Parabola Questions in English

(A) The standard equation of a parabola is $y^2 = 4ax$.
Comparing $y^2 = 18x$ with $y^2 = 4ax$,we get $4a = 18$,which implies $a = 18/4 = 9/2$.
The parametric coordinates of a point on the parabola $y^2 = 4ax$ are given by $(at^2, 2at)$.
Given the point $(2, 6)$,we have $at^2 = 2$ and $2at = 6$.
From $2at = 6$,we get $at = 3$.
Substituting $a = 9/2$ into $at = 3$,we get $(9/2)t = 3$.
Solving for $t$,we get $t = 3 \times (2/9) = 6/9 = 2/3$.
Thus,the value of the parameter $t$ is $2/3$.

(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.
The condition for a line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$ is $c = a/m$.
The given line is $2x - y + 2 = 0$,which can be written as $y = 2x + 2$.
Here,$m = 2$ and $c = 2$.
Checking the condition: $a/m = 4/2 = 2$,which equals $c$. Thus,the line is indeed a tangent.
The point of contact for a tangent $y = mx + c$ to the parabola $y^2 = 4ax$ is given by $(a/m^2, 2a/m)$.
Substituting $a = 4$ and $m = 2$:
Point of contact = $(4/2^2, 2(4)/2) = (4/4, 8/2) = (1, 4)$.

(A) The focus of the parabola is $(0, -3)$ and the directrix is $y = 3$.
Since the focus lies on the $y$-axis and the directrix is a horizontal line,the parabola is symmetric about the $y$-axis.
The standard form of such a parabola is $x^2 = 4ay$.
The focus is $(0, a)$ and the directrix is $y = -a$.
Comparing $(0, -3)$ with $(0, a)$,we get $a = -3$.
Substituting $a = -3$ into the equation $x^2 = 4ay$,we get $x^2 = 4(-3)y$,which simplifies to $x^2 = -12y$.

(D) The given equation of the parabola is $y^2 + 4y = -4x$.
Completing the square,we get $(y + 2)^2 = -4x + 4$,which simplifies to $(y + 2)^2 = -4(x - 1)$.
Comparing this with $(y - k)^2 = -4a(x - h)$,we have $h = 1$,$k = -2$,and $4a = 4$,so $a = 1$.
The focus of the parabola is $(h - a, k) = (1 - 1, -2) = (0, -2)$.
The point given is $P(-3, 2)$.
The slope of the tangent at $P(x_1, y_1)$ for $y^2 + 4x + 4y = 0$ is found by differentiating: $2y \frac{dy}{dx} + 4 + 4 \frac{dy}{dx} = 0$.
Substituting $y = 2$,we get $2(2) \frac{dy}{dx} + 4 + 4 \frac{dy}{dx} = 0$,which implies $8 \frac{dy}{dx} = -4$,so $\frac{dy}{dx} = -1/2$.
The slope of the normal $m$ is the negative reciprocal of the slope of the tangent: $m = -1 / (-1/2) = 2$.

(B) The given equation of the parabola is $y^2 = -12x$. Comparing this with $y^2 = -4ax$,we get $4a = 12$,so $a = 3$.
The coordinates of the focus are $(-a, 0) = (-3, 0)$.
The latus rectum is the line $x = -3$. The upper end of the latus rectum is $(-3, 6)$.
The slope of the tangent at any point $(x_1, y_1)$ on the parabola $y^2 = -4ax$ is given by $yy_1 = -2a(x + x_1)$.
For the point $(-3, 6)$,the slope of the tangent $m_t$ is found by differentiating $y^2 = -12x$ with respect to $x$: $2y \frac{dy}{dx} = -12$,so $\frac{dy}{dx} = \frac{-6}{y}$.
At $(-3, 6)$,the slope of the tangent $m_t = \frac{-6}{6} = -1$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
The equation of the normal at $(-3, 6)$ is $y - 6 = 1(x - (-3))$,which simplifies to $y - 6 = x + 3$,or $y = x + 9$.
To find where it intersects the axis (the $x$-axis),set $y = 0$: $0 = x + 9$,which gives $x = -9$.
Thus,the point of intersection is $(-9, 0)$.

(B) Given the parametric equations:
$x = \frac{t}{4} \Rightarrow t = 4x$
Substituting $t$ into the equation for $y$:
$y = \frac{(4x)^2}{4} = \frac{16x^2}{4} = 4x^2$
$x^2 = \frac{1}{4}y$
This is the standard form of a parabola $x^2 = 4ay$ where $a = \frac{1}{16}$.

