Parabola Questions in English

(A) The equation of the parabola is $y^2 = 4x$,so $a = 1$.
The slope of the line $y = 3x + 4$ is $m = 3$.
The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.
Substituting $a = 1$ and $m = 3$ into the formula:
$y = 3x - 2(1)(3) - (1)(3)^3$
$y = 3x - 6 - 27$
$y = 3x - 33$.
Thus,the equation of the normal is $y = 3x - 33$.

(A) The parabola is given by $y^2 = 2px$. The focus of this parabola is $S = (p/2, 0)$.
The directrix of the parabola is $x = -p/2$,or $x + p/2 = 0$.
The circle has its center at $(p/2, 0)$ and touches the directrix. The radius $r$ of the circle is the distance from the focus $(p/2, 0)$ to the directrix $x = -p/2$,which is $r = |p/2 - (-p/2)| = |p| = p$.
The equation of the circle is $(x - p/2)^2 + y^2 = p^2$.
Substitute $y^2 = 2px$ into the circle equation:
$(x - p/2)^2 + 2px = p^2$
$x^2 - px + p^2/4 + 2px = p^2$
$x^2 + px - 3p^2/4 = 0$
Solving for $x$ using the quadratic formula:
$x = \frac{-p \pm \sqrt{p^2 - 4(1)(-3p^2/4)}}{2} = \frac{-p \pm \sqrt{p^2 + 3p^2}}{2} = \frac{-p \pm 2p}{2}$
So,$x = p/2$ or $x = -3p/2$.
If $x = p/2$,$y^2 = 2p(p/2) = p^2$,so $y = \pm p$. The points are $(p/2, p)$ and $(p/2, -p)$.
If $x = -3p/2$,$y^2 = 2p(-3p/2) = -3p^2$,which has no real solution for $y$.
Thus,the points of intersection are $(p/2, p)$ and $(p/2, -p)$.

(D) Let the coordinates of the endpoints of the focal chord $BC$ be $B(at_{1}^{2}, 2at_{1})$ and $C(at_{2}^{2}, 2at_{2})$.
Since $BC$ is a focal chord,$t_{1}t_{2} = -1$.
The area of $\Delta ABC$ with vertices $A(0, 0)$,$B(at_{1}^{2}, 2at_{1})$,and $C(at_{2}^{2}, 2at_{2})$ is given by:
$\text{Area} = \frac{1}{2} |x_{A}(y_{B} - y_{C}) + x_{B}(y_{C} - y_{A}) + x_{C}(y_{A} - y_{B})|$
$\text{Area} = \frac{1}{2} |0 + at_{1}^{2}(2at_{2} - 0) + at_{2}^{2}(0 - 2at_{1})| = \frac{1}{2} |2a^{2}t_{1}^{2}t_{2} - 2a^{2}t_{1}t_{2}^{2}|$
$\text{Area} = a^{2} |t_{1}t_{2}(t_{1} - t_{2})| = a^{2} |-1(t_{1} - t_{2})| = a^{2} |t_{1} - t_{2}|$.
Given that the area is $5a^{2}$,we have $a^{2} |t_{1} - t_{2}| = 5a^{2}$,which implies $|t_{1} - t_{2}| = 5$.
The length of the focal chord is $L = a(t_{1} - t_{2})^{2}$.
Since $t_{2} = -1/t_{1}$,we have $t_{1} - t_{2} = t_{1} + 1/t_{1}$.
Thus,$(t_{1} - t_{2})^{2} = (t_{1} + 1/t_{1})^{2} = (t_{1} - 1/t_{1})^{2} + 4 = (t_{1} - t_{2})^{2} + 4 = 5^{2} + 4 = 29$.
Wait,the length of the focal chord is $a(t_{1} - t_{2})^{2} = a(t_{1} + 1/t_{1})^{2}$.
Using $|t_{1} - t_{2}| = 5$,we have $(t_{1} - t_{2})^{2} = 25$.
Since $t_{1}t_{2} = -1$,$(t_{1} + t_{2})^{2} = (t_{1} - t_{2})^{2} + 4t_{1}t_{2} = 25 - 4 = 21$.
The length of the focal chord $BC$ is $a(t_{1} - t_{2})\sqrt{(t_{1} + t_{2})^{2}} = a(5)\sqrt{21}$ is not correct.
The length of the focal chord is $a(t_{1} - t_{2})^{2} = a(t_{1} + 1/t_{1})^{2} = a(t_{1}^{2} + 1/t_{1}^{2} + 2) = a((t_{1} - 1/t_{1})^{2} + 4) = a(25 + 4) = 29a$.
Re-evaluating: The length of the focal chord is $a(t_{1} + 1/t_{1})^{2}$. Given $|t_{1} - t_{2}| = 5$,where $t_{2} = -1/t_{1}$,then $|t_{1} + 1/t_{1}| = 5$. Thus,length $= a(5)^{2} = 25a$.

