Parabola Questions in English

(A) The given parabola is $(y - 3)^2 = 12(x - 2)$.
Comparing this with $(y - k)^2 = 4a(x - h)$,we get $h = 2, k = 3$,and $4a = 12$,so $a = 3$.
The directrix of the parabola is $x = h - a = 2 - 3 = -1$.
It is a known property of parabolas that the locus of the intersection of two perpendicular tangents is the directrix.
Since the two tangents are perpendicular and meet at point $P$,point $P$ must lie on the directrix $x = -1$.
Substituting $x = -1$ into the equation of the first tangent $y = 3x - 2$:
$y = 3(-1) - 2 = -3 - 2 = -5$.
Therefore,the point $P$ is $(-1, -5)$.

(C) The given equation of the line is $m x - y + (10 + m^2) = 0$.
Rearranging this as a quadratic equation in $m$: $m^2 + m x + (10 - y) = 0$.
Since the line is a tangent to the parabola,the discriminant $D$ of this quadratic equation must be zero.
For $a m^2 + b m + c = 0$,$D = b^2 - 4ac = 0$.
Here,$a = 1$,$b = x$,and $c = (10 - y)$.
Substituting these values: $x^2 - 4(1)(10 - y) = 0$.
$x^2 - 40 + 4y = 0$.
$x^2 = -4y + 40$.
$x^2 = -4(y - 10)$.
Thus,the correct option is $C$.

(A) Let the coordinates of the end points of the double ordinate be $A(x_1, y_1)$ and $B(x_1, -y_1)$.
Given the length of the double ordinate $AB = 2y_1 = 16$,so $y_1 = 8$.
Thus,the coordinates are $A(x_1, 8)$ and $B(x_1, -8)$.
Since $A$ and $B$ lie on the parabola $y^2 = 8x$,we have $8^2 = 8x_1$,which gives $64 = 8x_1$,so $x_1 = 8$.
The coordinates of $A$ and $B$ are $(8, 8)$ and $(8, -8)$.
Let $\alpha$ be the angle made by $OA$ with the $X$-axis,where $O$ is the vertex $(0, 0)$.
Then $\tan \alpha = \frac{8}{8} = 1$,which implies $\alpha = \frac{\pi}{4}$.
The angle subtended by the double ordinate $AB$ at the vertex is $2\alpha = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$.

(B) The equation of the parabola is $y^2=4ax$. The focus is $(a, 0)$ and the directrix is $x=-a$.
The distance between the focus and the directrix is $2a$.
Given $2a = \frac{3}{2}$,so $a = \frac{3}{4}$.
The point on the parabola is $(4a, -4a) = (4 \times \frac{3}{4}, -4 \times \frac{3}{4}) = (3, -3)$.
The equation of the normal to the parabola $y^2=4ax$ at $(x_1, y_1)$ is $y-y_1 = -\frac{y_1}{2a}(x-x_1)$.
Substituting $x_1=3, y_1=-3$ and $a=\frac{3}{4}$:
$y - (-3) = -\frac{-3}{2(3/4)}(x-3)$
$y+3 = \frac{3}{3/2}(x-3)$
$y+3 = 2(x-3)$
$y+3 = 2x-6$
$2x-y=9$.

(C) For a parabola $y^2 = 4ax$,the tangent at point $t$ is given by $ty = x + at^2$. Here,$4a = 1$,so $a = 1/4$.
The intersection point of tangents at $t_i$ and $t_j$ is $(at_it_j, a(t_i + t_j))$.
Given $t_1 = 2, t_2 = -4, t_3 = 6$ and $a = 1/4$:
Point $L$ (intersection of $t_1, t_2$) = $(1/4(2)(-4), 1/4(2-4)) = (-2, -0.5)$.
Point $M$ (intersection of $t_2, t_3$) = $(1/4(-4)(6), 1/4(-4+6)) = (-6, 0.5)$.
Point $N$ (intersection of $t_3, t_1$) = $(1/4(6)(2), 1/4(6+2)) = (3, 2)$.
The area of triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area = $\frac{1}{2} |(-2)(0.5 - 2) + (-6)(2 - (-0.5)) + 3(-0.5 - 0.5)|$.
Area = $\frac{1}{2} |(-2)(-1.5) + (-6)(2.5) + 3(-1)|$.
Area = $\frac{1}{2} |3 - 15 - 3| = \frac{1}{2} |-15| = 7.5$.

(A) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Here,$4a = 8$,so $a = 2$.
The equation of the tangent is $y = mx + \frac{2}{m}$.
Since it passes through the point $(1, 3)$,we substitute $x = 1$ and $y = 3$:
$3 = m(1) + \frac{2}{m}$
$3 = m + \frac{2}{m}$
$m^2 - 3m + 2 = 0$
$(m - 1)(m - 2) = 0$
Thus,$m = 1$ or $m = 2$.
For $m = 1$,the tangent is $y = 1x + \frac{2}{1} \Rightarrow y = x + 2$.
For $m = 2$,the tangent is $y = 2x + \frac{2}{2} \Rightarrow y = 2x + 1$.
Comparing with the given options,$y = 2x + 1$ is present.

(C) The tangent at the vertex is given by $x-y+1=0 \dots (i)$.
The focus of the parabola is $S(0,0)$.
The distance from the focus to the tangent at the vertex is $a = \left|\frac{0-0+1}{\sqrt{1^2+(-1)^2}}\right| = \frac{1}{\sqrt{2}}$.
The directrix is parallel to the tangent at the vertex,so its equation is of the form $x-y+c=0$.
The distance from the focus to the directrix is $2a = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.
Thus,$\left|\frac{0-0+c}{\sqrt{1^2+(-1)^2}}\right| = \sqrt{2}$ $\Rightarrow \frac{|c|}{\sqrt{2}} = \sqrt{2}$ $\Rightarrow |c| = 2$.
Since the focus $(0,0)$ and the tangent at the vertex $x-y+1=0$ lie on the same side of the directrix,we choose $c=2$.
Therefore,the equation of the directrix is $x-y+2=0$.

