Parabola Questions in English

(B) Let the point on the $x$-axis be $(h, 0)$.
The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$.
Since the normal passes through $(h, 0)$,we have $0 = mh - 2am - am^3$.
Assuming $m \neq 0$,we get $h = 2a + am^2$,or $am^2 = h - 2a$.
For three real normals,the cubic equation in $m$ must have three real roots.
This occurs when the point $(h, 0)$ lies inside the evolute of the parabola,which implies $h > 2a$.
Thus,the $x$-coordinate must be greater than $2a$.

(D) The parabola is given by $y^2 = 4ax$. The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$.
For a point $(x_1, y_1)$ on the parabola $y^2 = 4ax$,the equation of the tangent is $yy_1 = 2a(x + x_1)$.
For the point $(a, 2a)$,the tangent is $y(2a) = 2a(x + a) \implies y = x + a \implies x - y + a = 0$.
For the point $(a, -2a)$,the tangent is $y(-2a) = 2a(x + a) \implies -y = x + a \implies x + y + a = 0$.
Thus,the equations of the tangents are $x - y + a = 0$ and $x + y + a = 0$.

(C) For a parabola $y^2 = 4ax$,the endpoints of a focal chord can be represented as $(x_1, y_1) = (at^2, 2at)$ and $(x_2, y_2) = (a/t^2, -2a/t)$.
Here,$x_1 = at^2$ and $x_2 = a/t^2$.
The Geometric Mean $(G.M.)$ of $x_1$ and $x_2$ is $\sqrt{x_1 x_2}$.
Therefore,the square of the $G.M.$ is $(\sqrt{x_1 x_2})^2 = x_1 x_2$.
Substituting the values,we get $x_1 x_2 = (at^2) \times (a/t^2) = a^2$.
Thus,the square of the $G.M.$ is $a^2$.

(B) The given equation of the parabola is $x^2 = -8y$.
Comparing this with the standard form $x^2 = -4ay$,we get:
$-4a = -8
\implies a = 2$.
The equation of the directrix for a parabola of the form $x^2 = -4ay$ is given by $y = a$.
Substituting the value of $a$,we get $y = 2$.
Thus,the equation of the directrix is $y = 2$.

(D) The equation of the parabola is $y^2 = 4a(x + a)$.
Let $X = x + a$,then $x = X - a$. Substituting this into the equation,we get $y^2 = 4aX$.
The condition for the line $y = mx + c$ to touch the parabola $y^2 = 4aX$ is $c' = \frac{a}{m}$,where $c'$ is the intercept on the $Y$-axis.
Substituting $x = X - a$ into $y = mx + c$,we get $y = m(X - a) + c = mX + (c - am)$.
Comparing this with $y = mX + c'$,we have $c' = c - am$.
Equating the two expressions for $c'$,we get $c - am = \frac{a}{m}$.
Therefore,$c = am + \frac{a}{m}$.

(C) The given equation of the parabola is $x^2 = -ay$.
Comparing this with the standard form $x^2 = -4Ay$,we get $4A = a$,which implies $A = a/4$.
The parabola $x^2 = -4Ay$ opens downwards with its vertex at $(0, 0)$ and focus at $(0, -A)$.
The equation of the directrix for the parabola $x^2 = -4Ay$ is given by $y = A$.
Substituting $A = a/4$,we get the equation of the directrix as $y = a/4$.

(C) Given the equation of the parabola is $y^2 = 4ax$.
Let $(x_1, y_1)$ be any point on the parabola.
Differentiating $y^2 = 4ax$ with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$.
At the point $(x_1, y_1)$,the slope of the tangent is $m = \frac{dy}{dx} = \frac{2a}{y_1}$.
The length of the subnormal is given by the formula $|y_1 \cdot \frac{dy}{dx}|$.
Substituting the value of the slope,we get the length of the subnormal $= |y_1 \cdot \frac{2a}{y_1}| = 2a$.

(B) The equation of the parabola is $y^2 = 12x$. Comparing this with $y^2 = 4ax$,we get $4a = 12$,so $a = 3$.
The given line is $x + y = K$,which can be written as $y = -x + K$. The slope of this line is $m = -1$.
The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.
Substituting $a = 3$ and $m = -1$ into the equation:
$y = (-1)x - 2(3)(-1) - (3)(-1)^3$
$y = -x + 6 + 3$
$y = -x + 9$
$x + y = 9$
Comparing this with $x + y = K$,we get $K = 9$.

