \frac{4a^2}{m^2}$
$m^2x^2 - 4amx + 4a^2 = 0$
Step $3$: Solve the quadratic equation for $x$.
$(mx - 2a)^2 = 0$
$mx = 2a \implies x = \frac{2a}{m}$.
Step $4$: Find the corresponding $y$ coordinate.
$y = \frac{x}{m} - \frac{a}{m^2} = \frac{2a}{m^2} - \frac{a}{m^2} = \frac{a}{m^2}$.
Thus,the point of contact is $\left( \frac{2a}{m}, \frac{a}{m^2} \right)$.

(C) For a parabola $y^2 = 4ax$,the semi-latus rectum is the harmonic mean of the segments of any focal chord.
Given $y^2 = 8x$,we have $4a = 8$,so $a = 2$.
The semi-latus rectum $l = 2a = 4$.
Let the segments of the focal chord be $SP$ and $SQ$. Then $SP, l, SQ$ are in harmonic progression.
Therefore,$l = \frac{2(SP)(SQ)}{SP + SQ}$.
Substituting the values $l = 4$ and $SP = 6$:
$4 = \frac{2(6)(SQ)}{6 + SQ}$
$4 = \frac{12(SQ)}{6 + SQ}$
$4(6 + SQ) = 12(SQ)$
$24 + 4(SQ) = 12(SQ)$
$24 = 8(SQ)$
$SQ = 3$.

(B) Given the equation of the parabola: $y^2 + 6x - 2y + 13 = 0$
Rearranging the terms to complete the square for $y$:
$y^2 - 2y = -6x - 13$
Adding $1$ to both sides:
$y^2 - 2y + 1 = -6x - 13 + 1$
$(y - 1)^2 = -6x - 12$
$(y - 1)^2 = -6(x + 2)$
Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex:
Here,$h = -2$ and $k = 1$.
Therefore,the vertex is $(-2, 1)$.

(A) The given parabola is $y^2 = 12x$. Comparing this with $y^2 = 4ax$,we get $4a = 12$,so $a = 3$.
The coordinates of the upper end of the latus rectum are $(a, 2a) = (3, 6)$.
The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.
Substituting $(x_1, y_1) = (3, 6)$ and $a = 3$,we get:
$y(6) = 2(3)(x + 3)$
$6y = 6(x + 3)$
$y = x + 3$.

(B) The parametric coordinates of any point on the parabola $y^2 = 4ax$ are given by $(at^2, 2at)$.
Comparing $y^2 = 9x$ with $y^2 = 4ax$,we get $4a = 9$,which implies $a = 9/4$.
The $y$-coordinate of the point is $2at = -6$.
Substituting the value of $a$:
$2(9/4)t = -6$
$(9/2)t = -6$
$t = -6 \times (2/9)$
$t = -12/9 = -4/3$.

(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to a parabola $S = 0$ is given by $SS_1 = T^2$.
Here,$S = y^2 - 12x = 0$ and the point is $(x_1, y_1) = (1, 4)$.
$S_1 = (4)^2 - 12(1) = 16 - 12 = 4$.
The tangent $T$ at $(x_1, y_1)$ is given by $yy_1 - 6(x + x_1) = 0$,which is $4y - 6(x + 1) = 4y - 6x - 6$.
Substituting into $SS_1 = T^2$:
$(y^2 - 12x)(4) = (4y - 6x - 6)^2$.
$4y^2 - 48x = 16y^2 + 36x^2 + 36 - 48xy - 48y + 72x$.
Rearranging the terms:
$36x^2 + 12y^2 - 48xy + 120x - 48y + 36 = 0$.
Dividing by $12$:
$3x^2 + y^2 - 4xy + 10x - 4y + 3 = 0$.