(D) Given the parametric equations $x-2=t^2$ and $y=2t$,we have $x=t^2+2$ and $y=2t$.
Substituting these into the parabola equation $y^2=a(x-b)$:
$(2t)^2 = a(t^2+2-b)$
$4t^2 = at^2 + a(2-b)$
Comparing the coefficients of $t^2$ and the constant terms on both sides:
$a = 4$
$a(2-b) = 0$ $\Rightarrow 4(2-b) = 0$ $\Rightarrow b = 2$
Therefore,$a+b = 4+2 = 6$.

(D) The equation of the parabola is $y^2 = 12x$,which is of the form $y^2 = 4ax$,where $4a = 12$,so $a = 3$.
Let the point of contact be $(x_1, y_1)$.
The equation of the tangent to the parabola $y^2 = 4ax$ at $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
Substituting $a = 3$,we get $yy_1 = 6(x + x_1)$,which simplifies to $6x - y_1y + 6x_1 = 0$.
We are given the tangent equation as $x - \sqrt{3}y + 9 = 0$.
Comparing the two equations: $\frac{6}{1} = \frac{-y_1}{-\sqrt{3}} = \frac{6x_1}{9}$.
From $\frac{6}{1} = \frac{y_1}{\sqrt{3}}$,we get $y_1 = 6\sqrt{3}$.
From $\frac{6}{1} = \frac{6x_1}{9}$,we get $x_1 = 9$.
Thus,the point of contact is $(9, 6\sqrt{3})$.

(C) The equation of the parabola is $y^2=16x$,so $4a=16$,which gives $a=4$.
The given line is $3x-4y+5=0$,which can be written as $y=\frac{3}{4}x+\frac{5}{4}$.
The slope of this line is $m_1=\frac{3}{4}$.
Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m \times m_1 = -1$.
Thus,$m = -\frac{4}{3}$.
The equation of a tangent to the parabola $y^2=4ax$ with slope $m$ is $y=mx+\frac{a}{m}$.
Substituting $a=4$ and $m=-\frac{4}{3}$,we get $y=-\frac{4}{3}x+\frac{4}{-4/3}$.
$y=-\frac{4}{3}x-3$.
Multiplying by $3$,we get $3y=-4x-9$,which simplifies to $4x+3y+9=0$.

(D) Given the curve $y^2 = 4x$ and the point $P(1, 2)$.
Differentiating both sides with respect to $x$:
$2y \cdot \frac{dy}{dx} = 4$
$\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}$
At the point $P(1, 2)$,the slope of the tangent $m$ is:
$m = \left. \frac{dy}{dx} \right|_{(1, 2)} = \frac{2}{2} = 1$
Since the slope $m = \tan \theta$,we have:
$\tan \theta = 1$
$\theta = 45^{\circ}$
Thus,the correct option is $D$.

(B) The equation of the tangent to the parabola $y^2=4ax$ at point $(x_1, y_1)$ is given by $yy_1 = 2a(x+x_1)$.
Here,$4a = 12$,so $a = 3$.
The point is $(x_1, y_1) = (3, -6)$.
Substituting these values into the formula:
$y(-6) = 2(3)(x+3)$
$-6y = 6(x+3)$
$-y = x+3$
$x+y+3 = 0$
Hence,option $B$ is correct.

(D) The equation of the tangent to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx+\frac{a}{m}$.
Here,the parabola is $y^2=8x$,so $4a=8$,which implies $a=2$.
The inclination is $30^{\circ}$,so the slope $m=\tan(30^{\circ})=\frac{1}{\sqrt{3}}$.
Substituting these values into the tangent equation:
$y=\frac{1}{\sqrt{3}}x+\frac{2}{1/\sqrt{3}}$
$y=\frac{x}{\sqrt{3}}+2\sqrt{3}$
Multiplying by $\sqrt{3}$ on both sides:
$\sqrt{3}y=x+6$
$x-\sqrt{3}y+6=0$.
Thus,the correct option is $D$.

(C) Given the parabola $y^2 = 6x$.
Differentiating with respect to $x$,we get:
$2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y}$.
The given equation of the tangent is $5x - 2y + k = 0$,which can be written as $2y = 5x + k$,or $y = \frac{5}{2}x + \frac{k}{2}$.
The slope of this tangent is $m = \frac{5}{2}$.
Equating the slope of the tangent to the derivative at the point of contact:
$\frac{3}{y} = \frac{5}{2} \Rightarrow y = \frac{6}{5}$.
Substituting the value of $y$ into the parabola equation $y^2 = 6x$:
$\left(\frac{6}{5}\right)^2 = 6x$ $\Rightarrow \frac{36}{25} = 6x$ $\Rightarrow x = \frac{6}{25}$.
Thus,the point of contact is $\left(\frac{6}{25}, \frac{6}{5}\right)$.

(C) The given parabolas are $y^2=32x$ $(i)$ and $2x^2=27y$ (ii).
To find the intersection points,substitute $x = \frac{y^2}{32}$ into (ii):
$2(\frac{y^2}{32})^2 = 27y \implies 2 \cdot \frac{y^4}{1024} = 27y \implies \frac{y^4}{512} = 27y \implies y(y^3 - 512 \cdot 27) = 0$.
Thus,$y=0$ or $y^3 = (8^3)(3^3) = 24^3$,so $y=24$.
For $y=24$,$x = \frac{24^2}{32} = \frac{576}{32} = 18$. So $P = (18, 24)$.
For $y^2=32x$,differentiating gives $2y \frac{dy}{dx} = 32 \implies \frac{dy}{dx} = \frac{16}{y}$. At $P(18, 24)$,$m_1 = \frac{16}{24} = \frac{2}{3}$.
For $2x^2=27y$,differentiating gives $4x = 27 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{4x}{27}$. At $P(18, 24)$,$m_2 = \frac{4(18)}{27} = \frac{72}{27} = \frac{8}{3}$.
The tangent of the angle $\theta$ is given by $\tan \theta = |\frac{m_2 - m_1}{1 + m_1 m_2}| = |\frac{8/3 - 2/3}{1 + (8/3)(2/3)}| = |\frac{6/3}{1 + 16/9}| = |\frac{2}{25/9}| = \frac{18}{25}$.
Therefore,$5\sqrt{\tan \theta} = 5\sqrt{\frac{18}{25}} = 5 \cdot \frac{3\sqrt{2}}{5} = 3\sqrt{2}$.