(B) Given curves are $y = x$ and $y^2 = 4x$.
For $y = x$,differentiating with respect to $x$,we get $m_1 = \frac{dy}{dx} = 1$.
For $y^2 = 4x$,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4$,which implies $m_2 = \frac{dy}{dx} = \frac{2}{y}$.
At the point $(4, 4)$,the slope $m_1 = 1$ and $m_2 = \frac{2}{4} = \frac{1}{2}$.
The angle of intersection $\theta$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{1 - 1/2}{1 + (1)(1/2)} \right| = \left| \frac{1/2}{3/2} \right| = \frac{1}{3}$.
Therefore,$\theta = \tan^{-1}\left(\frac{1}{3}\right)$.

(B) Given the curve $y^2 = 4ax$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$.
At the point $(a, 2a)$,the slope of the tangent $m_t = \frac{2a}{2a} = 1$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$.
The equation of the normal at $(a, 2a)$ is $y - y_1 = m_n(x - x_1)$.
Substituting the values,we get $y - 2a = -1(x - a)$.
$y - 2a = -x + a$.
Therefore,$x + y - 3a = 0$.

(B) Given the curve equation is $y^2 = 6x$.
Differentiating both sides with respect to $x$,we get:
$2y \frac{dy}{dx} = 6$
$\frac{dy}{dx} = \frac{6}{2y} = \frac{3}{y}$
Now,find the slope of the tangent at the point $(2, -3)$:
$m = \left( \frac{dy}{dx} \right)_{(2, -3)} = \frac{3}{-3} = -1$
The equation of the tangent line passing through $(x_1, y_1) = (2, -3)$ with slope $m = -1$ is given by:
$y - y_1 = m(x - x_1)$
$y - (-3) = -1(x - 2)$
$y + 3 = -x + 2$
$x + y + 1 = 0$

(A) Given the parametric equations of the curve: $x = at^2$ and $y = 2at$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.
For the tangent to be perpendicular to the $x$-axis,the slope must be undefined (i.e.,$\frac{dy}{dx} \to \infty$).
This occurs when the denominator $t = 0$.
Substituting $t = 0$ into the parametric equations:
$x = a(0)^2 = 0$
$y = 2a(0) = 0$
Therefore,the point of contact is $(0, 0)$.

(B) Given the curve equation: $y^2 = 16x$.
Differentiating with respect to $x$: $2y \frac{dy}{dx} = 16$,which gives $\frac{dy}{dx} = \frac{8}{y}$.
At the point $(1, 4)$,the slope of the tangent $m_t = \frac{8}{4} = 2$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.
The equation of the normal at $(1, 4)$ is given by $y - y_1 = m_n(x - x_1)$.
Substituting the values: $y - 4 = -\frac{1}{2}(x - 1)$.
Multiplying by $2$: $2y - 8 = -x + 1$,which simplifies to $x + 2y = 9$.

(A) Given the curve $y^2 = 4ax$. 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 $m = \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.
$1$. Length of Subtangent = $|\frac{y}{m}| = |\frac{2at}{1/t}| = 2at^2$.
$2$. Length of Tangent = $|y \sqrt{1 + \frac{1}{m^2}}| = |2at \sqrt{1 + t^2}| = 2at\sqrt{t^2 + 1}$.
$3$. Length of Subnormal = $|y \cdot m| = |2at \cdot \frac{1}{t}| = 2a$.
$4$. Length of Normal = $|y \sqrt{1 + m^2}| = |2at \sqrt{1 + \frac{1}{t^2}}| = |2at \frac{\sqrt{t^2 + 1}}{t}| = 2a\sqrt{t^2 + 1}$.
Thus,the lengths are $2at\sqrt{t^2 + 1}, 2at^2, 2a\sqrt{t^2 + 1}, 2a$.

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

(C) Given the parametric equations of the curve are $x = at^2$ and $y = 2at$.
First,we find the derivative $\frac{dy}{dx}$:
$\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.
Therefore,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.
The equation of the tangent at point $(at^2, 2at)$ is given by $(y - y_1) = m(x - x_1)$.
Substituting the values: $(y - 2at) = \frac{1}{t}(x - at^2)$.
Multiplying both sides by $t$,we get $ty - 2at^2 = x - at^2$.
Rearranging the terms,we get $ty = x + at^2$.

(D) The equations of the curves are $y^2 = 4x \dots (i)$ and $x^2 = 4y \dots (ii)$.
To find the intersection points,substitute $x = y^2/4$ from $(i)$ into $(ii)$:
$(y^2/4)^2 = 4y \implies y^4/16 = 4y \implies y^4 - 64y = 0 \implies y(y^3 - 64) = 0$.
This gives $y = 0$ or $y = 4$. If $y = 0$,$x = 0$. If $y = 4$,$x = 4$. The intersection points are $(0, 0)$ and $(4, 4)$.
Differentiating $(i)$ with respect to $x$: $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y} = m_1$.
Differentiating $(ii)$ with respect to $x$: $2x = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2} = m_2$.
At $(4, 4)$: $m_1 = \frac{2}{4} = 1/2$ and $m_2 = \frac{4}{2} = 2$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{2 - 1/2}{1 + (2)(1/2)} \right| = \left| \frac{3/2}{2} \right| = 3/4$.
Thus,$\theta = \tan^{-1}(3/4)$.