(B) The equation of the parabola is $y^2 = 6x$. Comparing this with $y^2 = 4ax$,we get $4a = 6$,so $a = 3/2$.
The vertex of the parabola is $(0, 0)$.
The coordinates of the ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$.
The negative end of the latus rectum is $(a, -2a) = (3/2, -2 \times 3/2) = (3/2, -3)$.
The chord passes through the vertex $(0, 0)$ and the point $(3/2, -3)$.
The slope $m$ of the line passing through $(0, 0)$ and $(3/2, -3)$ is $m = \frac{-3 - 0}{3/2 - 0} = \frac{-3}{3/2} = -2$.
The equation of the line is $y - 0 = -2(x - 0)$,which simplifies to $y = -2x$ or $y + 2x = 0$.

(D) Given the parametric equations:
$x - 2 = t^2$
$y = 2t$
From the second equation,we have $t = \frac{y}{2}$.
Substituting this value of $t$ into the first equation:
$x - 2 = (\frac{y}{2})^2$
$x - 2 = \frac{y^2}{4}$
$y^2 = 4(x - 2)$
Thus,the given equations represent the parabola $y^2 = 4(x - 2)$.

(D) For the parabola $y^2 = 4ax$,we have $a = 4$.
The equation of a normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.
Alternatively,the equation of a normal in terms of slope $m$ is $y = mx - 2am - am^3$.
Given the options,we test the form $y = m(x - c_1) - c_2$. Comparing with the options,let $m = \cos \theta$.
The equation of the normal is $y = m(x - 2a) - am^3$.
Substituting $a = 4$ and $m = \cos \theta$:
$y = \cos \theta (x - 2(4)) - 4 \cos^3 \theta$
$y = (x - 8) \cos \theta - 4 \cos^3 \theta$
Using the identity $\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta$,we have $4 \cos^3 \theta = \cos 3\theta + 3 \cos \theta$.
Substituting this into the equation:
$y = x \cos \theta - 8 \cos \theta - (\cos 3\theta + 3 \cos \theta)$
$y = x \cos \theta - 11 \cos \theta - \cos 3\theta$
$y = (x - 11) \cos \theta - \cos 3\theta$.
Thus,option $D$ is correct.

(C) Given the equation of the parabola: $x^2 - 4x - 3y + 10 = 0$
Rearranging the terms to complete the square for $x$:
$x^2 - 4x = 3y - 10$
Adding $4$ to both sides:
$x^2 - 4x + 4 = 3y - 10 + 4$
$(x - 2)^2 = 3y - 6$
$(x - 2)^2 = 3(y - 2)$
This is in the form $(x - h)^2 = 4a(y - k)$,where the axis of the parabola is given by $x - h = 0$.
Comparing,we get $h = 2$.
Therefore,the equation of the axis is $x - 2 = 0$.

(B) The standard equation of a parabola is $y^2 = 4ax$.
Comparing $y^2 = 8x$ with $y^2 = 4ax$,we get $4a = 8$,which implies $a = 2$.
The parametric equations for the parabola $y^2 = 4ax$ are given by $x = at^2$ and $y = 2at$.
Substituting $a = 2$ into these equations,we get:
$x = 2t^2$
$y = 2(2)t = 4t$
Thus,the parametric equations are $x = 2t^2$ and $y = 4t$.

(D) Since the point $(3, 2)$ lies on the parabola $y^2 = 4ax$,we have:
$2^2 = 4a(3)$ $\Rightarrow 4 = 12a$ $\Rightarrow a = \frac{1}{3}$.
Using the formula for the tangent $T = 0$ at point $(x_1, y_1)$ for $y^2 = 4ax$,which is $yy_1 = 2a(x + x_1)$:
$y(2) = 2(\frac{1}{3})(x + 3)$.
$2y = \frac{2}{3}(x + 3)$.
$3y = x + 3$.