(A) Let the tangent to the parabola $y^2=4ax$ at point $P(at^2, 2at)$ be $yt = x + at^2$ ... $(i)$.
Since the line $NM$ is perpendicular to the tangent and passes through the origin $O(0,0)$,its equation is $y = -tx$ ... (ii).
To find $N$,substitute $y = -tx$ into $(i)$: $t(-tx) = x + at^2 \implies -t^2x - x = at^2 \implies x = -\frac{at^2}{1+t^2}$.
Then $y = -t(-\frac{at^2}{1+t^2}) = \frac{at^3}{1+t^2}$. So,$N \equiv (-\frac{at^2}{1+t^2}, \frac{at^3}{1+t^2})$.
To find $M$,substitute $y = -tx$ into $y^2 = 4ax$: $(-tx)^2 = 4ax \implies t^2x^2 = 4ax$. Since $x \neq 0$,$x = \frac{4a}{t^2}$.
Then $y = -t(\frac{4a}{t^2}) = -\frac{4a}{t}$. So,$M \equiv (\frac{4a}{t^2}, -\frac{4a}{t})$.
Now,$ON = \sqrt{(-\frac{at^2}{1+t^2})^2 + (\frac{at^3}{1+t^2})^2} = \sqrt{\frac{a^2t^4(1+t^2)}{(1+t^2)^2}} = \frac{at^2}{\sqrt{1+t^2}}$.
And $OM = \sqrt{(\frac{4a}{t^2})^2 + (-\frac{4a}{t})^2} = \sqrt{\frac{16a^2}{t^4} + \frac{16a^2}{t^2}} = \frac{4a}{t^2} \sqrt{1+t^2}$.
Therefore,$ON \cdot OM = \frac{at^2}{\sqrt{1+t^2}} \cdot \frac{4a}{t^2} \sqrt{1+t^2} = 4a^2$.

(D) Let the coordinates of points $A$ and $B$ be $(x_1, \alpha_1)$ and $(x_2, \alpha_2)$ respectively. Since they lie on $y^2=4ax$,we have $\alpha_1^2=4ax_1$ and $\alpha_2^2=4ax_2$.
The tangent at point $(x_i, \alpha_i)$ is given by $y\alpha_i = 2a(x+x_i)$.
The intersection point $(x_3, \alpha_3)$ of the tangents at $A$ and $B$ is given by $\alpha_3 = \frac{\alpha_1+\alpha_2}{2}$.
From this,we have $2\alpha_3 = \alpha_1+\alpha_2$.
Rearranging the terms,we get $\alpha_3-\alpha_2 = \alpha_1-\alpha_3$.

(A) Let the point $P$ on the parabola $y^2 = 8x$ be $(2t^2, 4t)$.
Given $A = (-2, 0)$,let $Q = (h, k)$ be the midpoint of $\overline{AP}$.
Then $h = \frac{2t^2 - 2}{2} = t^2 - 1$ and $k = \frac{4t + 0}{2} = 2t$.
From $k = 2t$,we have $t = \frac{k}{2}$.
Substituting $t$ into the equation for $h$: $h = (\frac{k}{2})^2 - 1 = \frac{k^2}{4} - 1$.
Rearranging gives $k^2 = 4(h + 1)$.
The locus of $Q$ is $y^2 = 4(x + 1)$.
Comparing this with the standard form $Y^2 = 4aX$,where $Y = y$,$X = x + 1$,and $4a = 4$,we get $a = 1$.
The vertex of this parabola is $(-1, 0)$.
The focus is at $(X + a, Y) = (-1 + 1, 0) = (0, 0)$.

(D) Let the point of contact of the tangents from $P$ to the parabola $y^2 = 4ax$ be $A(at_1^2, 2at_1)$ and $B(at_2^2, 2at_2)$.
The point of intersection $P$ of the tangents at $A$ and $B$ is given by $P(at_1t_2, a(t_1 + t_2))$.
Let $P = (x, y)$,so $x = at_1t_2$ and $y = a(t_1 + t_2)$.
The slope of the tangent at $A(at_1^2, 2at_1)$ is $m_1 = \tan \theta_1 = \frac{1}{t_1}$.
The slope of the tangent at $B(at_2^2, 2at_2)$ is $m_2 = \tan \theta_2 = \frac{1}{t_2}$.
Given that $\tan \theta_1 + \tan \theta_2 = b$,we have $\frac{1}{t_1} + \frac{1}{t_2} = b$.
This simplifies to $\frac{t_1 + t_2}{t_1t_2} = b$.
Substituting the coordinates of $P$,we get $\frac{y/a}{x/a} = b$,which implies $\frac{y}{x} = b$.
Therefore,$y = bx$.

(D) Let the equation of the tangent to the parabola $y^2=4x$ be $y=mx+\frac{1}{m}$.
For the parabola $x^2=-32y$,the equation of the tangent with slope $m$ is $y=mx-a m^2$,where $x^2=4ay$ gives $a=-8$.
Thus,the tangent is $y=mx-(-8)m^2$,which simplifies to $y=mx+8m^2$.
Comparing the two tangent equations,we have $\frac{1}{m}=8m^2$.
This implies $m^3=\frac{1}{8}$,so $m=\frac{1}{2}$.
Substituting $m=\frac{1}{2}$ into the first tangent equation: $y=\frac{1}{2}x+\frac{1}{1/2} = \frac{1}{2}x+2$.
Multiplying by $2$,we get $2y=x+4$,or $x-2y+4=0$.