(C) Let $P(x, y)$ be a point on the parabola $y = x^2$ and $Q(0, c)$ be the given point.
Since $P$ lies on the parabola,$y = x^2$. The distance $PQ$ is given by $PQ = \sqrt{(x - 0)^2 + (y - c)^2}$.
To minimize the distance,we minimize $Z = PQ^2 = x^2 + (x^2 - c)^2$.
$Z = x^2 + x^4 - 2cx^2 + c^2 = x^4 + x^2(1 - 2c) + c^2$.
Taking the derivative with respect to $x$: $\frac{dZ}{dx} = 4x^3 + 2x(1 - 2c) = 2x(2x^2 + 1 - 2c)$.
Setting $\frac{dZ}{dx} = 0$,we get $x = 0$ or $x^2 = \frac{2c - 1}{2}$.
If $c \leq \frac{1}{2}$,then $x = 0$ is the only real solution,and the minimum distance is at $x = 0$,$PQ = \sqrt{0^2 + (0 - c)^2} = c$.
If $c > \frac{1}{2}$,then $x^2 = \frac{2c - 1}{2}$ gives the minimum. Substituting $x^2 = \frac{2c - 1}{2}$ into $Z$:
$Z = (\frac{2c - 1}{2})^2 + (\frac{2c - 1}{2} - c)^2 = \frac{(2c - 1)^2}{4} + (\frac{2c - 1 - 2c}{2})^2 = \frac{4c^2 - 4c + 1}{4} + \frac{1}{4} = \frac{4c^2 - 4c + 2}{4} = c^2 - c + \frac{1}{2}$.
Wait,re-evaluating $Z = x^2 + (x^2 - c)^2$ at $x^2 = c - \frac{1}{2}$:
$Z = (c - \frac{1}{2}) + (c - \frac{1}{2} - c)^2 = c - \frac{1}{2} + \frac{1}{4} = c - \frac{1}{4} = \frac{4c - 1}{4}$.
Thus,$PQ = \sqrt{\frac{4c - 1}{4}} = \frac{\sqrt{4c - 1}}{2}$.

(B) Let a point on the parabola be $(x, x^2)$. The square of the distance $d$ from $(0, c)$ is given by $z = d^2 = x^2 + (x^2 - c)^2$.
Let $t = x^2$. Since $0 \leq x \leq 5$,we have $0 \leq t \leq 25$. Then $z = t + (t - c)^2 = t^2 + t(1 - 2c) + c^2$.
To find the minimum,differentiate with respect to $t$: $\frac{dz}{dt} = 2t + 1 - 2c$.
Setting $\frac{dz}{dt} = 0$ gives $t = c - 1/2$.
If $0 \leq c - 1/2 \leq 25$ (i.e.,$1/2 \leq c \leq 51/2$),the minimum occurs at $t = c - 1/2$.
The minimum distance is $\sqrt{t + (t - c)^2} = \sqrt{(c - 1/2) + (-1/2)^2} = \sqrt{c - 1/2 + 1/4} = \sqrt{c - 1/4}$.

(D) Given the parametric equations $x = t^2$ and $y = 2t$.
First,find the derivative $\frac{dy}{dx}$:
$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$.
Therefore,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 $m' = -\frac{1}{m} = -1$.
At $t = 1$,the coordinates of the point are $x = (1)^2 = 1$ and $y = 2(1) = 2$.
The equation of the normal line passing through $(1, 2)$ with slope $-1$ is:
$y - 2 = -1(x - 1)$
$y - 2 = -x + 1$
$x + y - 3 = 0$.

(C) The given equation of the parabola is $y = \frac{a^3 x^2}{3} + \frac{a^2 x}{2} - 2a$.
To find the vertex,we complete the square for $x$:
$y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a} x \right) - 2a$
$y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a} x + \left(\frac{3}{4a}\right)^2 - \left(\frac{3}{4a}\right)^2 \right) - 2a$
$y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{a^3}{3} \cdot \frac{9}{16a^2} - 2a$
$y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{3a}{16} - 2a$
$y + \frac{35a}{16} = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2$.
The vertex $(h, k)$ is given by $h = -\frac{3}{4a}$ and $k = -\frac{35a}{16}$.
To find the locus,we eliminate $a$:
From $h = -\frac{3}{4a}$,we get $a = -\frac{3}{4h}$.
Substitute this into $k = -\frac{35a}{16}$:
$k = -\frac{35}{16} \left( -\frac{3}{4h} \right) = \frac{105}{64h}$.
Therefore,$hk = \frac{105}{64}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $xy = \frac{105}{64}$.

