Parabola Questions in English

(C) The given equation of the parabola is ${x^2} = 2x + 2y$.
Rearranging the terms,we get ${x^2} - 2x = 2y$.
Completing the square on the left side: ${x^2} - 2x + 1 = 2y + 1$.
This simplifies to ${(x - 1)^2} = 2\left( y + \frac{1}{2} \right)$.
Comparing this with the standard form ${(x - h)^2} = 4a(y - k)$,we have $4a = 2$,which gives $a = \frac{1}{2}$.
The vertex $(h, k)$ is $(1, -\frac{1}{2})$.
The focus of the parabola is given by $(h, k + a)$.
Substituting the values,we get $(1, -\frac{1}{2} + \frac{1}{2}) = (1, 0)$.

(A) Given equation of the parabola is ${y^2} - 4y - 2x - 8 = 0$.
Completing the square for the $y$ terms:
${y^2} - 4y = 2x + 8$
${y^2} - 4y + 4 = 2x + 8 + 4$
${(y - 2)^2} = 2x + 12$
${(y - 2)^2} = 2(x + 6)$.
Comparing this with the standard form ${(y - k)^2} = 4a(x - h)$,we get $4a = 2$,which implies $a = 0.5$.
The length of the latus rectum is given by $|4a|$.
Therefore,the length of the latus rectum is $2$.

(A) The definition of a parabola is the locus of a point $P(x, y)$ such that its distance from the focus $(a, b)$ is equal to its perpendicular distance from the directrix $\frac{x}{a} + \frac{y}{b} - 1 = 0$.
$(x - a)^2 + (y - b)^2 = \left( \frac{\frac{x}{a} + \frac{y}{b} - 1}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} \right)^2$
$(x - a)^2 + (y - b)^2 = \left( \frac{bx + ay - ab}{\sqrt{a^2 + b^2}} \right)^2$
$(x^2 - 2ax + a^2 + y^2 - 2by + b^2) = \frac{(bx + ay - ab)^2}{a^2 + b^2}$
$(a^2 + b^2)(x^2 + y^2 - 2ax - 2by + a^2 + b^2) = (bx + ay - ab)^2$
Expanding both sides and simplifying leads to:
$(ax - by)^2 - 2a^3x - 2b^3y + a^4 + a^2b^2 + b^4 = 0$.

(C) Given equation: $4y^2 - 20y + 2x + 17 = 0$
Divide by $4$: $y^2 - 5y + \frac{1}{2}x + \frac{17}{4} = 0$
Complete the square for $y$: $(y - \frac{5}{2})^2 - \frac{25}{4} + \frac{1}{2}x + \frac{17}{4} = 0$
$(y - \frac{5}{2})^2 = -\frac{1}{2}x + 2$
$(y - \frac{5}{2})^2 = -\frac{1}{2}(x - 4)$
Comparing with $(y - k)^2 = -4a(x - h)$,we get $4a = \frac{1}{2} = 0.5$.
Thus,the length of the latus rectum is $0.5$.

(B) The given equation is $x^2 - 4x - 4y + 4 = 0$.
By definition,a parabola is the locus of a point that moves such that its distance from a fixed point (focus) is equal to its distance from a fixed line (directrix).
Therefore,the eccentricity $e$ of any parabola is always equal to $1$.

(B) Given equation: $3x - 2y^2 - 4y + 7 = 0$
Rearranging the terms: $2y^2 + 4y = 3x + 7$
Divide by $2$: $y^2 + 2y = \frac{3}{2}x + \frac{7}{2}$
Complete the square: $y^2 + 2y + 1 = \frac{3}{2}x + \frac{7}{2} + 1$
$(y + 1)^2 = \frac{3}{2}x + \frac{9}{2}$
$(y + 1)^2 = \frac{3}{2}(x + 3)$
Comparing with the standard form $(y - k)^2 = 4a(x - h)$,the vertex $(h, k)$ is $(-3, -1)$.