(A) The given parabolas are $y^2=32x$ and $x^2=256y$.
The equation of a common tangent to the parabolas $y^2=4ax$ and $x^2=4by$ is given by $b^{1/3}y + a^{1/3}x + (a^2b^2)^{1/3} = 0$.
Comparing $y^2=32x$ with $y^2=4ax$,we get $4a=32 \Rightarrow a=8$.
Comparing $x^2=256y$ with $x^2=4by$,we get $4b=256 \Rightarrow b=64$.
Substituting these values into the formula:
$(64)^{1/3}y + (8)^{1/3}x + (8^2 \times 64^2)^{1/3} = 0$
$4y + 2x + (64 \times 4096)^{1/3} = 0$
$4y + 2x + (262144)^{1/3} = 0$
$4y + 2x + 64 = 0$
Dividing by $2$,we get $x + 2y + 32 = 0$. However,checking the options,the equation $2x+4y+64=0$ is equivalent to $x+2y+32=0$.

(A) For the parabola $y^2 = 4ax$,the tangent is $y = mx + \frac{a}{m}$. Here $4a = 32$,so $a = 8$. Thus,the tangent is $y = mx + \frac{8}{m}$.
For the parabola $x^2 = 4by$,the tangent is $x = my - bm^2$. Here $4b = 256$,so $b = 64$. Thus,the tangent is $x = my - 64m^2$.
Rewriting the first equation: $mx - y + \frac{8}{m} = 0$. Rewriting the second equation: $x - my + 64m^2 = 0$.
Since these represent the same line,the ratio of coefficients must be equal: $\frac{m}{1} = \frac{-1}{-m} = \frac{8/m}{64m^2}$.
From $\frac{m}{1} = \frac{1}{m}$,we get $m^2 = 1$,so $m = \pm 1$.
Checking the ratio $\frac{8/m}{64m^2} = \frac{1}{8m^3}$.
For $m = 1$,the ratio is $\frac{1}{8} \neq 1$. For $m = -1$,the ratio is $\frac{1}{-8} \neq -1$.
Let the tangent be $y = mx + c$. For $y^2 = 32x$,$c = \frac{8}{m}$. For $x^2 = 256y$,$c = -64m^2$.
Equating $c$: $\frac{8}{m} = -64m^2 \implies m^3 = -\frac{8}{64} = -\frac{1}{8} \implies m = -\frac{1}{2}$.
Then $c = -64(-\frac{1}{2})^2 = -64(\frac{1}{4}) = -16$.
The equation is $y = -\frac{1}{2}x - 16$,which simplifies to $2y = -x - 32$,or $x + 2y + 32 = 0$.

(D) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.
The point on the parabola at parameter $t$ is $(at^2, 2at) = (2t^2, 4t)$.
For $t = \frac{1}{\sqrt{2}}$,the point is $(2(\frac{1}{2}), 4(\frac{1}{\sqrt{2}})) = (1, 2\sqrt{2})$.
The equation of the normal to the parabola $y^2 = 4ax$ at parameter $t$ is $y + tx = 2at + at^3$.
Substituting $a = 2$ and $t = \frac{1}{\sqrt{2}}$,the equation of the normal $L$ is $y + \frac{1}{\sqrt{2}}x = 2(2)(\frac{1}{\sqrt{2}}) + 2(\frac{1}{2\sqrt{2}}) = \frac{4}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}}$.
Multiplying by $\sqrt{2}$,we get $x + \sqrt{2}y = 5$.
The focus of the parabola is $(a, 0) = (2, 0)$.
The foot of the perpendicular $(h, k)$ from $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $\frac{h - x_1}{A} = \frac{k - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$.
Here $A = 1, B = \sqrt{2}, C = -5$ and $(x_1, y_1) = (2, 0)$.
$\frac{h - 2}{1} = \frac{k - 0}{\sqrt{2}} = -\frac{1(2) + \sqrt{2}(0) - 5}{1^2 + (\sqrt{2})^2} = -\frac{2 - 5}{1 + 2} = -\frac{-3}{3} = 1$.
So,$h - 2 = 1 \implies h = 3$ and $\frac{k}{\sqrt{2}} = 1 \implies k = \sqrt{2}$.
The foot of the perpendicular is $(3, \sqrt{2})$.

(D) The equation of a normal to the parabola $y^2=4ax$ with slope $m$ is $y=mx-2am-am^3$.
Here,$a=1$. So,the normal equation is $y=mx-2m-m^3$.
Since the normal passes through $(5,2)$,we have $2=5m-2m-m^3$,which simplifies to $m^3-3m+2=0$.
Factoring the cubic equation,we get $(m-1)^2(m+2)=0$.
The roots are $m=1$ and $m=-2$.
The slope of the given normal $x-y-3=0$ is $m=1$.
Therefore,the slope of the other normal is $m=-2$.

(D) Given the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.
For a point $(at^2, 2at)$ on the parabola,we have $(2, -4) = (2t^2, 4t)$,which gives $t = -1$.
The normal at parameter $t$ meets the parabola again at parameter $t_2 = -t - \frac{2}{t}$.
Substituting $t = -1$,we get $t_2 = -(-1) - \frac{2}{-1} = 1 + 2 = 3$.
The coordinates of the point $(\alpha, \beta)$ are $(at_2^2, 2at_2) = (2(3)^2, 2(2)(3)) = (18, 12)$.
Thus,$\alpha + \beta = 18 + 12 = 30$.

(B) The equation of the parabola is $y^2=6x$. Comparing this with $y^2=4ax$,we get $4a=6$,so $a=1.5$.
The equation of the normal to the parabola $y^2=4ax$ at the point $(x_1, y_1)$ is given by $y-y_1 = -\frac{y_1}{2a}(x-x_1)$.
Substituting the values $x_1=24$,$y_1=12$,and $a=1.5$:
$y-12 = -\frac{12}{2(1.5)}(x-24)$
$y-12 = -\frac{12}{3}(x-24)$
$y-12 = -4(x-24)$
$y-12 = -4x+96$
$4x+y = 108$.

