Parabola Questions in English

(C) The equation of a tangent to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is given by $ty = x + at^2$,which can be rewritten as $x - ty + at^2 = 0$.
Given that the line $lx + my + n = 0$ is a tangent to the parabola,it must be identical to the equation $x - ty + at^2 = 0$.
Comparing the coefficients of the two equations,we have:
$\frac{1}{l} = \frac{-t}{m} = \frac{at^2}{n}$
From $\frac{1}{l} = \frac{-t}{m}$,we get $t = -\frac{m}{l}$.
From $\frac{1}{l} = \frac{at^2}{n}$,we get $t^2 = \frac{n}{al}$.
Substituting $t = -\frac{m}{l}$ into $t^2 = \frac{n}{al}$,we get:
$(-\frac{m}{l})^2 = \frac{n}{al} \implies \frac{m^2}{l^2} = \frac{n}{al} \implies m^2 = \frac{nl}{a}$.
The point of contact is $(at^2, 2at)$. Let $(x_1, y_1)$ be the point of contact.
Then $x_1 = at^2 = a(\frac{n}{al}) = \frac{n}{l}$ and $y_1 = 2at = 2a(-\frac{m}{l}) = -\frac{2am}{l}$.
Since the point of contact $(x_1, y_1)$ lies on the parabola $y^2 = 4ax$,the locus of the point of contact is the parabola itself.

(D) The line is $y = x + 2$. Substituting this into the parabola equation $y^2 = 8x$:
$(x + 2)^2 = 8x$
$x^2 + 4x + 4 = 8x$
$x^2 - 4x + 4 = 0$
$(x - 2)^2 = 0$
Thus,$x = 2$.
Substituting $x = 2$ into the line equation $y = x + 2$,we get $y = 2 + 2 = 4$.
Therefore,the point of contact is $(2, 4)$.

(B) The equation of the tangent to the parabola $y^2 = 4ax$ at the point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.
Substituting the point $(a, 2a)$,we get $y(2a) = 2a(x + a)$.
Dividing both sides by $2a$,we obtain $y = x + a$.
Comparing this with the slope-intercept form $y = mx + c$,the slope $m = 1$.
Since $m = \tan \theta$,we have $\tan \theta = 1$,which implies $\theta = \frac{\pi}{4}$.

(D) Given the line $lx + my + n = 0$ $(i)$ and the parabola $x^2 = y$ $(ii)$.
To find the intersection,substitute $y = x^2$ into the line equation: $lx + m(x^2) + n = 0$.
Rearranging gives the quadratic equation: $mx^2 + lx + n = 0$.
Since the line is tangent to the parabola,the discriminant of this quadratic equation must be zero $(D = 0)$.
$D = b^2 - 4ac = l^2 - 4(m)(n) = 0$.
Therefore,the condition of tangency is $l^2 = 4mn$.

(D) Given the parabola $y^2 = 9x$,we have $4a = 9$,so $a = \frac{9}{4}$.
The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Substituting $a = \frac{9}{4}$,the equation becomes $y = mx + \frac{9}{4m}$.
Since the tangent passes through the point $(4, 10)$,we substitute $x = 4$ and $y = 10$:
$10 = 4m + \frac{9}{4m}$
Multiply by $4m$ to get the quadratic equation:
$40m = 16m^2 + 9$
$16m^2 - 40m + 9 = 0$
Solving for $m$ using the quadratic formula or factoring:
$(4m - 9)(4m - 1) = 0$
$m = \frac{9}{4}$ or $m = \frac{1}{4}$.
For $m = \frac{9}{4}$,the equation is $y = \frac{9}{4}x + \frac{9}{4(9/4)}$ $\Rightarrow y = \frac{9}{4}x + 1$ $\Rightarrow 4y = 9x + 4$ $\Rightarrow 9x - 4y + 4 = 0$.
For $m = \frac{1}{4}$,the equation is $y = \frac{1}{4}x + \frac{9}{4(1/4)}$ $\Rightarrow y = \frac{1}{4}x + 9$ $\Rightarrow 4y = x + 36$ $\Rightarrow x - 4y + 36 = 0$.
Thus,both equations $(b)$ and $(c)$ are correct.

(B) The equation of the tangent to the parabola $y^2 = 4ax$ at point $t$ is given by $ty = x + at^2$.
Let the two tangents be at points $t_1$ and $t_2$. Their equations are $t_1y = x + at_1^2$ and $t_2y = x + at_2^2$.
The slopes of these tangents are $m_1 = \frac{1}{t_1}$ and $m_2 = \frac{1}{t_2}$.
Since the tangents are perpendicular,$m_1 \times m_2 = -1$,which implies $\frac{1}{t_1} \times \frac{1}{t_2} = -1$,so $t_1t_2 = -1$.
The point of intersection of these two tangents is $(at_1t_2, a(t_1 + t_2))$.
Substituting $t_1t_2 = -1$,the point of intersection becomes $(-a, a(t_1 + t_2))$.
Since the $x$-coordinate is always $-a$,the locus of the intersection point is the line $x = -a$,which can be written as $x + a = 0$.