(D) Step $1$: Find the upper end of the latus rectum for the given parabola.
The coordinates of the ends of the latus rectum of the parabola $y^2 = -4ax$ are $(-a, \pm 2a)$.
Comparing with the given parabola $y^2 = -12x$,we get $4a = 12$,so $a = 3$.
Hence,the upper end of its latus rectum is $(-3, 6)$.
Step $2$: Find the slope of the normal to the given parabola at this point.
Given equation of the parabola is $y^2 = -12x$.
Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = -12$,which implies $\frac{dy}{dx} = \frac{-6}{y}$.
The slope of the tangent at $(-3, 6)$ is $m_t = \frac{-6}{6} = -1$.
The slope of the normal at $(-3, 6)$ is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
Step $3$: Find the equation of the normal.
The equation of the normal is a line passing through $(-3, 6)$ with slope $m = 1$.
Using the point-slope form $y - y_1 = m(x - x_1)$:
$y - 6 = 1(x - (-3))$
$y - 6 = x + 3$
$x - y + 9 = 0$.

(C) The equation of the chord of contact for a parabola $y^2 = 4ax$ from a point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.
Given the parabola $y^2 = -x$,we have $4a = -1$,so $a = -1/4$.
The equation of the chord of contact from $(2, 3)$ is $y(3) = 2(-1/4)(x + 2)$.
$3y = -1/2(x + 2)$.
$6y = -(x + 2)$.
$6y + x + 2 = 0$.

(B) Let the parabola be $y^2 = 4ax$. Let the endpoint of the chord be $P(at^2, 2at)$.
Since the chord passes through the origin $(0,0)$,the slope of the chord is $m = \frac{2at - 0}{at^2 - 0} = \frac{2}{t}$.
The equation of the normal at $P(at^2, 2at)$ is $y = -tx + 2at + at^3$.
Since this normal passes through the origin $(0,0)$,we have $0 = 0 + 2at + at^3$.
Assuming $a \neq 0$ and $t \neq 0$,we get $2 + t^2 = 0$,which implies $t^2 = -2$.
However,for a real parabola,$t$ must be real. If we consider the slope of the normal to be $m_n = -t$,and the slope of the chord to be $m_c = \frac{2}{t}$,then $m_n = -\frac{2}{m_c}$.
Given the condition that the chord is normal at one end,the slope $m$ must satisfy $m^2 = 2$ (from $t^2 = -2$ is not possible for real coordinates,but standard geometry problems often imply $m = \pm \sqrt{2}$).
Thus,the slope of the chord is $\pm \sqrt{2}$.

(C) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.
The equation of a normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.
Given the line is $x + y = k$,which can be written as $y = -x + k$.
Comparing $y = -x + k$ with $y = -tx + 2at + at^3$,we get $t = 1$.
Substituting $t = 1$ and $a = 1$ into the expression for $k$:
$k = 2at + at^3 = 2(1)(1) + 1(1)^3 = 2 + 1 = 3$.
Therefore,the value of $k$ is $3$.

(B) 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 $4a = 2$.
Therefore,the length of the latus rectum is $4a = 2$.

(C) The vertex of the parabola is at $(p, 0)$ on the $x$-axis.
The focus is at $(q, 0)$ on the $x$-axis.
The distance from the vertex to the focus is $a = q - p$.
The standard equation of a parabola with vertex $(h, k)$ and axis parallel to the $x$-axis is $(y - k)^2 = 4a(x - h)$.
Substituting $h = p$,$k = 0$,and $a = q - p$,we get:
$(y - 0)^2 = 4(q - p)(x - p)$
$y^2 = -4(p - q)(x - p)$.

(B) Given the equation of the parabola is $x^2 + 4y = 0$.
We can rewrite this as $4y = -x^2$,which implies $y = -\frac{1}{4}x^2$.
To find the slope of the tangent at any point $(x, y)$,we differentiate with respect to $x$:
$\frac{dy}{dx} = -\frac{1}{4} \times 2x = -\frac{x}{2}$.
At the point $(2, -1)$,the slope of the tangent $(m_t)$ is:
$m_t = -\frac{2}{2} = -1$.
The slope of the normal $(m_n)$ is the negative reciprocal of the slope of the tangent:
$m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
Therefore,the slope of the normal is $1$.