(B) Given equation of the parabola: $4y^2 - 4y - 6x = 5$
Completing the square for $y$: $4(y^2 - y) = 6x + 5$
$4(y^2 - y + 1/4) = 6x + 5 + 1$
$4(y - 1/2)^2 = 6(x + 1)$
$(y - 1/2)^2 = \frac{3}{2}(x + 1)$
Comparing with the standard form $Y^2 = 4aX$,where $Y = y - 1/2$,$X = x + 1$,and $4a = 3/2$.
Thus,$a = 3/8$.
The focus in the $(X, Y)$ coordinate system is $(a, 0) = (3/8, 0)$.
Converting back to $(x, y)$ coordinates:
$x + 1 = 3/8 \Rightarrow x = -5/8$
$y - 1/2 = 0 \Rightarrow y = 1/2$
Therefore,the focus is $(-5/8, 1/2)$.

(A) Given equation of the parabola is $x^2 + 8x + 12y + 4 = 0$.
Completing the square for the $x$ terms:
$x^2 + 8x = -12y - 4$
$(x^2 + 8x + 16) = -12y - 4 + 16$
$(x + 4)^2 = -12y + 12$
$(x + 4)^2 = -12(y - 1)$
Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,where $(h, k)$ is the vertex:
$h = -4$ and $k = 1$.
Thus,the vertex is $(-4, 1)$.

(C) The standard equation of a parabola is given by ${(y - \beta)^2} = 4a(x - \alpha)$.
Comparing the given equation ${(y - 2)^2} = 20(x + 3)$ with the standard form,we identify the vertex $(\alpha, \beta) = (-3, 2)$ and $4a = 20$,which gives $a = 5$.
The focus of a parabola of the form ${(y - \beta)^2} = 4a(x - \alpha)$ is given by the coordinates $(\alpha + a, \beta)$.
Substituting the values,the focus is $(-3 + 5, 2) = (2, 2)$.

(C) The given equation of the parabola is ${x^2 - 4x - 8y + 12 = 0}$.
Rearranging the terms,we get:
${x^2 - 4x = 8y - 12}$
Completing the square on the left side:
${(x^2 - 4x + 4) = 8y - 12 + 4}$
${(x - 2)^2 = 8y - 8}$
${(x - 2)^2 = 8(y - 1)}$
Comparing this with the standard form ${(x - h)^2 = 4a(y - k)}$,we find that ${4a = 8}$.
Thus,the length of the latus rectum is $8$.

(D) Given equation of the parabola is $y^2 - x - 2y + 2 = 0$.
Rearranging the terms,we get $y^2 - 2y = x - 2$.
Completing the square on the left side: $(y - 1)^2 - 1 = x - 2$.
$(y - 1)^2 = x - 1$.
Let $Y = y - 1$ and $X = x - 1$. The equation becomes $Y^2 = X$.
Comparing this with the standard form $Y^2 = 4aX$,we have $4a = 1$,so $a = 1/4$.
The focus of $Y^2 = 4aX$ is $(a, 0)$,which is $(1/4, 0)$.
To find the focus in the $(x, y)$ coordinate system,we substitute back:
$X = 1/4 \implies x - 1 = 1/4 \implies x = 5/4$.
$Y = 0 \implies y - 1 = 0 \implies y = 1$.
Thus,the focus is $(5/4, 1)$.

(D) The standard equation of a parabola of the form $(y - k)^2 = 4a(x - h)$ has its vertex at $(h, k)$.
Comparing the given equation $(y - 2)^2 = 16(x - 1)$ with the standard form $(y - k)^2 = 4a(x - h)$,we identify $h = 1$ and $k = 2$.
Thus,the vertex of the parabola is $(1, 2)$.

(C) Given,the vertex of the parabola $(h, k) = (1, 1)$ and the focus $(h + a, k) = (3, 1)$.
Since the $y$-coordinates of the vertex and focus are the same,the axis of the parabola is parallel to the $x$-axis.
Comparing the coordinates,we have $h = 1$,$k = 1$,and $h + a = 3$.
Substituting $h = 1$ into $h + a = 3$,we get $1 + a = 3$,which implies $a = 2$.
The standard equation of a parabola opening to the right with vertex $(h, k)$ is $(y - k)^2 = 4a(x - h)$.
Substituting the values $h = 1$,$k = 1$,and $a = 2$,we get $(y - 1)^2 = 4(2)(x - 1)$.
Therefore,the equation is $(y - 1)^2 = 8(x - 1)$.