(D) The equation of the parabola is $y^2 = 8x$,which is of the form $y^2 = 4ax$,where $4a = 8$,so $a = 2$.
The directrix of the parabola $y^2 = 4ax$ is given by $x = -a$,which is $x = -2$.
It is a standard property of parabolas that the locus of the intersection of two perpendicular tangents is the directrix of the parabola.
Since the point lies on the given tangent line $y = x + 2$ and also on the directrix $x = -2$,we substitute $x = -2$ into the line equation:
$y = (-2) + 2 = 0$.
Thus,the required point is $(-2, 0)$.

(B) The focus of the parabola is $S = (0,0)$.
The directrix of the parabola is the line $x = 2$.
The axis of the parabola is the line passing through the focus and perpendicular to the directrix. Since the directrix is $x = 2$ (a vertical line),the axis is the $x$-axis $(y = 0)$.
The vertex of a parabola is the midpoint of the segment connecting the focus and the point of intersection of the axis and the directrix.
The point of intersection of the axis $(y = 0)$ and the directrix $(x = 2)$ is $(2,0)$.
The vertex is the midpoint of $(0,0)$ and $(2,0)$,which is $(\frac{0+2}{2}, \frac{0+0}{2}) = (1,0)$.
Thus,the vertex of the parabola is at $(1,0)$.

(D) The locus of the point of intersection of two perpendicular tangents to a parabola is its directrix.
For the parabola $y^2 = 4ax$,the equation of the directrix is $x = -a$.
Given the equation $y^2 = 4x$,we have $4a = 4$,which implies $a = 1$.
Therefore,the equation of the directrix is $x = -1$.
Thus,the locus of $P$ is $x = -1$.

(C) The equation of the first parabola is $y^2 = 4x$,which is of the form $y^2 = 4ax$ with $a = 1$.
The equation of the tangent to $y^2 = 4x$ with slope $m$ is $y = mx + \frac{a}{m} = mx + \frac{1}{m} \quad (1)$.
The equation of the second parabola is $x^2 = -32y$,which is of the form $x^2 = 4Ay$ with $4A = -32$,so $A = -8$.
The equation of the tangent to $x^2 = 4Ay$ with slope $m$ is $y = mx - Am^2$. Substituting $A = -8$,we get $y = mx - (-8)m^2 = mx + 8m^2 \quad (2)$.
Since the line is common to both parabolas,we equate the intercepts from $(1)$ and $(2)$:
$\frac{1}{m} = 8m^2$.
This simplifies to $8m^3 = 1$,which gives $m^3 = \frac{1}{8}$.
Taking the cube root,we get $m = \frac{1}{2}$.

(C) The minimum distance from a point to a curve occurs along the normal to the curve at that point.
For the parabola ${y^2} = 8x$,we have $4a = 8$,so $a = 2$.
The normal to the parabola at point $P(at^2, 2at) = (2t^2, 4t)$ is given by $y = -tx + 2at + at^3$.
Substituting $a = 2$,the normal is $y = -tx + 4t + 2t^3$.
Since this normal passes through the centre of the circle $C(0, -6)$,we have:
$-6 = -t(0) + 4t + 2t^3
$ $\Rightarrow 2t^3 + 4t + 6 = 0
$ $\Rightarrow t^3 + 2t + 3 = 0$.
By inspection,$t = -1$ is a root.
For $t = -1$,the point $P$ is $(2(-1)^2, 4(-1)) = (2, -4)$.
The distance $CP$ is the radius $r$ of the required circle:
$r^2 = CP^2 = (2 - 0)^2 + (-4 - (-6))^2 = 2^2 + 2^2 = 4 + 4 = 8$.
The equation of the circle with centre $P(2, -4)$ and radius squared $r^2 = 8$ is:
$(x - 2)^2 + (y + 4)^2 = 8
$ $\Rightarrow x^2 - 4x + 4 + y^2 + 8y + 16 = 8
$ $\Rightarrow x^2 + y^2 - 4x + 8y + 12 = 0$.