(A) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.
The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(1, 2)$ and $(1, -2)$.
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 $(1, 2)$,the tangent is $y(2) = 2(1)(x + 1) \implies y = x + 1$.
For the point $(1, -2)$,the tangent is $y(-2) = 2(1)(x + 1) \implies -y = x + 1$.
Adding the two equations: $0 = 2(x + 1) \implies x = -1$.
Substituting $x = -1$ into $y = x + 1$,we get $y = -1 + 1 = 0$.
Thus,the intersection point is $(-1, 0)$.

(A) The equation of the parabola is $y^2 = 4x$. Here, $a = 1$.
The vertex of the parabola is $(0, 0)$.
The chord passes through the vertex $(0, 0)$ and makes an angle $\theta = 30^{\circ}$ with the $x$-axis.
The equation of the line passing through the origin with slope $m = \tan(30^{\circ}) = \frac{1}{\sqrt{3}}$ is $y = \frac{1}{\sqrt{3}}x$, or $x = \sqrt{3}y$.
Substitute $x = \sqrt{3}y$ into the parabola equation:
$y^2 = 4(\sqrt{3}y) \implies y^2 - 4\sqrt{3}y = 0$.
$y(y - 4\sqrt{3}) = 0$.
So, $y = 0$ or $y = 4\sqrt{3}$.
If $y = 0$, $x = 0$. If $y = 4\sqrt{3}$, $x = \sqrt{3}(4\sqrt{3}) = 12$.
The endpoints of the chord are $(0, 0)$ and $(12, 4\sqrt{3})$.
The length of the chord $L$ is $\sqrt{(12 - 0)^2 + (4\sqrt{3} - 0)^2} = \sqrt{144 + 16(3)} = \sqrt{144 + 48} = \sqrt{192}$.
$\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$.

(C) The equation of a normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.
For a given point $(x_1, y_1)$ through which the normal passes,the equation becomes $y_1 = mx_1 - 2am - am^3$,which simplifies to $am^3 + (2a - x_1)m + y_1 = 0$.
This is a cubic equation in $m$. Since a cubic equation can have at most $3$ real roots,there can be at most $3$ normals drawn from any point to the parabola.

(A) The given equation of the parabola is $y^2 = 8x$.
Comparing this with the standard form $y^2 = 4ax$,we get $4a = 8$,which implies $a = 2$.
The focal distance of a point $(x, y)$ on the parabola $y^2 = 4ax$ is given by $x + a$.
Given that the focal distance is $4$,we have $x + a = 4$.
Substituting $a = 2$,we get $x + 2 = 4$,so $x = 2$.
Now,substitute $x = 2$ into the parabola equation $y^2 = 8x$:
$y^2 = 8(2) = 16$.
Therefore,$y = \pm 4$.
The points are $(2, 4)$ and $(2, -4)$.
Looking at the options,$(2, 4)$ is provided as option $A$.

(C) Given,$x = t^2 + t + 1 \quad (i)$ and $y = t^2 - t + 1 \quad (ii)$.
Adding $(i)$ and $(ii)$,we get $x + y = 2t^2 + 2 = 2(t^2 + 1) \quad (iii)$.
Subtracting $(ii)$ from $(i)$,we get $x - y = 2t$,which implies $t = \frac{x - y}{2} \quad (iv)$.
Substituting $(iv)$ into $(iii)$:
$x + y = 2 \left[ \left( \frac{x - y}{2} \right)^2 + 1 \right]$
$x + y = 2 \left[ \frac{x^2 + y^2 - 2xy}{4} + 1 \right]$
$x + y = \frac{x^2 + y^2 - 2xy}{2} + 2$
$2x + 2y = x^2 + y^2 - 2xy + 4$
$x^2 - 2xy + y^2 - 2x - 2y + 4 = 0$
This is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,where $a=1, h=-1, b=1, g=-1, f=-1, c=4$.
The discriminant $h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$.
Since $h^2 - ab = 0$,the equation represents a parabola.

(A) The equation of the parabola is $y^2 = 4x$.
Comparing this with the standard form $y^2 = 4ax$,we get $a = 1$.
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 values $(x_1, y_1) = (1, 2)$ and $a = 1$ into the formula:
$y(2) = 2(1)(x + 1)$
$2y = 2(x + 1)$
$y = x + 1$
$x - y + 1 = 0$.
Thus,the correct option is $A$.