(A) Let the points $P$ and $Q$ on the parabola $y^{2} = 4ax$ be represented by parameters $t_{1}$ and $t_{2}$ respectively,such that $P = (at_{1}^{2}, 2at_{1})$ and $Q = (at_{2}^{2}, 2at_{2})$.
The slope of the tangent at any point $t$ is $m = 1/t$.
Since the tangents at $P$ and $Q$ are perpendicular,the product of their slopes is $-1$:
$(1/t_{1}) \times (1/t_{2}) = -1 \implies t_{1}t_{2} = -1 \implies t_{2} = -1/t_{1}$.
Given $P(p, q) = (at_{1}^{2}, 2at_{1})$,we have $q = 2at_{1} \implies t_{1} = q/(2a)$.
Thus,$t_{2} = -1/(q/2a) = -2a/q$.
The coordinates of $Q$ are $(at_{2}^{2}, 2at_{2}) = (a(-2a/q)^{2}, 2a(-2a/q)) = (a(4a^{2}/q^{2}), -4a^{2}/q)$.
Since $P(p, q)$ lies on $y^{2} = 4ax$,we have $q^{2} = 4ap$. Substituting $q^{2} = 4ap$ into the $x$-coordinate of $Q$:
$x_{Q} = (4a^{3})/(4ap) = a^{2}/p$.
Therefore,the coordinates of $Q$ are $(a^{2}/p, -4a^{2}/q)$.

(A) For a parabola $y^{2} = 4ax$,if one endpoint of a focal chord is $(at_{1}^{2}, 2at_{1})$,then the other endpoint is $(at_{2}^{2}, 2at_{2})$ where $t_{1}t_{2} = -1$.
Given $y^{2} = 32x$,we have $4a = 32$,so $a = 8$.
The point $(2, -8)$ corresponds to $(at_{1}^{2}, 2at_{1})$.
$2at_{1} = -8 \implies 2(8)t_{1} = -8 \implies 16t_{1} = -8 \implies t_{1} = -\frac{1}{2}$.
Since $t_{1}t_{2} = -1$,we have $t_{2} = -\frac{1}{t_{1}} = -\frac{1}{-1/2} = 2$.
The other endpoint is $(at_{2}^{2}, 2at_{2}) = (8(2)^{2}, 2(8)(2)) = (8 \times 4, 32) = (32, 32)$.

(A) The coordinates of the ends of the latus rectum of the parabola $y^{2} = 4ax$ are $(a, 2a)$ and $(a, -2a)$.
The slope of the tangent at $(a, 2a)$ is given by $2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}$. At $(a, 2a)$,the slope of the tangent is $\frac{2a}{2a} = 1$.
Therefore,the slope of the normal at $(a, 2a)$ is $-1$.
The equation of the normal at $(a, 2a)$ is $y - 2a = -1(x - a)$,which simplifies to $x + y - 3a = 0$ $(i)$.
Similarly,the slope of the normal at $(a, -2a)$ is $1$,and the equation of the normal is $y - (-2a) = 1(x - a)$,which simplifies to $x - y - 3a = 0$ $(ii)$.
The combined equation of the normals is $(x + y - 3a)(x - y - 3a) = 0$.
This expands to $(x - 3a)^{2} - y^{2} = 0$,which is $x^{2} - 6ax + 9a^{2} - y^{2} = 0$ or $x^{2} - y^{2} - 6ax + 9a^{2} = 0$.