(B) The parabola is $y^2 = 4ax$. The point $P$ is $(2a, 2a\sqrt{2})$.
Comparing $P(2a, 2a\sqrt{2})$ with $(at^2, 2at)$,we get $t = \sqrt{2}$.
The normal at $t$ meets the parabola at $t_1 = -t - \frac{2}{t} = -\sqrt{2} - \frac{2}{\sqrt{2}} = -2\sqrt{2}$.
The coordinates of $Q$ are $(at_1^2, 2at_1) = (a(-2\sqrt{2})^2, 2a(-2\sqrt{2})) = (8a, -4a\sqrt{2})$.
The vertex is $O(0, 0)$.
The slope of $OP$ is $m_1 = \frac{2a\sqrt{2}}{2a} = \sqrt{2}$.
The slope of $OQ$ is $m_2 = \frac{-4a\sqrt{2}}{8a} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.
Since $m_1 \times m_2 = \sqrt{2} \times (-\frac{1}{\sqrt{2}}) = -1$,the lines $OP$ and $OQ$ are perpendicular.
Therefore,$\theta = 90^{\circ}$.

(C) The equation of a normal to the parabola $y^2 = 4px$ with slope $m$ is given by $y = mx - 2pm - pm^3$.
Given the normal equation $ax + by = 1$, we can rewrite it as $y = -\frac{a}{b}x + \frac{1}{b}$.
Comparing the two equations, we get $m = -\frac{a}{b}$ and the constant term $-2pm - pm^3 = \frac{1}{b}$.
Substituting $m = -\frac{a}{b}$ into the constant term equation:
$-2p(-\frac{a}{b}) - p(-\frac{a}{b})^3 = \frac{1}{b}$
$\frac{2pa}{b} + \frac{pa^3}{b^3} = \frac{1}{b}$
Multiplying both sides by $b^3$, we get $2pab^2 + pa^3 = b^2$, which can be rearranged as $pa^3 = b^2 - 2pab^2$.

(D) The equation of a normal to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx-2am-am^3$.
Comparing the given parabola $y^2=4x$ with $y^2=4ax$,we get $a=1$.
The given line is $y=2x+k$,so the slope $m=2$.
Substituting $a=1$ and $m=2$ into the normal equation:
$k = -2(1)(2) - (1)(2)^3$
$k = -4 - 8$
$k = -12$.
Thus,the correct option is $D$.

(D) The equation of a normal with slope $m$ to the parabola $y^2=4ax$ is given by $y=mx-2am-am^3$.
If this normal passes through a point $(h, k)$,then $k=mh-2am-am^3$,which is a cubic equation in $m$: $am^3 + (2a-h)m + k = 0$.
Let the roots of this equation be $m_1, m_2, m_3$.
Since the normals are perpendicular,let $m_1m_2 = -1$.
From the properties of roots,the product of roots is $m_1m_2m_3 = -k/a$.
Substituting $m_1m_2 = -1$,we get $(-1)m_3 = -k/a$,so $m_3 = k/a$.
Since $m_3$ is a root,it must satisfy the cubic equation: $a(k/a)^3 + (2a-h)(k/a) + k = 0$.
Simplifying this: $k^3/a^2 + 2k - hk/a + k = 0$.
Dividing by $k$ (assuming $k \neq 0$): $k^2/a^2 + 3 - h/a = 0$.
$k^2/a^2 = h/a - 3$.
$k^2 = ah - 3a^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = ax - 3a^2$,or $y^2 - ax + 3a^2 = 0$.

(A) The given equation of the parabola is $y^2=4x$,which implies $a=1$.
The equation of the normal to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx-2am-am^3$.
Substituting $a=1$,the equation of the normal is $y=mx-2m-m^3 \dots(I)$.
The given line is $x+3y+1=0$,which can be written as $y=-\frac{1}{3}x-\frac{1}{3}$. The slope of this line is $m_1 = -\frac{1}{3}$.
Since the normal is perpendicular to this line,the product of their slopes must be $-1$. Let $m$ be the slope of the normal,then $m \times (-\frac{1}{3}) = -1$,which gives $m=3$.
Substituting $m=3$ into equation $(I)$:
$y = 3x - 2(3) - (3)^3$
$y = 3x - 6 - 27$
$y = 3x - 33$
$3x - y = 33$.

(A) The axis of the parabola is the line passing through the focus $S(1,-1)$ and vertex $A(1,1)$. Since the $x$-coordinates are the same,the axis is the vertical line $x=1$.
The distance between the vertex $A(1,1)$ and focus $S(1,-1)$ is $a = \sqrt{(1-1)^2 + (1 - (-1))^2} = 2$.
Since the vertex is above the focus,the parabola opens downwards. The directrix is a horizontal line at a distance $a=2$ above the vertex $A(1,1)$.
The equation of the directrix is $y = 1 + 2 = 3$.
By the definition of a parabola,for any point $P(x,y)$ on it,the distance to the focus equals the distance to the directrix: $PS^2 = PM^2$.
$(x-1)^2 + (y+1)^2 = (y-3)^2$.
$(x-1)^2 + y^2 + 2y + 1 = y^2 - 6y + 9$.
$(x-1)^2 = -8y + 8 = 8(1-y)$.
Checking option $A$: For $\left(3, \frac{1}{2}\right)$,$(3-1)^2 = 2^2 = 4$ and $8(1 - \frac{1}{2}) = 8(\frac{1}{2}) = 4$.
Since $4=4$,the point $\left(3, \frac{1}{2}\right)$ lies on the parabola.

(A) The equation of the normal to the parabola $y^2 = 4ax$ at point $t_1$ is given by $y + t_1x = 2at_1 + at_1^3$.
Since this normal meets the parabola again at point $t_2 = (at_2^2, 2at_2)$,the point $(at_2^2, 2at_2)$ must satisfy the equation of the normal.
Substituting $x = at_2^2$ and $y = 2at_2$ into the normal equation:
$2at_2 + t_1(at_2^2) = 2at_1 + at_1^3$.
Dividing by $a$ (assuming $a \neq 0$):
$2t_2 + t_1t_2^2 = 2t_1 + t_1^3$.
Rearranging the terms:
$t_1(t_2^2 - t_1^2) + 2(t_2 - t_1) = 0$.
$t_1(t_2 - t_1)(t_2 + t_1) + 2(t_2 - t_1) = 0$.
Since $t_1 \neq t_2$ for the normal to meet the parabola at a distinct second point,we can divide by $(t_2 - t_1)$:
$t_1(t_2 + t_1) + 2 = 0$.
$t_1t_2 + t_1^2 + 2 = 0$.
Therefore,$t_1t_2 = -2 - t_1^2$.