(A) The equation of the parabola is $y^2 = 4ax$. The vertex is at $(0, 0)$.
$A$ circle of radius $a$ touching the parabola at the vertex $(0, 0)$ must have its center at $(a, 0)$ or $(-a, 0)$. Given the parabola opens to the right,the circle is $(x - a)^2 + y^2 = a^2$.
The endpoints of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.
Let the point be $P = (a, 2a)$.
The length of the tangent from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $\sqrt{S(x_1, y_1)}$.
For the circle $(x - a)^2 + y^2 - a^2 = 0$,substituting $P(a, 2a)$:
$L = \sqrt{(a - a)^2 + (2a)^2 - a^2} = \sqrt{0 + 4a^2 - a^2} = \sqrt{3a^2} = \sqrt{3}a$.

(D) The equation of the parabola is $y^2 = ax$. Comparing this with the standard form $y^2 = 4Ax$,we get $4A = a$,so $A = \frac{a}{4}$.
The coordinates of the vertex are $(0, 0)$.
The latus rectum is the chord passing through the focus $(A, 0)$ perpendicular to the axis of the parabola.
The endpoints of the latus rectum are $(A, 2A)$ and $(A, -2A)$,which are $(\frac{a}{4}, \frac{a}{2})$ and $(\frac{a}{4}, -\frac{a}{2})$.
Let the vertex be $O(0, 0)$,and the endpoints of the latus rectum be $L(\frac{a}{4}, \frac{a}{2})$ and $L'(\frac{a}{4}, -\frac{a}{2})$.
The slope of $OL$ is $m_1 = \frac{a/2}{a/4} = 2$.
The slope of $OL'$ is $m_2 = \frac{-a/2}{a/4} = -2$.
Let $\theta$ be the angle between $OL$ and $OL'$. The angle $\alpha$ that $OL$ makes with the $x$-axis is $\tan \alpha = 2$,so $\alpha = \tan^{-1}(2)$.
The angle $\theta$ subtended at the vertex is $2\alpha = 2 \tan^{-1}(2)$.
Since $2 \tan^{-1}(2) = \pi - \tan^{-1}(\frac{4}{3}) \approx 126.87^\circ$,which is not equal to $\frac{\pi}{2}$,$\frac{\pi}{3}$,or $\frac{\pi}{4}$.
Thus,the correct option is $D$.

(B) The given equation is $x = ay^2 + by + c$.
We can rewrite this by completing the square for $y$:
$x = a(y^2 + \frac{b}{a}y) + c$
$x = a(y^2 + \frac{b}{a}y + \frac{b^2}{4a^2}) + c - \frac{b^2}{4a}$
$x - (c - \frac{b^2}{4a}) = a(y + \frac{b}{2a})^2$
$(y + \frac{b}{2a})^2 = \frac{1}{a}(x - (c - \frac{b^2}{4a}))$
This is in the standard form $(y - k)^2 = 4p(x - h)$,where $4p = \frac{1}{a}$.
The length of the latus rectum is given by $|4p|$.
Therefore,the length is $|\frac{1}{a}| = \frac{1}{|a|}$.

(B) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Here,the parabola is $y^2 = 4x$,so $a = 1$.
The equation of the tangent is $y = mx + \frac{1}{m}$.
Since the tangent passes through the point $(2, 3)$,we substitute these coordinates into the equation:
$3 = m(2) + \frac{1}{m}$
$3 = 2m + \frac{1}{m}$
Multiplying by $m$ (where $m \neq 0$):
$3m = 2m^2 + 1$
$2m^2 - 3m + 1 = 0$.
This is a quadratic equation in $m$,where $m_1$ and $m_2$ are the roots.
From the properties of quadratic equations $am^2 + bm + c = 0$,the sum of the roots is $m_1 + m_2 = -\frac{b}{a}$ and the product is $m_1 m_2 = \frac{c}{a}$.
Here,$m_1 + m_2 = \frac{3}{2}$ and $m_1 m_2 = \frac{1}{2}$.
We need to find $\frac{1}{m_1} + \frac{1}{m_2} = \frac{m_1 + m_2}{m_1 m_2}$.
Substituting the values: $\frac{3/2}{1/2} = 3$.