(A) Let the point be $(h, k)$. The equation of a normal to the parabola $y^{2} = 4ax$ is given by $y = mx - 2am - am^{3}$.
Since it passes through $(h, k)$,we have $k = mh - 2am - am^{3}$,which simplifies to $am^{3} + m(2a - h) + k = 0$.
Let the slopes of the normals be $m_1, m_2, m_3$. The product of the roots is $m_1 m_2 m_3 = -k/a$.
Given that the normals make angles $\alpha$ and $\beta$ with the axis,their slopes are $m_1 = \tan \alpha$ and $m_2 = \tan \beta$. We are given $m_1 m_2 = 2$.
Substituting this into the product of roots,we get $2 m_3 = -k/a$,so $m_3 = -k/(2a)$.
Since $m_3$ is a root of the cubic equation $am^{3} + m(2a - h) + k = 0$,we substitute $m = -k/(2a)$:
$a(-k/(2a))^{3} + (-k/(2a))(2a - h) + k = 0$
$-k^{3}/(8a^{2}) - k + kh/(2a) + k = 0$
$-k^{3}/(8a^{2}) + kh/(2a) = 0$
$k^{2} = 4ah$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^{2} = 4ax$.

(C) Let the variable point on the parabola $y^{2} = 4ax$ be $P(at^{2}, 2at)$ and the focus be $S(a, 0)$.
Let $M(h, k)$ be the midpoint of $PS$.
Then $h = \frac{at^{2} + a}{2}$ and $k = \frac{2at + 0}{2} = at$.
From $k = at$,we get $t = \frac{k}{a}$.
Substituting $t$ into the equation for $h$: $h = \frac{a(k/a)^{2} + a}{2} = \frac{k^{2}/a + a}{2} = \frac{k^{2} + a^{2}}{2a}$.
Thus,$2ah = k^{2} + a^{2}$,which simplifies to $k^{2} = 2a(h - a/2)$.
The locus of $(h, k)$ is $y^{2} = 2a(x - a/2)$.
Comparing this with the standard form $Y^{2} = 4AX$,where $Y = y$,$X = x - a/2$,and $4A = 2a \Rightarrow A = a/2$.
The directrix is given by $X = -A$,so $x - a/2 = -a/2$.
Therefore,$x = 0$.

(B) Given parabola: $y^{2} = 4x + 4y$
Rearranging the terms: $y^{2} - 4y = 4x$
Completing the square: $y^{2} - 4y + 4 = 4x + 4$
$(y - 2)^{2} = 4(x + 1)$
Comparing with the standard form $Y^{2} = 4AX$,where $Y = y - 2$,$X = x + 1$,and $A = 1$.
Vertex: $(X = 0, Y = 0) \implies (x + 1 = 0, y - 2 = 0) \implies (-1, 2)$.
Focus: $(X = A, Y = 0) \implies (x + 1 = 1, y - 2 = 0) \implies (0, 2)$.
Axis: $Y = 0 \implies y - 2 = 0 \implies y = 2$.
Length of latus rectum: $4A = 4(1) = 4$.
Equation of directrix: $X = -A \implies x + 1 = -1 \implies x = -2$.

(B) Let the equation of the parabola be $y^{2} = 4ax$ and the endpoints of the chord $PQ$ be $P(at_1^{2}, 2at_1)$ and $Q(at_2^{2}, 2at_2)$.
The point of intersection $T$ of the tangents at $P$ and $Q$ has coordinates $(at_1t_2, a(t_1 + t_2))$.
The distance of a point $(at^{2}, 2at)$ from the focus $S(a, 0)$ is $a(1 + t^{2})$.
Thus,$SP = a(1 + t_1^{2})$ and $SQ = a(1 + t_2^{2})$.
The distance $ST$ from the focus $S(a, 0)$ to $T(at_1t_2, a(t_1 + t_2))$ is given by:
$ST^{2} = (at_1t_2 - a)^{2} + (a(t_1 + t_2) - 0)^{2}$
$ST^{2} = a^{2}(t_1^{2}t_2^{2} - 2t_1t_2 + 1 + t_1^{2} + 2t_1t_2 + t_2^{2})$
$ST^{2} = a^{2}(t_1^{2}t_2^{2} + t_1^{2} + t_2^{2} + 1)$
$ST^{2} = a^{2}(1 + t_1^{2})(1 + t_2^{2})$
$ST^{2} = SP \cdot SQ$
Since $ST^{2} = SP \cdot SQ$,the distances $SP, ST, SQ$ are in $G.P.$