(D) For a parabola $x^2=4ay$,the equation of the normal at point $(2at, at^2)$ is $y-at^2 = -t(x-2at)$,which simplifies to $x+ty = 2at+at^3$.
Here,$4a=8$,so $a=2$.
The equation of the normal is $x+ty = 4t+2t^3$.
Let the intersection point be $(h, k)$. Then $h+tk = 4t+2t^3$,or $2t^3 + (4-k)t - h = 0$.
This cubic equation in $t$ has three roots $t_1, t_2, t_3$. Since the normals are at right angles,the product of the slopes of the two normals is $-1$.
The slope of the normal is $m = -t$. Thus,$(-t_1)(-t_2) = -1$,which means $t_1 t_2 = -1$.
From the cubic equation,the product of the roots $t_1 t_2 t_3 = -h/2$.
Substituting $t_1 t_2 = -1$,we get $-t_3 = -h/2$,so $t_3 = h/2$.
Since $t_3$ is a root of the cubic equation,we substitute it back: $2(h/2)^3 + (4-k)(h/2) - h = 0$.
$h^3/4 + 2h - kh/2 - h = 0$.
$h^3/4 + h - kh/2 = 0$.
Dividing by $h$ (assuming $h \neq 0$),$h^2/4 + 1 - k/2 = 0$,which gives $h^2 + 4 - 2k = 0$,or $h^2 = 2k-4$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 = 2y-4$.
Wait,checking the options,the standard result for $x^2=4ay$ is $x^2 = a(y-3a)$. For $a=2$,$x^2 = 2(y-6) = 2y-12$.
Thus,the correct option is $D$.

(C) The equation of the parabola is $y^2 = 4x$,so $a = 1$.
The point $(-2, -1)$ lies outside the parabola because $(-1)^2 - 4(-2) = 1 + 8 = 9 > 0$.
The equation of a tangent to $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Substituting $a = 1$ and the point $(-2, -1)$,we get $-1 = -2m + \frac{1}{m}$.
Multiplying by $m$,we get $-m = -2m^2 + 1$,which simplifies to $2m^2 - m - 1 = 0$.
Factoring the quadratic,$(2m + 1)(m - 1) = 0$,so the slopes are $m_1 = 1$ and $m_2 = -1/2$.
The angle $\theta$ between the tangents is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
$\tan \theta = |\frac{1 - (-1/2)}{1 + (1)(-1/2)}| = |\frac{3/2}{1/2}| = 3$.
We need to find $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$.
Substituting $\tan \theta = 3$,we get $\tan 2\theta = \frac{2(3)}{1 - 3^2} = \frac{6}{1 - 9} = \frac{6}{-8} = -\frac{3}{4}$.

(D) Given the parabola $y^2 = 4x$, we have $a = 1$. The point is $P(-1, 2)$.
The equation of the chord of contact for $P(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
Substituting $x_1 = -1, y_1 = 2, a = 1$, we get $2y = 2(x - 1)$, which simplifies to $y = x - 1$ or $x - y - 1 = 0$.
The length of the chord of contact $L$ is given by $\frac{\sqrt{(y_1^2 - 4ax_1)^3}}{2a} = \frac{\sqrt{(2^2 - 4(1)(-1))^3}}{2(1)} = \frac{\sqrt{(4 + 4)^3}}{2} = \frac{\sqrt{8^3}}{2} = \frac{16\sqrt{2}}{2} = 8\sqrt{2}$.
The perpendicular distance $h$ from the point $(-1, 2)$ to the chord $x - y - 1 = 0$ is $h = \frac{|(-1) - (2) - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|-4|}{\sqrt{2}} = 2\sqrt{2}$.
The area of the triangle formed by the tangents and the chord of contact is given by $\frac{h^3}{2a} = \frac{(2\sqrt{2})^3}{2(1)} = \frac{8 \times 2\sqrt{2}}{2} = 8\sqrt{2}$.

(A) Given the parabola $y^2=12x$,we have $4a=12$,so $a=3$. Let the parameters of points $P$ and $Q$ be $t_1$ and $t_2$ respectively. The ordinates are $y_1=2at_1$ and $y_2=2at_2$. Given $y_1:y_2=1:2$,we have $t_1:t_2=1:2$,so $t_2=2t_1$. Let $t_1=t$,then $t_2=2t$.
The point of intersection $(x, y)$ of the normals at $t_1$ and $t_2$ is given by:
$x = 2a + a(t_1^2 + t_2^2 + t_1t_2) = 2(3) + 3(t^2 + 4t^2 + 2t^2) = 6 + 21t^2$
$y = -at_1t_2(t_1+t_2) = -3(t)(2t)(t+2t) = -6t^2(3t) = -18t^3$
From $x=6+21t^2$,we get $t^2 = \frac{x-6}{21}$,so $t = \pm \left(\frac{x-6}{21}\right)^{1/2}$.
Substituting $t$ into $y=-18t^3$,we get $y = -18 \left(\frac{x-6}{21}\right)^{3/2}$,which simplifies to $y+18\left(\frac{x-6}{21}\right)^{3/2}=0$.

(C) The given parabola is $y^2=12x$. Comparing with $y^2=4ax$,we get $a=3$. The focus is $(3, 0)$.
Let $Q(3t^2, 6t)$ be a point on the parabola.
Point $P(x, y)$ divides the segment joining $(3, 0)$ and $(3t^2, 6t)$ in the ratio $1:2$.
Using the section formula,$P(x, y) = \left(\frac{1(3t^2) + 2(3)}{1+2}, \frac{1(6t) + 2(0)}{1+2}\right) = \left(\frac{3t^2+6}{3}, \frac{6t}{3}\right) = (t^2+2, 2t)$.
Thus,$x = t^2+2$ and $y = 2t$.
From $y = 2t$,we get $t = \frac{y}{2}$.
Substituting $t$ into the equation for $x$: $x = (\frac{y}{2})^2 + 2 = \frac{y^2}{4} + 2$.
$x - 2 = \frac{y^2}{4} \Rightarrow y^2 = 4(x-2)$.