(C) Any tangent to the parabola $y^2 = 4x$ is of the form $y = mx + \frac{1}{m}$.
This line touches the circle $(x - 3)^2 + y^2 = 9$ (with center $(3, 0)$ and radius $3$) if the perpendicular distance from the center to the line equals the radius:
$3 = \left| \frac{m(3) - 0 + \frac{1}{m}}{\sqrt{m^2 + 1}} \right|$
Squaring both sides:
$9(m^2 + 1) = (3m + \frac{1}{m})^2$
$9m^2 + 9 = 9m^2 + 6 + \frac{1}{m^2}$
$3 = \frac{1}{m^2} \implies m^2 = \frac{1}{3} \implies m = \pm \frac{1}{\sqrt{3}}$.
Since the tangent is above the $x$-axis,we choose the positive slope $m = \frac{1}{\sqrt{3}}$.
Substituting $m$ into the tangent equation:
$y = \frac{1}{\sqrt{3}}x + \sqrt{3}$
Multiplying by $\sqrt{3}$:
$\sqrt{3}y = x + 3$.

(A) For the line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$,the condition is $c = \frac{a}{m}$.
Substituting $c = \frac{a}{m}$ into the line equation,we get $y = mx + \frac{a}{m}$.
To find the point of tangency,we substitute $y = mx + \frac{a}{m}$ into $y^2 = 4ax$:
$(mx + \frac{a}{m})^2 = 4ax$
$m^2x^2 + 2ax + \frac{a^2}{m^2} = 4ax$
$m^2x^2 - 2ax + \frac{a^2}{m^2} = 0$
$(mx - \frac{a}{m})^2 = 0$
This gives $x = \frac{a}{m^2}$.
Substituting $x = \frac{a}{m^2}$ into $y = mx + \frac{a}{m}$,we get $y = m(\frac{a}{m^2}) + \frac{a}{m} = \frac{a}{m} + \frac{a}{m} = \frac{2a}{m}$.
Thus,the point of contact is $\left( \frac{a}{m^2}, \frac{2a}{m} \right)$.

(D) Let the point $P$ on the parabola ${y^2} = 4ax$ be $(at^2, 2at)$ and the focus $S$ be $(a, 0)$.
The equation of the tangent at $P$ is $ty = x + at^2$.
The directrix of the parabola is $x = -a$.
To find the point $K$,substitute $x = -a$ into the tangent equation:
$ty = -a + at^2$
$y = \frac{a(t^2 - 1)}{t}$
So,$K = (-a, \frac{a(t^2 - 1)}{t})$.
Now,calculate the slopes of $SP$ and $SK$:
Slope of $SP$ $(m_1)$ = $\frac{2at - 0}{at^2 - a} = \frac{2at}{a(t^2 - 1)} = \frac{2t}{t^2 - 1}$.
Slope of $SK$ $(m_2)$ = $\frac{\frac{a(t^2 - 1)}{t} - 0}{-a - a} = \frac{a(t^2 - 1)}{t(-2a)} = -\frac{t^2 - 1}{2t}$.
Since $m_1 \times m_2 = (\frac{2t}{t^2 - 1}) \times (-\frac{t^2 - 1}{2t}) = -1$,the lines $SP$ and $SK$ are perpendicular.
Therefore,the angle $\angle PSK = 90^\circ$.

(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $t$ is $ty = x + at^2$.
Let the tangents at $t_1$ and $t_2$ be $t_1y = x + at_1^2$ and $t_2y = x + at_2^2$.
Subtracting the two equations: $(t_1 - t_2)y = a(t_1^2 - t_2^2) = a(t_1 - t_2)(t_1 + t_2)$.
Thus,$y = a(t_1 + t_2)$.
Substituting $y$ into the first equation: $t_1(a(t_1 + t_2)) = x + at_1^2$.
$at_1^2 + at_1t_2 = x + at_1^2$,which gives $x = at_1t_2$.
Therefore,the point of intersection is $(at_1t_2, a(t_1 + t_2))$.

(C) The given curves are $x^2 = 4(y + 1)$ and $x^2 = -4(y + 1)$.
To find the point of intersection,set the two equations equal: $4(y + 1) = -4(y + 1)$,which implies $8(y + 1) = 0$,so $y = -1$.
Substituting $y = -1$ into either equation gives $x^2 = 0$,so $x = 0$.
The point of intersection is $(0, -1)$.
For the first curve $x^2 = 4y + 4$,differentiating with respect to $x$ gives $2x = 4 \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x}{2}$. At $(0, -1)$,the slope $m_1 = \frac{0}{2} = 0$.
For the second curve $x^2 = -4y - 4$,differentiating with respect to $x$ gives $2x = -4 \frac{dy}{dx}$,so $\frac{dy}{dx} = -\frac{x}{2}$. At $(0, -1)$,the slope $m_2 = -\frac{0}{2} = 0$.
Since $m_1 = m_2 = 0$,the curves are tangent to each other at the point of intersection.
Therefore,the angle of intersection is $0$.