(A) Let $P(x, y)$ be any point on the parabola. By definition,the distance from $P$ to the focus $S(5, 3)$ is equal to the perpendicular distance from $P$ to the directrix $3x - 4y + 1 = 0$.
$PS^2 = PM^2$
$(x - 5)^2 + (y - 3)^2 = \left( \frac{3x - 4y + 1}{\sqrt{3^2 + (-4)^2}} \right)^2$
$(x^2 - 10x + 25) + (y^2 - 6y + 9) = \frac{(3x - 4y + 1)^2}{25}$
$25(x^2 + y^2 - 10x - 6y + 34) = 9x^2 + 16y^2 + 1 - 24xy + 6x - 8y$
$25x^2 + 25y^2 - 250x - 150y + 850 = 9x^2 + 16y^2 - 24xy + 6x - 8y + 1$
$16x^2 + 24xy + 9y^2 - 256x - 142y + 849 = 0$
$(4x + 3y)^2 - 256x - 142y + 849 = 0$

(D) To check which point lies on the parabola $x^2 = 4ay$,we substitute the given parametric coordinates into the equation.
For option $D$:
Given $x = 2at$ and $y = at^2$.
Substituting these into the equation $x^2 = 4ay$:
$(2at)^2 = 4a(at^2)$
$4a^2t^2 = 4a^2t^2$
Since the left-hand side equals the right-hand side,the point $(2at, at^2)$ lies on the parabola $x^2 = 4ay$.

(B) The vertex $V$ is $(h, k) = (2, -1)$ and the focus $S$ is $(2, -3)$.
Since the $x$-coordinates of the vertex and focus are the same,the axis of the parabola is vertical.
The distance $a$ between the vertex and the focus is $a = | -1 - (-3) | = 2$.
Since the focus is below the vertex,the parabola opens downwards.
The standard equation for a downward-opening parabola is $(x - h)^2 = -4a(y - k)$.
Substituting the values $h = 2$,$k = -1$,and $a = 2$:
$(x - 2)^2 = -4(2)(y - (-1))$
$(x - 2)^2 = -8(y + 1)$
Expanding the equation:
$x^2 - 4x + 4 = -8y - 8$
$x^2 - 4x + 8y + 12 = 0$.
Thus,the correct option is $B$.

(D) The given equation of the parabola is ${x^2 - 4x - 8y + 12 = 0}$.
Completing the square for the $x$ terms:
${x^2 - 4x + 4 = 8y - 12 + 4}$
${(x - 2)^2 = 8y - 8}$
${(x - 2)^2 = 8(y - 1)}$
Comparing this with the standard form ${X^2 = 4aY}$,where ${X = x - 2}$,${Y = y - 1}$,and ${4a = 8}$,we get ${a = 2}$.
The equation of the directrix for the parabola ${X^2 = 4aY}$ is ${Y = -a}$.
Substituting the values back:
${y - 1 = -2}$
${y = -1}$.

(A) Given: Vertex $A = (0, 6)$ and Focus $S = (0, 3)$.
Since the vertex and focus lie on the $y$-axis,the axis of the parabola is the $y$-axis.
The distance $a$ between the vertex and the focus is $a = 6 - 3 = 3$.
Since the focus is below the vertex,the parabola opens downwards.
The standard form of a downward-opening parabola with vertex $(h, k)$ is $(x - h)^2 = -4a(y - k)$.
Here,$(h, k) = (0, 6)$ and $a = 3$.
Substituting these values,we get $(x - 0)^2 = -4(3)(y - 6)$.
$x^2 = -12(y - 6)$.
$x^2 = -12y + 72$.
$x^2 + 12y = 72$.

(A) Given,the equation of the parabola is ${x^2} + 8y - 2x = 7$.
Rearranging the terms,we get ${x^2} - 2x = -8y + 7$.
Completing the square on the left side: ${x^2} - 2x + 1 = -8y + 7 + 1$.
This simplifies to ${(x - 1)^2} = -8y + 8$.
Factoring out $-8$,we get ${(x - 1)^2} = -8(y - 1)$.
Comparing this with the standard form ${(x - h)^2} = -4a(y - k)$,where $(h, k) = (1, 1)$ and $4a = 8$,we find $a = 2$.
The equation of the directrix for a downward-opening parabola is $y = k + a$.
Substituting the values,$y = 1 + 2 = 3$.
Therefore,the equation of the directrix is $y = 3$.