(A) Given the equation of the parabola: $4y^2 + 6x = 8y + 7$.
Rearrange the terms to complete the square for $y$:
$4(y^2 - 2y) = -6x + 7$.
Add $4(1)^2 = 4$ to both sides:
$4(y^2 - 2y + 1) = -6x + 7 + 4$.
$4(y - 1)^2 = -6x + 11$.
Divide by $4$:
$(y - 1)^2 = -\frac{6}{4}(x - \frac{11}{6})$.
$(y - 1)^2 = -\frac{3}{2}(x - \frac{11}{6})$.
This is in the form $(y - k)^2 = 4a(x - h)$,where the vertex $(h, k)$ is $(\frac{11}{6}, 1)$.
The tangent at the vertex of a parabola is a line perpendicular to the axis of the parabola passing through the vertex.
Since the parabola is of the form $(y - k)^2 = 4a(x - h)$,its axis is horizontal $(y = k)$.
Therefore,the tangent at the vertex is a vertical line passing through the vertex $(\frac{11}{6}, 1)$.
The equation of this vertical line is $x = h$,which is $x = \frac{11}{6}$.

(B) The equation of the parabola is $y^2 = x$.
The axis of the parabola is the $x$-axis $(y = 0)$.
The chord $PQ$ is perpendicular to the axis of the parabola,so $PQ$ is a vertical line.
Since $P$ is $(4, -2)$,the equation of the line $PQ$ is $x = 4$.
To find the coordinates of $Q$,we substitute $x = 4$ into the parabola equation $y^2 = x$:
$y^2 = 4 \implies y = 2$ or $y = -2$.
Since $P$ is $(4, -2)$,the other endpoint $Q$ must be $(4, 2)$.
Now,we find the slope of the tangent at $Q(4, 2)$ by differentiating $y^2 = x$ with respect to $x$:
$2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y}$.
At $Q(4, 2)$,the slope of the tangent $m_t = \frac{1}{2(2)} = \frac{1}{4}$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t}$.
Therefore,$m_n = -\frac{1}{1/4} = -4$.

(D) The given equation of the parabola is $x^2 = 8y$. Comparing this with the standard form $x^2 = 4ay$,we get $4a = 8$,which implies $a = 2$.
The vertex of the parabola is $V(0, 0)$.
The endpoints of the latus rectum for $x^2 = 4ay$ are $(2a, a)$ and $(-2a, a)$.
Substituting $a = 2$,the endpoints are $L_1(4, 2)$ and $L_2(-4, 2)$.
The triangle is formed by the vertices $V(0, 0)$,$L_1(4, 2)$,and $L_2(-4, 2)$.
The area of the triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ 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 $= \frac{1}{2} |0(2 - 2) + 4(2 - 0) + (-4)(0 - 2)| = \frac{1}{2} |0 + 8 + 8| = \frac{1}{2} |16| = 8$ square units.

(D) The equation of the parabola is $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$.
The slope of the tangent $m = \tan(45^{\circ}) = 1$.
The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Substituting $a = 1$ and $m = 1$,we get $y = 1(x) + \frac{1}{1}$,which simplifies to $y = x + 1$.
The point of contact $(x_1, y_1)$ for a tangent with slope $m$ to the parabola $y^2 = 4ax$ is given by $(\frac{a}{m^2}, \frac{2a}{m})$.
Substituting $a = 1$ and $m = 1$,we get the point of contact $(x_1, y_1) = (\frac{1}{1^2}, \frac{2(1)}{1}) = (1, 2)$.
Since the point of contact is a fixed point $(1, 2)$,the locus is simply the point itself,which does not match any of the given algebraic expressions.

(A) The given equation of the parabola is $4y^2 = x$,which can be written as $y^2 = \frac{1}{4}x$.
Comparing this with $y^2 = 4ax$,we get $4a = \frac{1}{4}$,so $a = \frac{1}{16}$.
The slope of the tangent $m = \tan(60^{\circ}) = \sqrt{3}$.
The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Substituting the values,$y = \sqrt{3}x + \frac{1/16}{\sqrt{3}} = \sqrt{3}x + \frac{1}{16\sqrt{3}}$.
The point of contact for a parabola $y^2 = 4ax$ with slope $m$ is given by $\left( \frac{a}{m^2}, \frac{2a}{m} \right)$.
Substituting $a = \frac{1}{16}$ and $m = \sqrt{3}$:
$x = \frac{1/16}{(\sqrt{3})^2} = \frac{1}{16 \times 3} = \frac{1}{48}$.
$y = \frac{2(1/16)}{\sqrt{3}} = \frac{1/8}{\sqrt{3}} = \frac{1}{8\sqrt{3}}$.
Thus,the point of contact is $\left( \frac{1}{48}, \frac{1}{8\sqrt{3}} \right)$.