(A) The equation of the parabola is ${y^2} = 16x$,so $4a = 16$,which implies $a = 4$.
For the point $P(16, 16)$,we have $y_1^2 = 16x_1$,so $16^2 = 16(16)$,which is satisfied.
The tangent at $P(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$,so $16y = 8(x + 16)$,which simplifies to $2y = x + 16$.
Setting $y = 0$ for the intersection with the $x$-axis,we get $x = -16$,so $A = (-16, 0)$.
The normal at $P(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{2a}(x - x_1)$,so $y - 16 = -\frac{16}{8}(x - 16)$,which simplifies to $y - 16 = -2(x - 16)$,or $y = -2x + 48$.
Setting $y = 0$ for the intersection with the $x$-axis,we get $2x = 48$,so $x = 24$,thus $B = (24, 0)$.
The circle passes through $A(-16, 0)$,$B(24, 0)$,and $P(16, 16)$. Since $A$ and $B$ lie on the $x$-axis,the center $C$ must have an $x$-coordinate of $\frac{-16 + 24}{2} = 4$.
Let $C = (4, k)$. Since $CA = CP$,we have $(4 - (-16))^2 + (k - 0)^2 = (4 - 16)^2 + (k - 16)^2$.
$20^2 + k^2 = (-12)^2 + (k - 16)^2 \Rightarrow 400 + k^2 = 144 + k^2 - 32k + 256$.
$400 = 400 - 32k \Rightarrow k = 0$. Thus,$C = (4, 0)$.
Slope of $PC$ $(m_1)$ = $\frac{16 - 0}{16 - 4} = \frac{16}{12} = \frac{4}{3}$.
Slope of $PB$ $(m_2)$ = $\frac{16 - 0}{16 - 24} = \frac{16}{-8} = -2$.
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{4}{3} - (-2)}{1 + (\frac{4}{3})(-2)} \right| = \left| \frac{\frac{10}{3}}{1 - \frac{8}{3}} \right| = \left| \frac{\frac{10}{3}}{-\frac{5}{3}} \right| = |-2| = 2$.

(A) The length of the latus rectum of a parabola is given by $L.R. = 2 \times d$,where $d$ is the perpendicular distance from the focus $(x_1, y_1)$ to the directrix $ax + by + c = 0$.
The distance $d$ is calculated as $d = \left| \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \right|$.
Given focus $(3, 3)$ and directrix $3x - 4y - 2 = 0$,we have:
$d = \left| \frac{3(3) - 4(3) - 2}{\sqrt{3^2 + (-4)^2}} \right| = \left| \frac{9 - 12 - 2}{\sqrt{9 + 16}} \right| = \left| \frac{-5}{5} \right| = 1$.
Therefore,the length of the latus rectum is $L.R. = 2 \times 1 = 2$.

(C) Given equation: ${y^2} - 2x - 2y + 5 = 0$
Rearranging the terms: ${y^2} - 2y = 2x - 5$
Adding $1$ to both sides to complete the square: ${y^2} - 2y + 1 = 2x - 5 + 1$
${(y - 1)^2} = 2x - 4$
${(y - 1)^2} = 2(x - 2)$
This is of the form ${(y - k)^2} = 4a(x - h)$,where $4a = 2$,so $a = \frac{1}{2}$,vertex $(h, k) = (2, 1)$.
The focus is $(h + a, k) = (2 + \frac{1}{2}, 1) = (\frac{5}{2}, 1)$.
The directrix is $x = h - a = 2 - \frac{1}{2} = \frac{3}{2}$.

(C) Given $x = t^2 + t + 1$ and $y = t^2 - t + 1$.
Adding the two equations: $x + y = 2t^2 + 2 = 2(t^2 + 1)$.
Subtracting the two equations: $x - y = 2t$,which implies $t = \frac{x - y}{2}$.
Substituting $t$ into the expression for $x + y$:
$x + y = 2\left(\left(\frac{x - y}{2}\right)^2 + 1\right) = 2\left(\frac{(x - y)^2}{4} + 1\right) = \frac{(x - y)^2}{2} + 2$.
Multiplying by $2$: $2(x + y) = (x - y)^2 + 4$.
Expanding: $2x + 2y = x^2 - 2xy + y^2 + 4$.
Rearranging: $x^2 - 2xy + y^2 - 2x - 2y + 4 = 0$.
Comparing with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a = 1, h = -1, b = 1$.
Since $h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$,the curve represents a parabola.

(B) The equation of the tangent at point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.
For the given parabola $y^2 = 4x$,we have $a = 1$.
The coordinates of the ends of the latus rectum are $L(1, 2)$ and $L'(1, -2)$.
The equation of the tangent at $L(1, 2)$ is $2y = 2(x + 1)$,which simplifies to $y = x + 1$.
The equation of the tangent at $L'(1, -2)$ is $-2y = 2(x + 1)$,which simplifies to $y = -(x + 1)$.
Solving these two equations simultaneously:
$x + 1 = -(x + 1)$
$2(x + 1) = 0$
$x = -1$
Substituting $x = -1$ into $y = x + 1$,we get $y = 0$.
Thus,the point of intersection is $(-1, 0)$.