(A) Given equation of the parabola is $2x^2 + 5y - 3x + 4 = 0$.
Rearranging the terms,we get $2x^2 - 3x = -5y - 4$.
Dividing by $2$,we have $x^2 - \frac{3}{2}x = -\frac{5}{2}y - 2$.
Completing the square on the left side: $x^2 - \frac{3}{2}x + (\frac{3}{4})^2 = -\frac{5}{2}y - 2 + \frac{9}{16}$.
$(x - \frac{3}{4})^2 = -\frac{5}{2}y - \frac{23}{16}$.
The axis of a parabola of the form $(x - h)^2 = 4a(y - k)$ is given by $x - h = 0$.
Therefore,the equation of the axis is $x - \frac{3}{4} = 0$,which implies $x = \frac{3}{4}$.

(A) Given equation of the parabola is ${x^2} + 6x + 20y - 51 = 0$.
Rearranging the terms,we get:
${x^2} + 6x = -20y + 51$.
Completing the square on the left side:
${x^2} + 6x + 9 = -20y + 51 + 9$.
This simplifies to:
${(x + 3)^2} = -20y + 60$.
Factoring out $-20$ on the right side:
${(x + 3)^2} = -20(y - 3)$.
Comparing this with the standard form ${(x - h)^2} = 4a(y - k)$,the axis of the parabola is given by $x - h = 0$.
Here,$h = -3$,so the axis is $x + 3 = 0$.

(D) Given the parabola equation $y = x^2 - x$.
At $x = 1$,the $y$-coordinate is $y = (1)^2 - 1 = 0$. So,the point of tangency is $(1, 0)$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 2x - 1$.
Evaluating the derivative at $x = 1$: $\left( \frac{dy}{dx} \right)_{x=1} = 2(1) - 1 = 1$.
Using the point-slope form of the line equation $y - y_1 = m(x - x_1)$,where $m = 1$ and $(x_1, y_1) = (1, 0)$:
$y - 0 = 1(x - 1)$
$y = x - 1$.

(A) The point of intersection of the latus rectum and the axis of a parabola is its focus.
Given the equation of the parabola: ${y^2} + 4x + 2y - 8 = 0$.
Rearranging the terms to complete the square for $y$:
${y^2} + 2y = -4x + 8$
${y^2} + 2y + 1 = -4x + 8 + 1$
${(y + 1)^2} = -4x + 9$
${(y + 1)^2} = -4(x - 9/4)$.
Comparing this with the standard form ${(y - k)^2} = 4a(x - h)$,we get $h = 9/4$,$k = -1$,and $4a = -4$,which implies $a = -1$.
The focus of the parabola is given by $(h + a, k)$.
Substituting the values: $(9/4 - 1, -1) = (5/4, -1)$.

(D) Let the point of contact be $(h, k)$.
Since the point $(h, k)$ lies on the parabola $y^2 = 2x$,we have $k^2 = 2h$.
The equation of the tangent to the parabola $y^2 = 4ax$ at $(h, k)$ is $ky = 2a(x + h)$.
Here,$4a = 2$,so $a = \frac{1}{2}$.
Thus,the tangent equation is $ky = 1(x + h)$,or $x - ky + h = 0$.
Comparing this with the given tangent $18x - 6y + 1 = 0$,we get:
$\frac{1}{18} = \frac{-k}{-6} = \frac{h}{1}$.
From $\frac{1}{18} = \frac{k}{6}$,we get $k = \frac{6}{18} = \frac{1}{3}$.
From $\frac{1}{18} = h$,we get $h = \frac{1}{18}$.
Thus,the point of contact is $\left( \frac{1}{18}, \frac{1}{3} \right)$.

(C) The given line is $lx + my + n = 0$,which can be rewritten as $y = -\frac{l}{m}x - \frac{n}{m}$.
The condition for a line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$ is $c = \frac{a}{m}$.
Here,the slope of the line is $M = -\frac{l}{m}$ and the intercept is $C = -\frac{n}{m}$.
Substituting these into the condition $C = \frac{a}{M}$,we get:
$-\frac{n}{m} = \frac{a}{-l/m}$
$-\frac{n}{m} = -\frac{am}{l}$
$ln = am^2$.