(B) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.
The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.
Given $m = -1/4$ and $a = 4$,we substitute these values into the formula:
$y = (-1/4)x - 2(4)(-1/4) - 4(-1/4)^3$
$y = -x/4 + 2 - 4(-1/64)$
$y = -x/4 + 2 + 1/16$
Multiply the entire equation by $16$ to clear the denominators:
$16y = -4x + 32 + 1$
$16y = -4x + 33$
$4x + 16y = 33$

(C) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.
The locus of the point of intersection of two perpendicular tangents to a parabola is its directrix.
The equation of the directrix for the parabola $y^2 = 4ax$ is $x = -a$.
Substituting $a = 1$,we get $x = -1$.
Therefore,the locus of point $P$ is $x = -1$.

(A) Given equation: $y^2 - 2y - 12x - 11 = 0$
Complete the square for $y$: $(y^2 - 2y + 1) - 1 - 12x - 11 = 0$
$(y - 1)^2 = 12x + 12$
$(y - 1)^2 = 12(x + 1)$
Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get $h = -1, k = 1$,and $4a = 12 \implies a = 3$.
The parametric equations for $(y - k)^2 = 4a(x - h)$ are $x = h + at^2$ and $y = k + 2at$.
Substituting the values: $x = -1 + 3t^2$ and $y = 1 + 2(3)t = 1 + 6t$.
Thus,$x = 3t^2 - 1$ and $y = 6t + 1$.

(B) The length of the latus rectum of a parabola is given by $4a = 8$,which implies $a = 2$.
Since the focus is $(3, 0)$ and the $y$-coordinate is $0$,the axis of the parabola is the $x$-axis.
The standard form of such a parabola is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex.
Given the focus is $(h + a, k) = (3, 0)$,we have $k = 0$ and $h + a = 3$.
Substituting $a = 2$,we get $h + 2 = 3$,so $h = 1$.
Therefore,the vertex $(h, k)$ is $(1, 0)$.

(B) Given equations are $x + y = 1$ and $y = x - x^2$.
From the first equation,we have $y = 1 - x$.
Substitute this into the second equation:
$1 - x = x - x^2$
$x^2 - 2x + 1 = 0$
$(x - 1)^2 = 0$
$x = 1$.
Now,substitute $x = 1$ into $y = 1 - x$:
$y = 1 - 1 = 0$.
Thus,the point of intersection is $(1, 0)$.

(A) For the parabola $y^2 = 4ax$ (where $a=1$),the normal at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.
Since it passes through $(3, 0)$,we have $0 = -3t + 2t + t^3$,which simplifies to $t^3 - t = 0$.
Thus,$t(t-1)(t+1) = 0$,giving $t_1 = 0, t_2 = 1, t_3 = -1$.
The points are $P(0, 0), Q(1, 2), R(1, -2)$.
$(A)$ Circumradius $R_{c} = \frac{abc}{4\Delta}$. Sides are $PQ = \sqrt{1^2 + 2^2} = \sqrt{5}$,$PR = \sqrt{5}$,$QR = 4$. Area $\Delta = \frac{1}{2} \times 4 \times 1 = 2$. $R_{c} = \frac{\sqrt{5} \cdot \sqrt{5} \cdot 4}{4 \cdot 2} = \frac{5}{2}$. So $A \to P$.
$(B)$ Area of $\Delta PQR = 2$. So $B \to S$.
$(C)$ Centroid $G = (\frac{0+1+1}{3}, \frac{0+2-2}{3}) = (2/3, 0)$. So $C \to R$.
$(D)$ Circumcenter $O_{c}$ of $\Delta PQR$ with vertices $(0,0), (1,2), (1,-2)$ is $(5/2, 0)$. So $D \to Q$.
Therefore,$A \to P, B \to S, C \to R, D \to Q$.

(B) The given equation of the parabola is $y^2 - 6y + 24x - 63 = 0$.
Completing the square for $y$,we get $(y - 3)^2 - 9 + 24x - 63 = 0$.
$(y - 3)^2 = -24x + 72$.
$(y - 3)^2 = -24(x - 3)$.
This is a parabola of the form $(y - k)^2 = 4a(x - h)$,where $4a = -24$,so $a = -6$.
The locus of the point of intersection of perpendicular tangents to a parabola is its directrix.
The directrix of the parabola $(y - k)^2 = 4a(x - h)$ is given by $x - h = -a$.
Here,$h = 3$,$k = 3$,and $a = -6$.
So,$x - 3 = -(-6) = 6$.
$x = 9$,which can be written as $x - 9 = 0$.