(C) For a parabola $y^2 = 4ax$,the length of the subnormal at any point $(x_1, y_1)$ is given by $|2a|$.
Given the equation $y^2 = 16x$,we have $4a = 16$,which implies $a = 4$.
The length of the subnormal is constant for all points on the parabola and is equal to $|2a| = |2 \times 4| = 8$.
Thus,at $x = 4$,the length of the subnormal is $8$.

(A) For a parabola,the length of the semi-latus rectum is the harmonic mean of the segments of any focal chord.
Let $l$ be the semi-latus rectum.
Then,$\frac{1}{SP} + \frac{1}{SQ} = \frac{2}{l}$.
Given $SP = 3$ and $SQ = 2$,we have:
$\frac{1}{3} + \frac{1}{2} = \frac{2}{l}$
$\frac{5}{6} = \frac{2}{l}$
$l = \frac{12}{5}$.
The length of the latus rectum is $2l = 2 \times \frac{12}{5} = \frac{24}{5}$.

(C) The equation of the parabola is $y^2 = 24x$,so $4a = 24$,which gives $a = 6$.
The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$.
Since the normal is parallel to the tangent $y = 2x + 3$,its slope $m$ must be $2$.
Substituting $a = 6$ and $m = 2$ into the normal equation:
$y = 2x - 2(6)(2) - 6(2)^3$
$y = 2x - 24 - 48$
$y = 2x - 72$.
The distance between two parallel lines $y = mx + c_1$ and $y = mx + c_2$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{1 + m^2}}$.
Here,$c_1 = 3$,$c_2 = -72$,and $m = 2$.
$d = \frac{|3 - (-72)|}{\sqrt{1 + 2^2}} = \frac{75}{\sqrt{5}} = \frac{75\sqrt{5}}{5} = 15\sqrt{5}$.

(D) Given equation: $x^2 + 4x + 2y - 7 = 0$
Rearranging the terms: $x^2 + 4x = -2y + 7$
Completing the square for $x$: $(x + 2)^2 - 4 = -2y + 7$
$(x + 2)^2 = -2y + 11$
$(x + 2)^2 = -2(y - 11/2)$
Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,the vertex $(h, k)$ is $(-2, 11/2)$.

(A) The standard equation of a parabola is $y^2 = 4ax$.
Comparing $y^2 = 12x$ with $y^2 = 4ax$,we get $4a = 12$,which implies $a = 3$.
The equation of the directrix for the parabola $y^2 = 4ax$ is given by $x = -a$.
Substituting the value of $a$,we get $x = -3$,which can be written as $x + 3 = 0$.

(D) For the parabola $y^2 = 4ax$, where $a = 1$, the chord of contact from point $(x_1, y_1) = (-1, 2)$ is given by $yy_1 = 2a(x + x_1)$.
Substituting the values, we get $y(2) = 2(1)(x - 1)$, which simplifies to $y = x - 1$ or $x - y - 1 = 0$.
The points of contact are found by solving $y^2 = 4x$ and $x = y + 1$, leading to $y^2 - 4y - 4 = 0$.
The length of the chord of contact $L$ is given by $\frac{\sqrt{(y_1^2 - 4ax_1)^3}}{a} = \frac{\sqrt{(2^2 - 4(1)(-1))^3}}{1} = \sqrt{8^3} = 16\sqrt{2}$.
The distance $d$ from the point $(-1, 2)$ to the line $x - y - 1 = 0$ is $d = \frac{|-1 - 2 - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
The area of the triangle formed by the tangents and the chord of contact is given by $\frac{1}{2} \times L \times d = \frac{1}{2} \times 16\sqrt{2} \times 2\sqrt{2} = 8 \times 2 \times 2 = 32$ (Wait, re-evaluating: The formula for the area of the triangle formed by tangents from $(x_1, y_1)$ to $y^2=4ax$ is $\frac{(y_1^2 - 4ax_1)^{3/2}}{2a} = \frac{(4+4)^{3/2}}{2} = \frac{8\sqrt{8}}{2} = 4(2\sqrt{2}) = 8\sqrt{2}$).