(A) The given parabolas are $y^2 = 4ax$ and $x^2 = 4ay$.
Solving these,we substitute $y = x^2 / (4a)$ into $y^2 = 4ax$,giving $(x^2 / (4a))^2 = 4ax$,which simplifies to $x^4 = 64a^3x$.
This yields $x(x^3 - 64a^3) = 0$,so $x = 0$ or $x = 4a$.
The intersection points are $(0, 0)$ and $(4a, 4a)$.
The line $2bx + 3cy + 4d = 0$ passes through both points.
For $(0, 0)$: $2b(0) + 3c(0) + 4d = 0 \Rightarrow d = 0$.
For $(4a, 4a)$: $2b(4a) + 3c(4a) + 4d = 0$.
Since $d = 0$,this simplifies to $8ab + 12ac = 0$,which implies $4a(2b + 3c) = 0$.
Assuming $a \neq 0$,we get $2b + 3c = 0$.
Combining $d = 0$ and $2b + 3c = 0$,we get $d^2 + (2b + 3c)^2 = 0$.

(B) The equation of the parabola is $y^2 = 4(x+4)$.
Let a point on the parabola be $P(x, y) = (t^2-4, 2t)$.
The distance squared from the origin $(0,0)$ to any point on the parabola is $d^2 = x^2 + y^2 = (t^2-4)^2 + (2t)^2$.
$d^2 = t^4 - 8t^2 + 16 + 4t^2 = t^4 - 4t^2 + 16$.
For the circle $x^2+y^2=k^2$ to lie inside the parabola,we must have $k^2 \leq d^2$ for all $t$.
Let $u = t^2$,where $u \geq 0$. Then $f(u) = u^2 - 4u + 16$.
The minimum value of $f(u)$ occurs at $u = -(-4)/(2 \times 1) = 2$.
Since $u=2 \geq 0$,the minimum value is $f(2) = 2^2 - 4(2) + 16 = 4 - 8 + 16 = 12$.
Thus,$k^2 \leq 12$,which implies $k \leq \sqrt{12} = 2\sqrt{3}$.
The largest value of $k$ is $2\sqrt{3}$.

(B) The equation of the parabola is $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$. The focus $S$ is $(a, 0) = (1, 0)$.
Since $P = (4, 4)$ lies on the parabola,we can find the parameter $t_1$ for point $P$ using $x = at_1^2$ and $y = 2at_1$. Thus,$4 = 1 \cdot t_1^2 \implies t_1 = 2$.
For a focal chord,the parameters $t_1$ and $t_2$ of the endpoints $P$ and $Q$ satisfy $t_1 t_2 = -1$. Therefore,$t_2 = -\frac{1}{t_1} = -\frac{1}{2}$.
The focal distance of a point with parameter $t$ on the parabola $y^2 = 4ax$ is given by $a(1 + t^2)$.
For point $Q$ with parameter $t_2 = -\frac{1}{2}$,the focal distance $SQ = a(1 + t_2^2) = 1 \cdot (1 + (-\frac{1}{2})^2) = 1 + \frac{1}{4} = \frac{5}{4}$.

(D) For a parabola $y^2 = 4ax$,the length of a focal chord making an angle $\theta$ with the axis is $L = 4a \csc^2 \theta$.
Given $4a = 16$,so $a = 4$. The length $L = 25$.
Thus,$25 = 16 \csc^2 \theta \implies \csc^2 \theta = \frac{25}{16} \implies \sin^2 \theta = \frac{16}{25} \implies \sin \theta = \pm \frac{4}{5}$.
Let the two chords be at angles $\theta$ and $\pi - \theta$ (or $-\theta$) with the axis. The area of the quadrilateral formed by the endpoints of two focal chords is given by the formula $Area = \frac{1}{2} |(x_1 - x_2)(y_1 + y_2)|$ or more simply $Area = 2a^2 \sin(2\theta) \csc^4 \theta$ is not standard,but the area of a quadrilateral formed by two focal chords is $Area = \frac{1}{2} L_1 L_2 \sin \phi$,where $\phi$ is the angle between the chords.
Here $L_1 = L_2 = 25$. The angle between the chords is $2\theta$. Since $\sin \theta = \frac{4}{5}$,$\cos \theta = \frac{3}{5}$.
Then $\sin(2\theta) = 2 \sin \theta \cos \theta = 2 \times \frac{4}{5} \times \frac{3}{5} = \frac{24}{25}$.
Area $= \frac{1}{2} \times 25 \times 25 \times \frac{24}{25} = \frac{1}{2} \times 25 \times 24 = 300$ sq. units.

(B) The equation of the parabola is $y^2 = 15x$,so $4a = 15$,which gives $a = \frac{15}{4}$.
The point $P$ is $\left(\frac{15}{2}, \frac{15}{\sqrt{2}}\right)$. Let $P = (at^2, 2at)$.
$at^2 = \frac{15}{2} \implies \frac{15}{4}t^2 = \frac{15}{2} \implies t^2 = 2 \implies t = \sqrt{2}$.
The normal at $t$ meets the parabola again at $t_1 = -t - \frac{2}{t} = -\sqrt{2} - \frac{2}{\sqrt{2}} = -2\sqrt{2}$.
The vertex is $V(0,0)$. The angle $\theta$ subtended by the chord $PQ$ at the vertex is given by the angle between vectors $\vec{VP}$ and $\vec{VQ}$.
$P = (at^2, 2at) = (\frac{15}{2}, \frac{15}{\sqrt{2}})$ and $Q = (at_1^2, 2at_1) = (\frac{15}{4}(8), 2(\frac{15}{4})(-2\sqrt{2})) = (30, -15\sqrt{2})$.
The slopes are $m_1 = \frac{2at}{at^2} = \frac{2}{t} = \frac{2}{\sqrt{2}} = \sqrt{2}$ and $m_2 = \frac{2}{t_1} = \frac{2}{-2\sqrt{2}} = -\frac{1}{\sqrt{2}}$.
$\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{\sqrt{2} + \frac{1}{\sqrt{2}}}{1 + \sqrt{2}(-\frac{1}{\sqrt{2}})}| = |\frac{3/\sqrt{2}}{0}| = \infty$.
Thus,$\theta = 90^\circ$ or $\frac{\pi}{2}$.
Substituting $\theta = 90^\circ$ into the expression: $\sin(30^\circ) + \cos(60^\circ) - \sec(120^\circ) = \frac{1}{2} + \frac{1}{2} - (-2) = 1 + 2 = 3$.