(B) The given equations are $y^2 = 4(x + 1)$ and $x^2 = 4(y + 1)$.
These represent two parabolas.
The first parabola $y^2 = 4(x + 1)$ has its vertex at $(-1, 0)$ and opens to the right along the $x$-axis.
The second parabola $x^2 = 4(y + 1)$ has its vertex at $(0, -1)$ and opens upwards along the $y$-axis.
Since the axis of the first parabola is the $x$-axis and the axis of the second parabola is the $y$-axis,and these axes are perpendicular to each other,the angle of intersection between these two curves at their point of intersection is $90^\circ$.

(D) The given parabola is $y^2 = ax$,which can be written as $y^2 = 4 \left( \frac{a}{4} \right) x$.
Comparing this with the standard form $y^2 = 4Ax$,we have $A = \frac{a}{4}$.
The slope of the tangent $m$ is given by $\tan(45^{\circ}) = 1$.
For a parabola $y^2 = 4Ax$,the slope of the tangent at any point $(x_1, y_1)$ is $m = \frac{2A}{y_1}$.
Substituting the values,$1 = \frac{2(a/4)}{y_1} = \frac{a/2}{y_1}$.
Thus,$y_1 = \frac{a}{2}$.
Since the point $(x_1, y_1)$ lies on the parabola $y^2 = ax$,we have $y_1^2 = ax_1$.
Substituting $y_1 = \frac{a}{2}$,we get $\left( \frac{a}{2} \right)^2 = ax_1$,which implies $\frac{a^2}{4} = ax_1$.
Therefore,$x_1 = \frac{a}{4}$.
The point of contact is $\left( \frac{a}{4}, \frac{a}{2} \right)$.

(B) Let the parabola be $y^2 = 4ax$. The extremities of a focal chord are $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$,where $t_1t_2 = -1$.
The tangents at $P$ and $Q$ intersect at the point $(at_1t_2, a(t_1+t_2))$.
Substituting $t_1t_2 = -1$,the point of intersection is $(-a, a(t_1+t_2))$.
Since the directrix of the parabola $y^2 = 4ax$ is $x = -a$,the intersection point lies on the directrix.

(B) Let the chord pass through the origin $(0,0)$. The equation of such a line is $y = mx$.
Substituting $y = mx$ into the parabola equation $y^2 = 4ax$,we get $(mx)^2 = 4ax$,which simplifies to $m^2x^2 - 4ax = 0$.
This gives $x(m^2x - 4a) = 0$,so the intersection points are $x_1 = 0$ and $x_2 = \frac{4a}{m^2}$.
The corresponding $y$-coordinates are $y_1 = 0$ and $y_2 = m(\frac{4a}{m^2}) = \frac{4a}{m}$.
The midpoint $(h, k)$ of the chord is given by $h = \frac{0 + \frac{4a}{m^2}}{2} = \frac{2a}{m^2}$ and $k = \frac{0 + \frac{4a}{m}}{2} = \frac{2a}{m}$.
From $k = \frac{2a}{m}$,we have $m = \frac{2a}{k}$.
Substituting this into $h = \frac{2a}{m^2}$,we get $h = \frac{2a}{(2a/k)^2} = \frac{2a \cdot k^2}{4a^2} = \frac{k^2}{2a}$.
Thus,$k^2 = 2ah$. Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2ax$.

(B) The equation of the parabola is $y^2 = 8x$. Comparing with $y^2 = 4ax$,we get $a = 2$.
The slope of the given line $x - 2y + 5 = 0$ is $m = 1/2$.
Since the normal is parallel to this line,the slope of the normal is $m_n = 1/2$.
The slope of the normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $-t$.
Equating the slopes: $-t = 1/2$,so $t = -1/2$.
The coordinates of the point are $(at^2, 2at) = (2(-1/2)^2, 2(2)(-1/2)) = (2(1/4), -2) = (1/2, -2)$.
Thus,the required point is $(1/2, -2)$.

(D) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.
For a given point $(x_1, y_1)$ through which the normal passes,we have $y_1 = mx_1 - 2am - am^3$.
Rearranging this,we get the cubic equation in $m$: $am^3 + (2a - x_1)m + y_1 = 0$.
Since this is a cubic equation in $m$,it can have at most $3$ real roots.
Therefore,a maximum of $3$ normals can be drawn from a point to a parabola.

(A) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.
The slope of the normal to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is given by $m = -\frac{y_1}{2a}$.
The normal is inclined at $60^\circ$ to the $x$-axis,so its slope $m = \tan(60^\circ) = \sqrt{3}$.
Equating the slopes: $\sqrt{3} = -\frac{y_1}{2(2)} = -\frac{y_1}{4}$.
Thus,$y_1 = -4\sqrt{3}$.
Since the point lies on the parabola $y^2 = 8x$,we have $(-4\sqrt{3})^2 = 8x_1$.
$16 \times 3 = 8x_1 \implies 48 = 8x_1 \implies x_1 = 6$.
Therefore,the required point is $(6, -4\sqrt{3})$.