(B) Let $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ be two points on the parabola $y^2 = 4ax$.
Then the tangents at $P$ and $Q$ intersect at $T(at_1t_2, a(t_1 + t_2))$.
Now,$SP = \sqrt{(at_1^2 - a)^2 + (2at_1 - 0)^2} = a(t_1^2 + 1)$.
$SQ = a(t_2^2 + 1)$.
$ST = \sqrt{(at_1t_2 - a)^2 + (a(t_1 + t_2) - 0)^2} = a\sqrt{(1 + t_1^2)(1 + t_2^2)}$.
Therefore,$ST^2 = a^2(1 + t_1^2)(1 + t_2^2) = SP \cdot SQ$.
Since $ST^2 = SP \cdot SQ$,the terms $SP, ST, SQ$ are in $G.P.$

(D) The equation of the parabola is $y^2 = 4ax$. The focus of this parabola is at $(a, 0)$.
Substituting the coordinates of the focus $(a, 0)$ into the line equation $x - y - a = 0$,we get $a - 0 - a = 0$,which is $0 = 0$.
Since the line passes through the focus,it is a focal chord of the parabola.
It is a standard property of parabolas that the tangents drawn at the endpoints of any focal chord intersect at a right angle $(\frac{\pi}{2})$ on the directrix.
Therefore,the angle between the tangents is $\frac{\pi}{2}$.

(A) For the parabola ${y^2} = 4ax$,we have $4a = 12$,so $a = 3$.
The coordinates of the ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(3, 6)$ and $(3, -6)$.
The equation of the tangent to the parabola ${y^2} = 4ax$ at point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
For point $(3, 6)$,the tangent is $6y = 6(x + 3) \implies y = x + 3$.
For point $(3, -6)$,the tangent is $-6y = 6(x + 3) \implies y = -(x + 3)$.
Solving these two equations: $x + 3 = -(x + 3) \implies 2x + 6 = 0 \implies x = -3$.
Since the equation of the directrix for ${y^2} = 12x$ is $x = -a$,which is $x = -3$,the tangents meet at the directrix.

(C) Let $P(at^2, 2at)$ be any point on the parabola $y^2 = 4ax$. The equation of the tangent at $P$ is $ty = x + at^2$. It meets the axis $(y=0)$ at $T(-at^2, 0)$.
The equation of the normal at $P$ is $y = -tx + 2at + at^3$. It meets the axis $(y=0)$ at $G(2a + at^2, 0)$.
The focus $S$ is at $(a, 0)$.
$SP = \sqrt{(at^2 - a)^2 + (2at - 0)^2} = \sqrt{a^2(t^2-1)^2 + 4a^2t^2} = \sqrt{a^2(t^4 - 2t^2 + 1 + 4t^2)} = \sqrt{a^2(t^2+1)^2} = a(t^2+1)$.
$ST = |x_S - x_T| = |a - (-at^2)| = a(1 + t^2)$.
$SG = |x_G - x_S| = |(2a + at^2) - a| = a(1 + t^2)$.
Thus,$ST = SG = SP$.

(C) The equation of the normal to the parabola $y^2 = 4ax$ in slope form is $y = mx - 2am - am^3$.
For the given parabola $y^2 = x$,we have $4a = 1$,which implies $a = \frac{1}{4}$.
Substituting $a$ into the normal equation,we get $y = mx - \frac{1}{2}m - \frac{1}{4}m^3$.
Since the normal passes through the point $(C, 0)$,we substitute $x = C$ and $y = 0$:
$0 = mC - \frac{1}{2}m - \frac{1}{4}m^3$.
This simplifies to $m(C - \frac{1}{2} - \frac{1}{4}m^2) = 0$.
One solution is $m = 0$. For the other two normals to be real and distinct,the quadratic equation $C - \frac{1}{2} - \frac{1}{4}m^2 = 0$ must have two distinct real roots for $m^2$.
This requires $C - \frac{1}{2} > 0$,which implies $C > \frac{1}{2}$.

(B) The equation of the line passing through the points $(au^2, 2au)$ and $(av^2, 2av)$ is given by:
$y - 2au = \frac{2av - 2au}{av^2 - au^2}(x - au^2)$
$y - 2au = \frac{2a(v - u)}{a(v - u)(v + u)}(x - au^2)$
$y - 2au = \frac{2}{v + u}(x - au^2)$
Since this is a focal chord,it must pass through the focus $(a, 0)$.
Substituting $(a, 0)$ into the equation:
$0 - 2au = \frac{2}{v + u}(a - au^2)$
$-2au = \frac{2a(1 - u^2)}{v + u}$
$-u(v + u) = 1 - u^2$
$-uv - u^2 = 1 - u^2$
$-uv = 1$
$uv + 1 = 0$