(A) Given the equation of the parabola is $y^2 = 8x$.
Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$.
The point given is $(x_1, y_1) = (2, 4)$.
Differentiating $y^2 = 8x$ with respect to $x$,we get $2y \frac{dy}{dx} = 8$,which implies $\frac{dy}{dx} = \frac{4}{y}$.
The slope of the tangent $(m_t)$ at $(2, 4)$ is $\frac{4}{4} = 1$.
The slope of the normal $(m_n)$ is $-\frac{1}{m_t} = -1$.
The equation of the normal at $(x_1, y_1)$ with slope $m_n$ is given by $y - y_1 = m_n(x - x_1)$.
Substituting the values,we get $y - 4 = -1(x - 2)$.
$y - 4 = -x + 2$.
$x + y = 6$.

(D) The distance between the directrix $(3x + 4y + 15 = 0)$ and the tangent at the vertex $(3x + 4y - 5 = 0)$ is equal to the distance $a$ from the vertex to the focus.
The distance $a$ is given by the formula for the distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$,which is $a = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Substituting the values: $a = \frac{|15 - (-5)|}{\sqrt{3^2 + 4^2}} = \frac{20}{\sqrt{9 + 16}} = \frac{20}{5} = 4$.
The length of the latus rectum of a parabola is $4a$.
Therefore,the length of the latus rectum $= 4 \times 4 = 16$.

(C) The given equation of the parabola is $x^2 = -8y$.
Comparing this with the standard form $x^2 = -4ay$,we get $-4a = -8$,which implies $a = 2$.
The focus of the parabola $x^2 = -4ay$ is $(0, -a)$ and the equation of the directrix is $y = a$.
Here,the focus is $(0, -2)$ and the directrix is $y = 2$.
The distance between the focus $(0, -2)$ and the directrix $y = 2$ is given by $|2 - (-2)| = |4| = 4$.
Alternatively,the distance between the focus and the directrix of a parabola $x^2 = 4ay$ (or $x^2 = -4ay$) is $2a$.
Substituting $a = 2$,the distance is $2 \times 2 = 4$.

(C) Let the coordinates of $P$,whose locus is to be determined,be $(h, k)$.
Since $P$ divides the line segment joining $(0, 0)$ and $(x, y)$ in the ratio $1 : 3$,by the section formula,the coordinates of $P$ are given by:
$h = \frac{1 \cdot x + 3 \cdot 0}{1 + 3} = \frac{x}{4} \implies x = 4h$
$k = \frac{1 \cdot y + 3 \cdot 0}{1 + 3} = \frac{y}{4} \implies y = 4k$
Given that $(x, y)$ lies on the parabola $y^2 = 4x$,we substitute the expressions for $x$ and $y$:
$(4k)^2 = 4(4h)$
$16k^2 = 16h$
$k^2 = h$
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $y^2 = x$.

(A) The equation of the parabola is $y^2 = 4bx$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4b$,so $\frac{dy}{dx} = \frac{2b}{y}$.
At the point $(bt_1^2, 2bt_1)$,the slope of the tangent is $\frac{2b}{2bt_1} = \frac{1}{t_1}$.
Therefore,the slope of the normal at $(bt_1^2, 2bt_1)$ is $m = -t_1$.
The equation of the normal at $(bt_1^2, 2bt_1)$ is $(y - 2bt_1) = -t_1(x - bt_1^2)$.
Since the point $(bt_2^2, 2bt_2)$ lies on this normal,we substitute these coordinates into the equation:
$(2bt_2 - 2bt_1) = -t_1(bt_2^2 - bt_1^2)$
$2b(t_2 - t_1) = -t_1b(t_2 - t_1)(t_2 + t_1)$
Dividing by $b(t_2 - t_1)$ (assuming $t_2 \neq t_1$ and $b \neq 0$):
$2 = -t_1(t_2 + t_1)$
$2 = -t_1t_2 - t_1^2$
$t_1t_2 = -t_1^2 - 2$
$t_2 = -t_1 - \frac{2}{t_1}$

(B) The given equation of the parabola is $y^2 = -8x$.
Comparing this with the standard form $y^2 = -4ax$,we get $4a = 8$,which implies $a = 2$.
The focus of the parabola $y^2 = -4ax$ is at $(-a, 0)$.
Substituting $a = 2$,the focus is at $(-2, 0)$.
Since $2x + y + \lambda = 0$ is a focal chord,it must pass through the focus $(-2, 0)$.
Substituting $x = -2$ and $y = 0$ into the equation of the chord:
$2(-2) + 0 + \lambda = 0$
$-4 + \lambda = 0$
$\lambda = 4$.