(A) The equation of the line is $x \cos \alpha + y \sin \alpha = p$,which can be written as $x \cos \alpha + y \sin \alpha - p = 0$.
The equation of the parabola is $y^2 = 4a(x + a)$.
The condition for the line $y = mx + c$ to touch the parabola $y^2 = 4ax'$ (where $x' = x+a$) is $c = \frac{a'}{m}$.
Rewriting the line equation: $y \sin \alpha = -x \cos \alpha + p \implies y = -x \cot \alpha + p \csc \alpha$.
Here,$m = -\cot \alpha$ and $c = p \csc \alpha$.
The parabola is $y^2 = 4a(x+a)$. Comparing with $Y^2 = 4aX$,we have $Y = y$,$X = x+a$,and the parameter $a$ remains $a$.
The condition for tangency is $c = \frac{a}{m} + am$ (since the vertex is at $(-a, 0)$).
Substituting the values: $p \csc \alpha = \frac{a}{-\cot \alpha} + a(-\cot \alpha)$.
$p \csc \alpha = -a \tan \alpha - a \cot \alpha = -a (\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}) = -a (\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}) = -\frac{a}{\sin \alpha \cos \alpha}$.
$p \frac{1}{\sin \alpha} = -\frac{a}{\sin \alpha \cos \alpha} \implies p = -\frac{a}{\cos \alpha}$.
Thus,$p \cos \alpha = -a$,or $p \cos \alpha + a = 0$.

(C) The slope of the tangent line is $m = \tan \theta$.
The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx + \frac{a}{m}$.
Substituting $m = \tan \theta$ into the equation,we get:
$y = x \tan \theta + \frac{a}{\tan \theta}$.
Since $\frac{1}{\tan \theta} = \cot \theta$,the equation becomes:
$y = x \tan \theta + a \cot \theta$.

(B) The given parabola is $y^2 = 4(x + \frac{5}{4})$.
Comparing this with $Y^2 = 4aX$,we have $a = 1$,$Y = y$,and $X = x + \frac{5}{4}$.
The slope of the line $y = 2x + 7$ is $m = 2$.
The equation of a tangent to the parabola $Y^2 = 4aX$ with slope $m$ is $Y = mX + \frac{a}{m}$.
Substituting the values,we get $y = 2(x + \frac{5}{4}) + \frac{1}{2}$.
$y = 2x + \frac{5}{2} + \frac{1}{2}$.
$y = 2x + 3$.
Thus,the equation of the tangent is $2x - y + 3 = 0$.

(A) The slope of the tangent $m = \tan(60^\circ) = \sqrt{3}$.
The equation of the tangent to the parabola $y^2 = 4ax$ at the point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
The slope of this tangent is $m = \frac{2a}{y_1}$.
Equating the slopes: $\sqrt{3} = \frac{2a}{y_1} \implies y_1 = \frac{2a}{\sqrt{3}}$.
Since the point $(x_1, y_1)$ lies on the parabola $y^2 = 4ax$,we have $y_1^2 = 4ax_1$.
Substituting $y_1$: $\left( \frac{2a}{\sqrt{3}} \right)^2 = 4ax_1 \implies \frac{4a^2}{3} = 4ax_1 \implies x_1 = \frac{a}{3}$.
Thus,the point of contact is $\left( \frac{a}{3}, \frac{2a}{\sqrt{3}} \right)$.

(B) The given line is $y = 2x + \lambda$ and the parabola is $y^2 = 2x$.
Substituting $y = 2x + \lambda$ into the parabola equation: $(2x + \lambda)^2 = 2x$.
$4x^2 + 4x\lambda + \lambda^2 = 2x$.
$4x^2 + (4\lambda - 2)x + \lambda^2 = 0$.
For the line not to meet the parabola,the discriminant $D$ must be less than $0$.
$D = (4\lambda - 2)^2 - 4(4)(\lambda^2) < 0$.
$16\lambda^2 - 16\lambda + 4 - 16\lambda^2 < 0$.
$-16\lambda + 4 < 0$.
$16\lambda > 4$.
$\lambda > \frac{4}{16} = \frac{1}{4}$.

(A) The coordinates of any point $P$ on the parabola $y^2 = 4ax$ in terms of parameter $t$ are given by $(at^2, 2at)$.
The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
Substituting $x_1 = at^2$ and $y_1 = 2at$ into the equation:
$y(2at) = 2a(x + at^2)$
Dividing both sides by $2a$:
$yt = x + at^2$.