(C) The equation of the parabola is $y^2 = 4ax$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$.
At the point $(at^2, 2at)$,the slope of the tangent is $m_T = \frac{2a}{2at} = \frac{1}{t}$.
The slope of the normal $m_N$ is given by $m_N = -\frac{1}{m_T}$.
Therefore,$m_N = -\frac{1}{1/t} = -t$.

(B) For the parabola $y^2 = 4ax$,the slope of the tangent at any point $(x_1, y_1)$ is given by $2y y_1 = 4a(x + x_1) \implies y = \frac{2a}{y_1}(x + x_1)$.
At the point $\left( \frac{a}{4}, a \right)$,the slope of the tangent $m_t = \frac{2a}{a} = 2$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.
The equation of the normal at $\left( \frac{a}{4}, a \right)$ is given by $y - y_1 = m_n(x - x_1)$.
$y - a = -\frac{1}{2}(x - \frac{a}{4})$.
$2(y - a) = -(x - \frac{a}{4})$.
$2y - 2a = -x + \frac{a}{4}$.
$x + 2y = 2a + \frac{a}{4} = \frac{9a}{4}$.
$4x + 8y = 9a$.
$4x + 8y - 9a = 0$.

(C) The equation of the parabola is $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y}$.
At the point $P\left( \frac{a}{m^2}, \frac{2a}{m} \right)$,the slope of the tangent is $m_t = \frac{2a}{2a/m} = m$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{m}$.
The equation of the normal is $y - y_1 = m_n(x - x_1)$.
$y - \frac{2a}{m} = -\frac{1}{m} \left( x - \frac{a}{m^2} \right)$.
Multiplying by $m^3$,we get $m^3y - 2am^2 = -m^2x + a$.
Rearranging the terms,we get $m^3y + m^2x = 2am^2 + a$.

(C) The equation of the parabola is $y^2 = -8x$. Comparing this with $y^2 = -4ax$,we get $a = 2$.
The equation of the line is $y = -2x - k$,which is in the form $y = mx + c$,where $m = -2$ and $c = -k$.
The condition for the line $y = mx + c$ to be a normal to the parabola $y^2 = -4ax$ is $c = -2am - am^3$.
Substituting the values $a = 2$,$m = -2$,and $c = -k$ into the condition:
$-k = -2(2)(-2) - 2(-2)^3$
$-k = 8 - 2(-8)$
$-k = 8 + 16$
$-k = 24$
$k = -24$.

(D) The equation of the normal to the parabola $y^2 = 4ax$ at the point $t_1$ is $y = -t_1x + 2at_1 + at_1^3$.
For the point $(a, 2a)$,we have $2a = t_1(2a) \implies t_1^2 = 1$,so $t_1 = 1$ (since $y = 2a > 0$).
The relation between the parameters of two points on the parabola where the normal at one point meets the parabola at another is given by $t_2 = -t_1 - \frac{2}{t_1}$.
Substituting $t_1 = 1$ and $t_2 = t$:
$t = -1 - \frac{2}{1} = -3$.

(B) The equation of the parabola is $y^2 = 6x$. Comparing this with $y^2 = 4ax$,we get $4a = 6$,so $a = \frac{3}{2}$.
The vertex of the parabola is $(0, 0)$.
The coordinates of the endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$.
The negative end of the latus rectum is $(a, -2a) = (\frac{3}{2}, -2 \times \frac{3}{2}) = (\frac{3}{2}, -3)$.
The equation of the line passing through the vertex $(0, 0)$ and the point $(\frac{3}{2}, -3)$ is given by $y - 0 = \frac{-3 - 0}{3/2 - 0}(x - 0)$.
$y = \frac{-3}{3/2}x = -2x$.
Therefore,$y + 2x = 0$.

(C) The equation of the chord of contact of tangents drawn from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.
Here,the parabola is $y^2 = 8x$,so $4a = 8$,which means $a = 2$. The point is $(2, 5)$.
Substituting these values,the equation of the chord of contact is $5y = 2 \times 2(x + 2)$,which simplifies to $5y = 4x + 8$,or $x = \frac{5y - 8}{4}$.
Substituting $x$ into the parabola equation $y^2 = 8x$:
$y^2 = 8 \left( \frac{5y - 8}{4} \right) = 2(5y - 8) = 10y - 16$.
$y^2 - 10y + 16 = 0$,which factors as $(y - 2)(y - 8) = 0$.
Thus,$y = 2$ or $y = 8$.
If $y = 2$,$x = \frac{5(2) - 8}{4} = \frac{2}{4} = 0.5$. If $y = 8$,$x = \frac{5(8) - 8}{4} = \frac{32}{4} = 8$.
The points of intersection are $(0.5, 2)$ and $(8, 8)$.
The length of the chord is $\sqrt{(8 - 0.5)^2 + (8 - 2)^2} = \sqrt{(7.5)^2 + 6^2} = \sqrt{56.25 + 36} = \sqrt{92.25} = \sqrt{\frac{369}{4}} = \frac{3}{2}\sqrt{41}$.