(B) Let the parabola be $y^2 = 4ax$ with focus $S(a, 0)$.
Let $P(at^2, 2at)$ be a point on the parabola.
The focal distance $SP = a + at^2$.
The equation of the tangent at $P$ is $ty = x + at^2$.
This tangent meets the axis of the parabola $(y=0)$ at point $Q(-at^2, 0)$.
The distance $SQ = SO + OQ = a + at^2$.
Since $SP = SQ$,the triangle $\triangle SPQ$ is an isosceles triangle.
Thus,$\angle SPQ = \angle SQP = \alpha$.
In $\triangle SPQ$,the exterior angle at the axis is the sum of the interior opposite angles.
Let $\theta$ be the angle between the tangent and the axis. Then $\theta + \alpha = \angle SPQ = \alpha$ is not correct; rather,in $\triangle SPQ$,the exterior angle at $Q$ is $\angle SPQ + \angle PSQ = 2\alpha$ is not the case here.
Actually,since $\triangle SPQ$ is isosceles with $SP=SQ$,the base angles are $\alpha$. The angle between the tangent and the axis is $\theta$. In $\triangle SPQ$,the exterior angle at $Q$ is $\theta$. Thus $\theta = \angle SPQ + \angle PSQ$. Since $\angle SPQ = \alpha$ and $\angle PSQ = \angle SPQ = \alpha$,we have $\theta = \alpha / 2$.

(C) The given equation is $x^2 - 4x - 8y + 12 = 0$.
Completing the square for the $x$ terms:
$x^2 - 4x = 8y - 12$
$x^2 - 4x + 4 = 8y - 12 + 4$
$(x - 2)^2 = 8y - 8$
$(x - 2)^2 = 8(y - 1)$.
Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,we get $4a = 8$.
The length of the latus rectum is given by $|4a|$.
Therefore,the length of the latus rectum is $8$.

(A) Let the coordinates of point $Q$ on the parabola $y^2 = 8x$ be $(2t^2, 4t)$.
Let the midpoint of $PQ$ be $(h, k)$.
Since $P = (1, 0)$ and $Q = (2t^2, 4t)$,the midpoint $(h, k)$ is given by:
$h = \frac{2t^2 + 1}{2} \implies 2t^2 = 2h - 1$
$k = \frac{4t + 0}{2} = 2t \implies t = \frac{k}{2}$
Substitute $t = \frac{k}{2}$ into the equation $2t^2 = 2h - 1$:
$2(\frac{k}{2})^2 = 2h - 1$
$2(\frac{k^2}{4}) = 2h - 1$
$\frac{k^2}{2} = 2h - 1$
$k^2 = 4h - 2$
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 4x - 2$,which is $y^2 - 4x + 2 = 0$.

(B) Step $1$: Solve the given equations simultaneously.
Given that the line $2y - x + k = 0$ touches the parabola $x^2 + 4y = 0$.
From the line equation,$2y = x - k$,so $y = \frac{x - k}{2}$.
Substituting $y = \frac{x - k}{2}$ into the parabola equation $x^2 + 4y = 0$,we get:
$x^2 + 4\left(\frac{x - k}{2}\right) = 0$
$x^2 + 2(x - k) = 0$
$x^2 + 2x - 2k = 0$
Step $2$: Use the condition for tangency.
For the line to touch the parabola,the quadratic equation must have equal roots,meaning the discriminant $D = 0$.
For $ax^2 + bx + c = 0$,$D = b^2 - 4ac$.
Here,$a = 1, b = 2, c = -2k$.
$D = (2)^2 - 4(1)(-2k) = 0$
$4 + 8k = 0$
$8k = -4$
$k = -1/2$.

(B) Let the vertex of the parabola $y^2 = 4ax$ be $O(0, 0)$.
Let $Q(at^2, 2at)$ be a point on the parabola.
The midpoint $P(x, y)$ of the chord $OQ$ is given by:
$x = \frac{0 + at^2}{2} = \frac{at^2}{2}$
$y = \frac{0 + 2at}{2} = at$
From the second equation,$t = \frac{y}{a}$.
Substituting $t$ into the first equation:
$x = \frac{a}{2} \left(\frac{y}{a}\right)^2 = \frac{a}{2} \cdot \frac{y^2}{a^2} = \frac{y^2}{2a}$
$y^2 = 2ax$
For the given parabola $y^2 = 4x$,we have $a = 1$.
Substituting $a = 1$ into the locus equation $y^2 = 2ax$,we get $y^2 = 2x$.