(D) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,which implies $a = 4$.
The line $y = mx + c$ is a tangent to the parabola $y^2 = 4ax$ if $c = \frac{a}{m}$.
Here,$m = 2$ and $a = 4$.
Substituting these values,we get $c = \frac{4}{2} = 2$.

(A) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.
The condition for the line $y = mx + c$ to be a tangent to the parabola $y^2 = 4ax$ is $c = \frac{a}{m}$.
Given the line $y = mx + 1$,we have $c = 1$.
Substituting the values into the condition: $1 = \frac{1}{m}$.
Therefore,$m = 1$.

(A) Given curves are $y^2 = 4x$ and $x^2 = 32y$.
For $y^2 = 4x$,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(16, 8)$,$m_1 = \frac{2}{8} = \frac{1}{4}$.
For $x^2 = 32y$,differentiating with respect to $x$,we get $2x = 32 \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x}{16}$. At $(16, 8)$,$m_2 = \frac{16}{16} = 1$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{1 - 1/4}{1 + (1)(1/4)} \right| = \left| \frac{3/4}{5/4} \right| = \frac{3}{5}$.
Therefore,$\theta = \tan^{-1}\left(\frac{3}{5}\right)$.

(A) The equation of any tangent to the parabola $y^2 = 4ax$ is given by $y = mx + \frac{a}{m}$,where $m$ is the slope.
The line perpendicular to this tangent passing through the focus $(a, 0)$ has a slope of $-\frac{1}{m}$.
The equation of this perpendicular line is $y - 0 = -\frac{1}{m}(x - a)$,which simplifies to $y = -\frac{x}{m} + \frac{a}{m}$.
To find the locus of the foot of the perpendicular,we eliminate the parameter $m$ from the two equations:
$1) \ y = mx + \frac{a}{m} \implies my = m^2x + a$
$2) \ y = -\frac{x}{m} + \frac{a}{m} \implies my = -x + a$
Equating the two expressions for $my$:
$m^2x + a = -x + a$
$m^2x = -x$
$x(m^2 + 1) = 0$
Since $m^2 + 1 \neq 0$ for real $m$,we must have $x = 0$.
Thus,the locus is the tangent at the vertex,which is the $y$-axis,represented by $x = 0$.

(C) The slope of the tangent line $x + y = 1$ is $m = -1$.
For the parabola $y^2 - y + x = 0$,we differentiate with respect to $x$ to find the slope of the tangent at any point $(h, k)$:
$2y \frac{dy}{dx} - \frac{dy}{dx} + 1 = 0$
$\frac{dy}{dx}(2y - 1) = -1$
$\frac{dy}{dx} = \frac{-1}{2y - 1}$
At the point of contact $(h, k)$,the slope of the tangent must equal the slope of the line:
$\frac{-1}{2k - 1} = -1$
$2k - 1 = 1$
$2k = 2$
$k = 1$
Since the point $(h, k)$ lies on the line $x + y = 1$,we substitute $k = 1$:
$h + 1 = 1$
$h = 0$
Thus,the point of contact is $(0, 1)$.

(A) The equation of the tangent to the standard parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx + \frac{a}{m}$.
The given parabola is $y^2 = 4a(x + a)$.
By shifting the origin to $(-a, 0)$,the equation becomes $Y^2 = 4aX$,where $Y = y$ and $X = x + a$.
The tangent to $Y^2 = 4aX$ is $Y = mX + \frac{a}{m}$.
Substituting back $Y = y$ and $X = x + a$,we get $y = m(x + a) + \frac{a}{m}$.
Expanding this,we have $y = mx + ma + \frac{a}{m}$.
Comparing this with the given line $y = mx + c$,we get $c = ma + \frac{a}{m}$.
Therefore,$ma + \frac{a}{m} = c$.