(C) The semi-latus rectum of a parabola is the harmonic mean between the segments of any focal chord.
Let the segments of the focal chord be $a$ and $c$,and the semi-latus rectum be $b$.
By the property of the focal chord,we have $b = \frac{2ac}{a + c}$.
This implies that $\frac{1}{b} = \frac{a + c}{2ac} = \frac{1}{2} \left( \frac{1}{c} + \frac{1}{a} \right)$.
Since the reciprocals of $a, b, c$ are in arithmetic progression,$a, b, c$ are in harmonic progression $(H.P.)$.

(A) The equation of the line is $lx + my + n = 0$,which can be written as $\frac{lx + my}{-n} = 1$.
The equation of the parabola is $y^2 = 4ax$,which can be written as $y^2 = 4ax(1)$.
Substituting the value of $1$ from the line equation into the parabola equation,we get the homogeneous equation of the pair of lines passing through the vertex:
$y^2 = 4ax \left( \frac{lx + my}{-n} \right)$
$-ny^2 = 4alx^2 + 4amxy$
$4alx^2 + 4amxy + ny^2 = 0$.
Since these lines subtend a right angle at the vertex,the sum of the coefficients of $x^2$ and $y^2$ must be zero.
Therefore,$4al + n = 0$.

(C) Let the equation of the parallel chords be $y = mx + c$,where $m$ is constant and $c$ is a variable parameter.
Substituting $x = \frac{y - c}{m}$ into the parabola equation $y^2 = 4ax$,we get:
$y^2 = 4a \left( \frac{y - c}{m} \right)$
$my^2 - 4ay + 4ac = 0$
$y^2 - \frac{4a}{m}y + \frac{4ac}{m} = 0$
Let the points of intersection be $(x_1, y_1)$ and $(x_2, y_2)$. The sum of the roots is $y_1 + y_2 = \frac{4a}{m}$.
The $y$-coordinate of the mid-point is $Y = \frac{y_1 + y_2}{2} = \frac{2a}{m}$.
Since $Y = \frac{2a}{m}$ is a constant,the mid-points lie on a line parallel to the axis of the parabola $(y = 0)$.

(D) Let the points $P$ and $Q$ on the parabola $y^2 = 4ax$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ respectively.
The equation of the normal at point $t$ is $y = -tx + 2at + at^3$.
If the normals at $t_1$ and $t_2$ intersect at a point $R(at_3^2, 2at_3)$ on the parabola,then $t_3$ must satisfy the relation $t_1t_2 + t_3(t_1 + t_2 + t_3) + 2 = 0$ is not the standard form,but rather for the intersection of normals at $t_1, t_2$ meeting at $t_3$,the condition is $t_1t_2 = 2$ and $t_3 = -(t_1 + t_2)$.
The ordinates of $P$ and $Q$ are $y_1 = 2at_1$ and $y_2 = 2at_2$.
The product of the ordinates is $y_1y_2 = (2at_1)(2at_2) = 4a^2(t_1t_2)$.
Since $t_1t_2 = 2$,the product is $4a^2(2) = 8a^2$.

(A) The equation of the parabola is ${x^2} = 4ay$.
Comparing this with the standard form ${x^2} = 4ay$,we have the vertex at the origin and the axis along the $y$-axis.
The equation of a normal to the parabola ${x^2} = 4ay$ with slope $m$ is given by $x = my - 2am - am^3$.
Comparing this with the given equation $x = my + c$,we get $c = - 2am - am^3$.

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

(D) The equation of a normal to the parabola $y^2 = 4ax$ at the point $(am^2, -2am)$ is $y = mx - 2am - am^3$.
For the parabola $y^2 = 4x$,we have $a = 1$. Thus,the normal at $(m^2, -2m)$ is $y = mx - 2m - m^3$.
The slope of this normal is $m$.
If the normal makes equal angles with the coordinate axes,its inclination $\theta$ must be $45^{\circ}$ or $135^{\circ}$,so the slope $m = \tan(45^{\circ}) = 1$ or $m = \tan(135^{\circ}) = -1$.
Case $1$: If $m = 1$,the point is $(1^2, -2(1)) = (1, -2)$.
Case $2$: If $m = -1$,the point is $((-1)^2, -2(-1)) = (1, 2)$.
Comparing with the given options,the correct point is $(1, -2)$.

(D) The given parabola is ${y^2} = 4a(x - a)$.
Let the point on the parabola be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $y{y_1} = 2a(x + x_1 - 2a)$.
The slope of the tangent is $m_t = \frac{2a}{y_1}$.
The slope of the normal is $m = -\frac{1}{m_t} = -\frac{y_1}{2a}$,which implies $y_1 = -2am$.
Since the point $(x_1, y_1)$ lies on the parabola,${y_1^2} = 4a(x_1 - a)$.
Substituting $y_1 = -2am$,we get $(-2am)^2 = 4a(x_1 - a)$,which simplifies to $4a^2m^2 = 4a(x_1 - a)$,so $x_1 - a = am^2$,or $x_1 = a + am^2$.
The equation of the normal at $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$.
Substituting $x_1$ and $y_1$,we get $y - (-2am) = m(x - (a + am^2))$.
$y + 2am = mx - am - am^3$.
$y = m(x - a) - 2am - am^3$.