(D) Let the coordinates of point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.
The slope of the tangent at $P$ is given by differentiating $y^2 = 4ax$ with respect to $x$:
$2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
The slope of the normal at $P$ is $-t$.
The equation of the normal at $P(at^2, 2at)$ is:
$y - 2at = -t(x - at^2)$
$y - 2at = -tx + at^3$
$tx + y = 2at + at^3$.
To find the $x$-intercept $N$,set $y = 0$:
$tx = 2at + at^3 \implies x = 2a + at^2$.
Let $M$ be the projection of $P$ on the $x$-axis,so $M$ is $(at^2, 0)$.
The length of the subnormal is the distance $MN = |x_N - x_M| = |(2a + at^2) - at^2| = 2a$.

(B) The eccentricity $e$ of a conic section is defined as the ratio of the distance of a point from the focus to its distance from the directrix.
For a parabola,the distance from the focus is equal to the distance from the directrix.
Therefore,the eccentricity $e = 1$.

(B) The slope of the tangent at any point $(x, y)$ is $\frac{dy}{dx}$.
The slope of the normal at $(x, y)$ is $-\frac{dx}{dy}$.
The equation of the normal at $(x, y)$ is given by:
$Y - y = -\frac{dx}{dy}(X - x)$
Since the normal intersects the $x$-axis at $(x + 1, 0)$,we substitute $X = x + 1$ and $Y = 0$:
$0 - y = -\frac{dx}{dy}(x + 1 - x)$
$-y = -\frac{dx}{dy}(1)$
$y = \frac{dx}{dy}$
$y \, dy = dx$
Integrating both sides,we get:
$\int y \, dy = \int dx$
$\frac{y^2}{2} = x + C$
Since the conic passes through the origin $(0, 0)$,we have $0 = 0 + C$,so $C = 0$.
Thus,the equation of the conic is $y^2 = 2x$.
Comparing this with the standard form $y^2 = 4ax$,we get $4a = 2$,which is the length of the latus rectum.
Therefore,the length of the latus rectum is $2$.

(B) Option $(A): x = 3 \cos t, y = 4 \sin t$
Eliminating $t$,we get $\frac{x^2}{9} + \frac{y^2}{16} = 1$,which is an ellipse.
Option $(B): x^2 - 2 = -2 \cos t \Rightarrow x^2 = 2 - 2 \cos t = 2(1 - \cos t)$.
Also,$y = 4 \cos^2 \frac{t}{2} = 2(1 + \cos t)$.
From the first,$\cos t = 1 - \frac{x^2}{2}$.
Substituting into $y$: $y = 2(1 + 1 - \frac{x^2}{2}) = 2(2 - \frac{x^2}{2}) = 4 - x^2$.
This is $x^2 = -(y - 4)$,which is a parabola.
Option $(C): \sqrt{x} = \tan t, \sqrt{y} = \sec t$
Eliminating $t$ using $\sec^2 t - \tan^2 t = 1$,we get $y - x = 1$,which is a straight line.
Option $(D): x = \sqrt{1 - \sin t}, y = \sin \frac{t}{2} + \cos \frac{t}{2}$
$x^2 = 1 - \sin t$ and $y^2 = 1 + \sin t$.
Adding them,$x^2 + y^2 = 2$,which is a circle.

(B) For a parabola $y^2 = 4ax$,the chord of contact from a point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.
Here,$a = 1$ and the point is $(2, 3)$,so the chord of contact is $3y = 2(x + 2)$,which simplifies to $3y = 2x + 4$,or $x = \frac{3y - 4}{2}$.
Substituting this into the parabola equation $y^2 = 4x$:
$y^2 = 4(\frac{3y - 4}{2})$
$y^2 = 2(3y - 4)$
$y^2 - 6y + 8 = 0$
$(y - 2)(y - 4) = 0$
Thus,$y = 2$ or $y = 4$.
If $y = 2$,$x = \frac{3(2) - 4}{2} = \frac{2}{2} = 1$.
If $y = 4$,$x = \frac{3(4) - 4}{2} = \frac{8}{2} = 4$.
The points of contact are $(1, 2)$ and $(4, 4)$.

(C) The equation of a tangent to the parabola $y^2 = 4x$ with slope $m$ is $y = mx + \frac{1}{m}$.
Since the tangent passes through $P(h, k)$,we have $k = mh + \frac{1}{m}$,which simplifies to $m^2h - mk + 1 = 0$.
Let $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$ be the roots of this quadratic equation.
From the properties of roots,$m_1 + m_2 = \frac{k}{h}$ and $m_1 m_2 = \frac{1}{h}$.
Given $\theta_1 + \theta_2 = \frac{\pi}{4}$,we have $\tan(\theta_1 + \theta_2) = \tan(\frac{\pi}{4}) = 1$.
Using the formula $\tan(\theta_1 + \theta_2) = \frac{m_1 + m_2}{1 - m_1 m_2} = 1$.
Substituting the values,$\frac{k/h}{1 - 1/h} = 1$.
$\frac{k}{h-1} = 1 \Rightarrow k = h - 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y = x - 1$,or $x - y - 1 = 0$.