(C) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.
The slope of the line $y = 3x + 5$ is $m_1 = 3$.
Let the slope of the tangent be $m$. Since the angle between the tangent and the line is $45^\circ$,we have $\tan 45^\circ = |\frac{m - m_1}{1 + m \cdot m_1}|$.
$1 = |\frac{m - 3}{1 + 3m}|$.
This gives two cases:
Case $1$: $\frac{m - 3}{1 + 3m} = 1 \implies m - 3 = 1 + 3m \implies -2m = 4 \implies m = -2$.
Case $2$: $\frac{m - 3}{1 + 3m} = -1 \implies m - 3 = -1 - 3m \implies 4m = 2 \implies m = \frac{1}{2}$.
The equation of a tangent to $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
For $m = -2$: $y = -2x + \frac{2}{-2} \implies y = -2x - 1 \implies 2x + y + 1 = 0$.
For $m = \frac{1}{2}$: $y = \frac{1}{2}x + \frac{2}{1/2} \implies y = \frac{1}{2}x + 4 \implies x - 2y + 8 = 0$.
Comparing with the given options,$2x + y + 1 = 0$ is the correct equation.

(D) The end points of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.
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 the point $(a, 2a)$,the tangent is $y(2a) = 2a(x + a)$,which simplifies to $y = x + a$. The slope $m_1 = 1$.
For the point $(a, -2a)$,the tangent is $y(-2a) = 2a(x + a)$,which simplifies to $y = -(x + a)$. The slope $m_2 = -1$.
Since $m_1 \times m_2 = 1 \times (-1) = -1$,the tangents are perpendicular to each other.
Therefore,the angle between them is $\frac{\pi}{2}$.

(C) Given the line $y = mx + c$ and the parabola $x^2 = 4ay$.
Substituting $y$ from the line equation into the parabola equation:
$x^2 = 4a(mx + c)$
$x^2 - 4amx - 4ac = 0$
Since the line touches the parabola,the discriminant of this quadratic equation must be zero $(D = 0)$:
$B^2 - 4AC = 0$
$(-4am)^2 - 4(1)(-4ac) = 0$
$16a^2m^2 + 16ac = 0$
Dividing by $16a$ (assuming $a \neq 0$):
$am^2 + c = 0$
$c = -am^2$

(B) For a parabola of the form $x^2 = 4ay$,the tangents drawn at the ends of a focal chord are perpendicular to each other.
Conversely,the locus of the point of intersection of two perpendicular tangents to any parabola is its directrix.
For the parabola $x^2 = 4ay$,the directrix is given by the equation $y = -a$.

(A) Any line passing through the origin is given by $y = mx$.
Since this line is a tangent to the parabola $y^2 = 4a(x - a)$,substituting $y = mx$ into the equation gives:
$(mx)^2 = 4a(x - a)$
$m^2x^2 - 4ax + 4a^2 = 0$
For the line to be a tangent,the roots of this quadratic equation must be equal,so the discriminant $D = 0$:
$(-4a)^2 - 4(m^2)(4a^2) = 0$
$16a^2 - 16a^2m^2 = 0$
$16a^2(1 - m^2) = 0$
Since $a \neq 0$,we have $1 - m^2 = 0$,which implies $m^2 = 1$,so $m = 1$ or $m = -1$.
The slopes of the two tangents are $m_1 = 1$ and $m_2 = -1$.
Since $m_1 \times m_2 = 1 \times (-1) = -1$,the tangents are perpendicular to each other.
Therefore,the angle between the tangents is $90^\circ$.

(A) The given equation of the line is $x = my + k$,which can be rewritten as $x - my = k$.
Substituting $x = my + k$ into the parabola equation $x^2 = 4ay$:
$(my + k)^2 = 4ay$
$m^2y^2 + 2mky + k^2 = 4ay$
$m^2y^2 + (2mk - 4a)y + k^2 = 0$
Since the line touches the parabola,the discriminant of this quadratic equation must be zero:
$D = (2mk - 4a)^2 - 4(m^2)(k^2) = 0$
$4m^2k^2 - 16amk + 16a^2 - 4m^2k^2 = 0$
$-16amk + 16a^2 = 0$
$16amk = 16a^2$
$k = a/m$