(D) Let the ends of the focal chord be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.
Since $PQ$ is a focal chord,$t_1 t_2 = -1$.
The intersection point of the tangents at $P$ and $Q$ is $(at_1 t_2, a(t_1 + t_2))$.
Substituting $t_1 t_2 = -1$,the $x$-coordinate becomes $x = a(-1) = -a$.
Thus,the tangents intersect on the line $x = -a$,which is the directrix of the parabola.
Therefore,the equation is $x + a = 0$.

(A) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.
If the normal passes through a point $(h, k)$,then $k = mh - 2am - am^3$,which rearranges to $am^3 + (2a - h)m + k = 0$.
This cubic equation in $m$ has three roots $m_1, m_2, m_3$,corresponding to the three normals drawn from the point $(h, k)$ to the parabola.
The coordinates of the feet of these normals are $(am_i^2, -2am_i)$ for $i = 1, 2, 3$.
The centroid $(x_c, y_c)$ of the triangle formed by these three points is given by:
$x_c = \frac{a(m_1^2 + m_2^2 + m_3^2)}{3}$ and $y_c = \frac{-2a(m_1 + m_2 + m_3)}{3}$.
From the cubic equation $am^3 + 0m^2 + (2a - h)m + k = 0$,we have the sum of roots $m_1 + m_2 + m_3 = 0$.
Since $m_1 + m_2 + m_3 = 0$,it follows that $y_c = 0$.
The condition $y_c = 0$ implies that the centroid lies on the axis of the parabola.

(D) The equation of a circle with the points $(x_1, y_1)$ and $(x_2, y_2)$ as the endpoints of its diameter is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Given the endpoints of the normal chord are $(3, 6)$ and $(27, -18)$,the equation of the circle is:
$(x - 3)(x - 27) + (y - 6)(y + 18) = 0$
Expanding the terms:
$(x^2 - 27x - 3x + 81) + (y^2 + 18y - 6y - 108) = 0$
${x^2} + {y^2} - 30x + 12y + 81 - 108 = 0$
${x^2} + {y^2} - 30x + 12y - 27 = 0$.

(A) For the parabola $y^2 = 4ax$ (where $a=1$),the normal at point $P(t_1^2, 2t_1)$ meets the parabola again at $Q(t_2^2, 2t_2)$,where $t_2 = -t_1 - \frac{2}{t_1}$.
If the chord $PQ$ subtends a right angle at the vertex $(0,0)$,then the product of the slopes of $OP$ and $OQ$ is $-1$.
Slope of $OP = \frac{2t_1}{t_1^2} = \frac{2}{t_1}$ and slope of $OQ = \frac{2t_2}{t_2^2} = \frac{2}{t_2}$.
So,$(\frac{2}{t_1}) \times (\frac{2}{t_2}) = -1 \Rightarrow t_1 t_2 = -4$.
Substituting $t_2 = -\frac{4}{t_1}$ into the normal condition: $-t_1 - \frac{2}{t_1} = -\frac{4}{t_1}$ $\Rightarrow t_1 = \frac{2}{t_1}$ $\Rightarrow t_1^2 = 2$.
Thus,$t_1 = \sqrt{2}$ and $t_2 = -2\sqrt{2}$.
The coordinates are $P(2, 2\sqrt{2})$ and $Q(8, -4\sqrt{2})$.
The length $PQ = \sqrt{(8-2)^2 + (-4\sqrt{2} - 2\sqrt{2})^2} = \sqrt{6^2 + (-6\sqrt{2})^2} = \sqrt{36 + 72} = \sqrt{108} = 6\sqrt{3}$.

(B) The equation of the parabola is ${y^2} = 12x$,which is of the form ${y^2} = 4ax$,where $a = 3$.
The equation of a normal to the parabola ${y^2} = 4ax$ at point $t$ is given by $y + tx = 2at + at^3$.
Substituting $a = 3$,the normal equation becomes $y + tx = 6t + 3t^3$.
Comparing this with the given normal $x + y = k$,we have $tx + y = 6t + 3t^3$ and $x + y = k$.
Since these represent the same line,the coefficients must be proportional:
$\frac{t}{1} = \frac{1}{1} = \frac{6t + 3t^3}{k}$.
From $\frac{t}{1} = 1$,we get $t = 1$.
Substituting $t = 1$ into the ratio $\frac{1}{1} = \frac{6(1) + 3(1)^3}{k}$,we get $1 = \frac{6 + 3}{k}$,which implies $k = 9$.