(C) Let $P(t, -1/t)$ be a point on the curve $xy = -1$. The equation of the tangent at $P$ is given by $x(-1/t) + y(t) = -2$,which simplifies to $y = x/t^2 + 2/t$.
For this line to be a tangent to the parabola $y^2 = 8x$ (where $a = 2$),the condition $c = a/m$ must be satisfied.
Here,$m = 1/t^2$ and $c = 2/t$.
Substituting these into the condition: $2/t = 2 / (1/t^2) \implies 2/t = 2t^2 \implies t^3 = 1 \implies t = 1$.
Substituting $t = 1$ into the tangent equation $y = x/t^2 + 2/t$,we get $y = x + 2$.

(A) The parabolas $y^2 = 4ax$ and $x^2 = 4ay$ are symmetric with respect to the line $y = x$.
Therefore,the line $y = x$ passes through their points of intersection.
Solving $y^2 = 4ax$ and $y = x$:
$x^2 = 4ax \Rightarrow x(x - 4a) = 0 \Rightarrow x = 0$ or $x = 4a$.
When $x = 0$,$y = 0$. When $x = 4a$,$y = 4a$.
Thus,the points of intersection are $(0, 0)$ and $(4a, 4a)$.
Since the line $2bx + 3cy + 4d = 0$ passes through $(0, 0)$:
$2b(0) + 3c(0) + 4d = 0 \Rightarrow 4d = 0 \Rightarrow d = 0$.
Since it also passes through $(4a, 4a)$:
$2b(4a) + 3c(4a) + 4d = 0 \Rightarrow 8ab + 12ac + 4d = 0$.
Since $d = 0$ and $a \neq 0$,we have $8ab + 12ac = 0 \Rightarrow 4a(2b + 3c) = 0 \Rightarrow 2b + 3c = 0$.
Therefore,$d^2 + (2b + 3c)^2 = 0^2 + 0^2 = 0$.

(D) Let the point of intersection be $(x_1, y_1)$.
For the curve $y^2 = 4ax$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y}$.
Thus,the slope of the tangent at $(x_1, y_1)$ is $m_1 = \frac{2a}{y_1}$.
For the curve $xy = c^2$,differentiating with respect to $x$ gives $x \frac{dy}{dx} + y = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
Thus,the slope of the tangent at $(x_1, y_1)$ is $m_2 = -\frac{y_1}{x_1}$.
Since the curves cut orthogonally,$m_1 \times m_2 = -1$.
$\frac{2a}{y_1} \times (-\frac{y_1}{x_1}) = -1 \Rightarrow \frac{2a}{x_1} = 1 \Rightarrow x_1 = 2a$.
Since $(x_1, y_1)$ lies on $y^2 = 4ax$,$y_1^2 = 4a(2a) = 8a^2$.
Since $(x_1, y_1)$ lies on $xy = c^2$,$x_1 y_1 = c^2$,so $(x_1 y_1)^2 = c^4$.
Substituting $x_1 = 2a$ and $y_1^2 = 8a^2$,we get $c^4 = x_1^2 y_1^2 = (2a)^2 (8a^2) = 4a^2 \times 8a^2 = 32a^4$.

(B) Given curves are $x^2=4y$ $(i)$ and $y^2=4x$ $(ii)$.
To find the point of intersection,substitute $y = \frac{x^2}{4}$ into $(ii)$:
$\left(\frac{x^2}{4}\right)^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 = 64x \implies x(x^3 - 64) = 0$.
So,$x=0$ or $x=4$. For $x=4$,$y = \frac{16}{4} = 4$. Thus,the point of intersection other than the origin is $(4,4)$.
Now,differentiate $(i)$ with respect to $x$: $2x = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2}$.
At $(4,4)$,the slope $m_1 = \frac{4}{2} = 2$.
Differentiate $(ii)$ with respect to $x$: $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$.
At $(4,4)$,the slope $m_2 = \frac{2}{4} = \frac{1}{2}$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{2 - \frac{1}{2}}{1 + 2 \times \frac{1}{2}} \right| = \left| \frac{\frac{3}{2}}{2} \right| = \frac{3}{4}$.
Since $\tan \theta = \frac{3}{4}$,we have $\sin \theta = \frac{3}{5}$,so $\theta = \sin^{-1}\left(\frac{3}{5}\right)$.

(C) Given the parabola $y^2 = 6x$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 6$,which implies $\frac{dy}{dx} = \frac{3}{y}$.
Given the tangent line $5x - 2y + k = 0$,we can rewrite it as $y = \frac{5}{2}x + \frac{k}{2}$.
The slope of this line is $m = \frac{5}{2}$.
At the point of contact,the slope of the tangent to the parabola must equal the slope of the line: $\frac{3}{y} = \frac{5}{2}$.
Solving for $y$,we get $y = \frac{6}{5}$.
Substituting $y = \frac{6}{5}$ into the parabola equation $y^2 = 6x$:
$(\frac{6}{5})^2 = 6x$
$\frac{36}{25} = 6x$
$x = \frac{36}{25 \times 6} = \frac{6}{25}$.
Thus,the point of contact is $(\frac{6}{25}, \frac{6}{5})$.