(B) Let the coordinates of points $P$ and $Q$ on the parabola $y^2 = 4ax$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ respectively.
Thus,the ordinates are ${y_1 = 2at_1}$ and ${y_2 = 2at_2}$.
The point of intersection of the tangents at $P$ and $Q$ is given by $(at_1t_2, a(t_1 + t_2))$.
Therefore,the ordinate of the intersection point is ${y_3 = a(t_1 + t_2)}$.
Substituting $t_1 = \frac{y_1}{2a}$ and $t_2 = \frac{y_2}{2a}$ into the expression for $y_3$:
${y_3 = a \left( \frac{y_1}{2a} + \frac{y_2}{2a} \right) = \frac{y_1 + y_2}{2}}$.
This implies that $2y_3 = y_1 + y_2$,which means ${y_1, y_3, y_2}$ are in $A.P.$

(C) Solving $x^2 = 4y$ and $y^2 = 4x$,we get $x = 0, y = 0$ and $x = 4, y = 4$.
Since the abscissa is not zero,the point $P$ is $(4, 4)$.
The equation of the tangent to $y^2 = 4x$ at $(4, 4)$ is $y(4) = 2(x + 4)$,which simplifies to $2x - y + 4 = 0$ (Slope $m_1 = 2$).
The equation of the tangent to $x^2 = 4y$ at $(4, 4)$ is $x(4) = 2(y + 4)$,which simplifies to $x - 2y + 4 = 0$ (Slope $m_2 = 1/2$).
Let $\theta_1$ and $\theta_2$ be the angles the tangents make with the $x$-axis. Then $\tan \theta_1 = 2$ and $\tan \theta_2 = 1/2$.
Since $\tan \theta_1 = \cot \theta_2 = \tan(90^\circ - \theta_2)$,we have $\theta_1 = 90^\circ - \theta_2$,or $\theta_1 + \theta_2 = 90^\circ$.
Thus,the tangents make complementary angles with the $x$-axis.

(B) The equation of the parabola is $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$.
The equation of the line is $y = 2x + c$,which is in the form $y = mx + c$,where $m = 2$.
The condition for the line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$ is $c = \frac{a}{m}$.
Substituting the values $a = 1$ and $m = 2$,we get $c = \frac{1}{2}$.

(A) The equation of the parabola is $y^2 = 4ax$.
Substituting $y = mx + c$ into the parabola equation:
$(mx + c)^2 = 4ax$
$m^2x^2 + 2mcx + c^2 - 4ax = 0$
$m^2x^2 + (2mc - 4a)x + c^2 = 0$.
For the line to touch the parabola,the discriminant of this quadratic equation must be zero:
$D = (2mc - 4a)^2 - 4(m^2)(c^2) = 0$
$4m^2c^2 - 16amc + 16a^2 - 4m^2c^2 = 0$
$-16amc + 16a^2 = 0$
$16a^2 = 16amc$
$a = mc$
$c = a/m$.
Thus,the correct condition is $c = a/m$.

(C) Given the parabola equation is $y^2 = 4ax$.
Since it passes through the point $(1, -2)$,we substitute $x = 1$ and $y = -2$ into the equation:
$(-2)^2 = 4a(1)$
$4 = 4a$
$\Rightarrow a = 1$.
Now,the equation of the parabola is $y^2 = 4x$.
The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.
Substituting $a = 1$,$x_1 = 1$,and $y_1 = -2$:
$y(-2) = 2(1)(x + 1)$
$-2y = 2(x + 1)$
$-y = x + 1$
$x + y + 1 = 0$.
Thus,the required tangent is $x + y + 1 = 0$.

(D) The given line is $y = 3x + 7$,which has a slope $m_1 = 3$.
Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m \times m_1 = -1$,so $m = -\frac{1}{3}$.
The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Comparing $y^2 = 16x$ with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.
Substituting $a = 4$ and $m = -\frac{1}{3}$ into the tangent equation:
$y = -\frac{1}{3}x + \frac{4}{-1/3}$
$y = -\frac{1}{3}x - 12$
Multiplying by $3$ gives $3y = -x - 36$,which simplifies to $x + 3y + 36 = 0$.

(D) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.
Given point $(x_1, y_1) = (a/t^2, 2a/t)$.
Substituting these values into the tangent equation:
$y(2a/t) = 2a(x + a/t^2)$
Dividing both sides by $2a$:
$y/t = x + a/t^2$
Multiplying both sides by $t$:
$y = tx + a/t$
Wait,let us re-evaluate the simplification:
$y/t = (t^2x + a)/t^2$
$y = (t^2x + a)/t$
$ty = t^2x + a$.