(A) The equation of the normal to the parabola $y^2 = 4bx$ at the point $P(bt_1^2, 2bt_1)$ is given by $y + t_1x = 2bt_1 + bt_1^3$.
Since this normal meets the parabola again at $Q(bt_2^2, 2bt_2)$,the point $Q$ must satisfy the equation of the normal.
Substituting $x = bt_2^2$ and $y = 2bt_2$ into the normal equation:
$2bt_2 + t_1(bt_2^2) = 2bt_1 + bt_1^3$.
Dividing by $b$ (assuming $b \neq 0$):
$2t_2 + t_1t_2^2 = 2t_1 + t_1^3$.
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 $t_1 \neq t_2$ (as the points are distinct),we can divide by $(t_2 - t_1)$:
$t_1(t_2 + t_1) + 2 = 0$.
$t_1t_2 + t_1^2 + 2 = 0$.
$t_1t_2 = -t_1^2 - 2$.
$t_2 = -t_1 - \frac{2}{t_1}$.

(A) The parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus is $(a, 0) = (4, 0)$.
Any focal chord passing through $(4, 0)$ with slope $m$ has the equation $y - 0 = m(x - 4)$,or $mx - y - 4m = 0$.
This line is tangent to the circle $(x - 6)^2 + y^2 = 2$,which has center $(6, 0)$ and radius $r = \sqrt{2}$.
The perpendicular distance from the center $(6, 0)$ to the line $mx - y - 4m = 0$ must be equal to the radius $\sqrt{2}$.
Using the distance formula: $\frac{|m(6) - 0 - 4m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$.
$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$.
Squaring both sides: $\frac{4m^2}{m^2 + 1} = 2$.
$4m^2 = 2m^2 + 2$.
$2m^2 = 2$,so $m^2 = 1$,which gives $m = \pm 1$.
Thus,the possible values of the slope are $\{-1, 1\}$.

(D) The equation of the parabola is ${y^2 = 8x}$,which is of the form ${y^2 = 4ax}$,so ${a = 2}$.
The point $(2, 4)$ corresponds to the parameter ${t_1}$ where ${at_1^2 = 2}$ and ${2at_1 = 4}$.
Substituting ${a = 2}$,we get ${2t_1^2 = 2 \implies t_1^2 = 1 \implies t_1 = 1}$ (since ${y > 0}$).
If the normal at ${t_1}$ meets the parabola at ${t_2}$,then ${t_2 = -t_1 - \frac{2}{t_1}}$.
Substituting ${t_1 = 1}$,we get ${t_2 = -1 - \frac{2}{1} = -3}$.
The coordinates of the point are $(at_2^2, 2at_2) = (2(-3)^2, 2(2)(-3)) = (18, -12)$.

(C) For a parabola $y^2 = 4ax$,the focus is at $(a, 0)$.
The equation of the polar of a point $(x_1, y_1)$ with respect to the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.
Substituting the focus $(a, 0)$ into the equation,we get $y(0) = 2a(x + a)$,which simplifies to $0 = 2a(x + a)$.
Since $a \neq 0$,we have $x + a = 0$,or $x = -a$.
This is the equation of the directrix of the parabola.
Therefore,the polar of the focus is the directrix.

(B) The equation of the diameter of a parabola $y^2 = 4ax$ corresponding to a chord with slope $m$ is given by $y = \frac{2a}{m}$.
Given the parabola $y^2 = x$,we have $4a = 1$,so $a = \frac{1}{4}$.
The given chord is $x - y + 1 = 0$,which can be written as $y = x + 1$. Thus,the slope $m = 1$.
Substituting these values into the formula:
$y = \frac{2 \times (1/4)}{1}$
$y = \frac{1/2}{1}$
$y = \frac{1}{2}$
$2y = 1$.

(C) The equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,so $a = 3$.
The vertex of the parabola is at $(0, 0)$.
The latus rectum is the line $y = a = 3$.
The ends of the latus rectum are found by substituting $y = 3$ into $x^2 = 12y$,which gives $x^2 = 36$,so $x = \pm 6$.
Thus,the coordinates of the ends of the latus rectum are $L_1(6, 3)$ and $L_2(-6, 3)$.
The triangle is formed by the vertices $(0, 0)$,$(6, 3)$,and $(-6, 3)$.
The base of the triangle is the length of the latus rectum,which is $6 - (-6) = 12$.
The height of the triangle is the $y$-coordinate of the latus rectum,which is $3$.
The area of the triangle is $\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 3 = 18 \ sq. \ unit$.

(D) Given the parabola $y^2 = 4x$. The ordinates of the vertices are $y_1 = 1, y_2 = 2, y_3 = 4$.
Since the points lie on the parabola,their $x$-coordinates are given by $x = \frac{y^2}{4}$.
For $y_1 = 1$,$x_1 = \frac{1^2}{4} = \frac{1}{4}$.
For $y_2 = 2$,$x_2 = \frac{2^2}{4} = 1$.
For $y_3 = 4$,$x_3 = \frac{4^2}{4} = 4$.
The vertices of the triangle are $(\frac{1}{4}, 1), (1, 2),$ and $(4, 4)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |\frac{1}{4}(2 - 4) + 1(4 - 1) + 4(1 - 2)|$.
Area $= \frac{1}{2} |\frac{1}{4}(-2) + 3 + 4(-1)|$.
Area $= \frac{1}{2} |-\frac{1}{2} + 3 - 4| = \frac{1}{2} |-\frac{1}{2} - 1| = \frac{1}{2} |-\frac{3}{2}| = \frac{3}{4}$.