(B) The equation of the normal to the parabola $y^2 = 4ax$ (where $a=1$) at parameter $t$ is $y + tx = 2at + at^3$,which simplifies to $y + tx = 2t + t^3$.
The slope of this normal is $-t$. Given that the normal makes an angle of $\frac{\pi}{4}$ with the $x$-axis,its slope is $\tan(\frac{\pi}{4}) = 1$.
Therefore,$-t = 1$,which gives $t = -1$.
The coordinates of point $P$ are $(at^2, 2at) = (1, -2)$.
If the normal at $t_1$ meets the parabola again at $t_2$,then $t_2 = -t_1 - \frac{2}{t_1}$.
Substituting $t_1 = -1$,we get $t_2 = -(-1) - \frac{2}{-1} = 1 + 2 = 3$.
The coordinates of point $Q$ are $(at_2^2, 2at_2) = (3^2, 2(3)) = (9, 6)$.
The length of the normal chord $PQ$ is the distance between $P(1, -2)$ and $Q(9, 6)$:
$PQ = \sqrt{(9-1)^2 + (6 - (-2))^2} = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$.

(B) Let the parameters of the endpoints of the focal chord be $t_1$ and $t_2$.
For a parabola $y^2 = 4ax$,the coordinates are $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$.
Since it is a focal chord,the product of the parameters is $t_1t_2 = -1$.
We have $x_1 = at_1^2$ and $x_2 = at_2^2$,so $x_1x_2 = a^2(t_1t_2)^2 = a^2(-1)^2 = a^2$.
We have $y_1 = 2at_1$ and $y_2 = 2at_2$,so $y_1y_2 = 4a^2(t_1t_2) = 4a^2(-1) = -4a^2$.
Therefore,$x_1x_2 + y_1y_2 = a^2 - 4a^2 = -3a^2$.

(D) The equation of the normal to the parabola $y^2 = 4ax$ at parameter $t_1$ is $y + t_1x = 2at_1 + at_1^3$.
Since this normal cuts the parabola at $t_2$,we have the relation $t_2 = -(t_1 + \frac{2}{t_1})$.
Squaring both sides,we get $t_2^2 = (t_1 + \frac{2}{t_1})^2 = (t_1 - \frac{2}{t_1})^2 + 8$.
Since $(t_1 - \frac{2}{t_1})^2 \geq 0$,it follows that $t_2^2 \geq 8$.

(A) Let the common focus be $(h, k)$.
The equation of a parabola with focus $(h, k)$ and directrix $L = 0$ is $(x - h)^2 + (y - k)^2 = L^2$.
For the first parabola with directrix $y = 0$ (the $x$-axis),the equation is $(x - h)^2 + (y - k)^2 = y^2$.
For the second parabola with directrix $x = 0$ (the $y$-axis),the equation is $(x - h)^2 + (y - k)^2 = x^2$.
To find the common chord,we subtract the two equations:
$((x - h)^2 + (y - k)^2 - y^2) - ((x - h)^2 + (y - k)^2 - x^2) = 0$
This simplifies to $x^2 - y^2 = 0$,which implies $x^2 = y^2$,or $y = \pm x$.
Thus,the slopes of the common chords are $m = 1$ and $m = -1$,which can be written as $\pm 1$.

(B) Let the point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.
The equation of the tangent at $P$ is $ty = x + at^2$ $....(1)$
The focus of the parabola is $S(a, 0)$.
The line perpendicular to the tangent $(1)$ passing through $S(a, 0)$ has the slope $-t$. Its equation is $y - 0 = -t(x - a)$,which simplifies to $tx + y = at$ $....(2)$
The line joining the vertex $O(0, 0)$ to $P(at^2, 2at)$ has the slope $\frac{2at}{at^2} = \frac{2}{t}$. Its equation is $y = \frac{2}{t}x$,or $2x - ty = 0$ $....(3)$
To find the locus of $R(h, k)$,we eliminate $t$ from $(2)$ and $(3)$.
From $(3)$,$t = \frac{2h}{k}$.
Substituting this into $(2)$: $(\frac{2h}{k})h + k = a(\frac{2h}{k})$.
Multiplying by $k$: $2h^2 + k^2 = 2ah$.
Replacing $(h, k)$ with $(x, y)$,the locus is $2x^2 + y^2 - 2ax = 0$.