(B) Let the vertices of the equilateral triangle be $O(0, 0)$,$P(x, y)$,and $Q(x, -y)$.
Since the triangle is equilateral and symmetric about the $x$-axis,the angle made by the side $OP$ with the $x$-axis is $30^\circ$.
The equation of the line $OP$ is $y = x \tan(30^\circ) = \frac{x}{\sqrt{3}}$,or $x = y\sqrt{3}$.
Substituting this into the parabola equation $y^2 = 4ax$,we get $y^2 = 4a(y\sqrt{3}) = 4\sqrt{3}ay$.
Since $y \neq 0$,we have $y = 4\sqrt{3}a$.
Then $x = (4\sqrt{3}a)\sqrt{3} = 12a$.
The side length $L$ of the triangle is the distance $PQ$,which is $2y = 2(4\sqrt{3}a) = 8a\sqrt{3}$.

(C) The coordinates of the vertices of the triangle on the parabola $y^2 = 4ax$ are given by $\left( \frac{y_1^2}{4a}, y_1 \right)$,$\left( \frac{y_2^2}{4a}, y_2 \right)$,and $\left( \frac{y_3^2}{4a}, y_3 \right)$.
The area $\Delta$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the coordinates:
$\Delta = \frac{1}{2} \left| \frac{y_1^2}{4a}(y_2 - y_3) + \frac{y_2^2}{4a}(y_3 - y_1) + \frac{y_3^2}{4a}(y_1 - y_2) \right|$
$\Delta = \frac{1}{8a} |y_1^2(y_2 - y_3) + y_2^2(y_3 - y_1) + y_3^2(y_1 - y_2)|$
Using the identity $y_1^2(y_2 - y_3) + y_2^2(y_3 - y_1) + y_3^2(y_1 - y_2) = -(y_1 - y_2)(y_2 - y_3)(y_3 - y_1)$,we get:
$\Delta = \frac{1}{8a} |(y_1 - y_2)(y_2 - y_3)(y_3 - y_1)|$.

(B) The equation of the chord of contact for a point $(x_1, y_1)$ with respect to the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.
Given the parabola $y^2 = 4x$,we have $4a = 4$,so $a = 1$.
The point is $(x_1, y_1) = (-1, 2)$.
Substituting these values into the formula:
$y(2) = 2(1)(x + (-1))$
$2y = 2(x - 1)$
$y = x - 1$.
Thus,the equation of the chord of contact is $y = x - 1$.

(C) For a 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 $2y = 2(x - 1)$,which simplifies to $y = x - 1$ or $x - y - 1 = 0$.
To find the points of contact $P$ and $Q$,we substitute $x = y + 1$ into $y^2 = 4x$:
$y^2 = 4(y + 1) \implies y^2 - 4y - 4 = 0$.
The roots are $y = \frac{4 \pm \sqrt{16 + 16}}{2} = 2 \pm 2\sqrt{2}$.
Thus,the points are $P(3 + 2\sqrt{2}, 2 + 2\sqrt{2})$ and $Q(3 - 2\sqrt{2}, 2 - 2\sqrt{2})$.
The length of the chord of contact $PQ = \sqrt{(4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = 8$.
The perpendicular distance $p$ from the point $(-1, 2)$ to the line $x - y - 1 = 0$ is $p = \frac{|-1 - 2 - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
The area of the triangle formed by the tangents and the chord of contact is given by $\frac{1}{2} \times PQ \times p = \frac{1}{2} \times 8 \times 2\sqrt{2} = 8\sqrt{2}$.

(C) Let the point on the parabola be $(x, y)$. Since $2y = x^2$,we have $y = x^2/2$.
The distance $D$ between $(x, x^2/2)$ and $(0, 3)$ is given by $D^2 = (x - 0)^2 + (x^2/2 - 3)^2$.
Let $f(x) = D^2 = x^2 + (x^4/4 - 3x^2 + 9) = x^4/4 - 2x^2 + 9$.
To find the minimum distance,we set $f'(x) = 0$:
$f'(x) = x^3 - 4x = x(x^2 - 4) = 0$.
This gives $x = 0$ or $x = \pm 2$.
For $x = 0$,$y = 0$,distance is $\sqrt{0^2 + (0-3)^2} = 3$.
For $x = \pm 2$,$y = 2^2/2 = 2$,distance is $\sqrt{(\pm 2 - 0)^2 + (2 - 3)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236$.
Since $\sqrt{5} < 3$,the points $(\pm 2, 2)$ are the nearest points.
Thus,option $(c)$ is correct.