(B) Let $P(at_1^2, 2at_1)$ be one endpoint of the normal chord,then the other endpoint is $Q(at_2^2, 2at_2)$.
For a normal chord,the relation between parameters is $t_2 = -t_1 - \frac{2}{t_1} \dots (1)$.
The slope of $OP$ is $m_1 = \frac{2at_1}{at_1^2} = \frac{2}{t_1}$.
The slope of $OQ$ is $m_2 = \frac{2at_2}{at_2^2} = \frac{2}{t_2}$.
Since the chord subtends a right angle at the vertex $O(0,0)$,the product of slopes is $-1$:
$m_1 \times m_2 = -1$ $\Rightarrow \frac{2}{t_1} \times \frac{2}{t_2} = -1$ $\Rightarrow t_1t_2 = -4 \dots (2)$.
Substituting $(2)$ into $(1)$:
$-4/t_1 = -t_1 - 2/t_1$ $\Rightarrow -4/t_1 = -(t_1^2 + 2)/t_1$ $\Rightarrow 4 = t_1^2 + 2$ $\Rightarrow t_1^2 = 2$ $\Rightarrow t_1 = \pm \sqrt{2}$.
The slope of the line joining the vertex and the endpoint $P$ is $m_1 = \frac{2}{t_1} = \frac{2}{\pm \sqrt{2}} = \pm \sqrt{2}$.

(D) The equation of the parabola is $y^2 = 8(x - 3)$.
Comparing this with $(y - k)^2 = 4a(x - h)$,we get $4a = 8$,so $a = 2$.
The vertex is $(h, k) = (3, 0)$ and the focus $S$ is $(h + a, k) = (3 + 2, 0) = (5, 0)$.
The directrix is $x = h - a = 3 - 2 = 1$.
Let $P$ be a point on the parabola. By definition,the distance $SP$ is equal to the perpendicular distance from $P$ to the directrix,which is $PM$.
Since $\triangle SPM$ is an equilateral triangle,$SP = PM = SM$.
Let $M$ be the point $(1, y_P)$ on the directrix. The distance $SM$ is the distance between $(5, 0)$ and $(1, y_P)$,so $SM = \sqrt{(5 - 1)^2 + (0 - y_P)^2} = \sqrt{16 + y_P^2}$.
Also,in the right-angled triangle formed by the projection of $P$ onto the directrix,the angle $\angle SMP = 60^\circ$ (as it is equilateral).
Let $Z$ be the point $(1, 0)$ on the directrix. In $\triangle SZM$,$\angle SZM = 90^\circ$ and $SZ = 5 - 1 = 4$.
Since $\angle SMZ = 30^\circ$,we have $\cos(30^\circ) = \frac{SZ}{SM}$.
$\frac{\sqrt{3}}{2} = \frac{4}{SM} \Rightarrow SM = \frac{8}{\sqrt{3}}$.
Wait,looking at the geometry,$PM$ is horizontal. The distance $PM$ is the horizontal distance from $x=1$ to $P(x, y)$,so $PM = x - 1$.
Since $SP = PM$,$SP = x - 1$. Also $SP = x + a - 2h = x + 2 - 6 = x - 4$ is incorrect; $SP = x + a = x + 2$ relative to vertex. Actually $SP = x - 3 + 2 = x - 1$.
In $\triangle SPM$,$PM = SP = SM = L$. The height of the triangle from $S$ to $PM$ is $L \sin(60^\circ) = \frac{L\sqrt{3}}{2}$.
The vertical distance from $S(5, 0)$ to the line $PM$ is $y_P$. So $y_P = \frac{L\sqrt{3}}{2}$.
Also $PM = x_P - 1 = L$. Since $y_P^2 = 8(x_P - 3)$,we have $(\frac{L\sqrt{3}}{2})^2 = 8(L + 1 - 3) = 8(L - 2)$.
$\frac{3L^2}{4} = 8L - 16 \Rightarrow 3L^2 - 32L + 64 = 0$.
$(3L - 8)(L - 8) = 0$. Since $L$ must be $8$ for the geometry to hold,$L = 8$.

(C) Let the point on the parabola be $P(at^2, 2at)$.
The tangent at $P$ is $ty = x + at^2$.
The directrix is $x = -a$.
To find the intersection point $Q$ of the tangent and the directrix,substitute $x = -a$ into the tangent equation:
$ty = -a + at^2 \implies y = \frac{a(t^2 - 1)}{t}$.
So,$Q = (-a, \frac{a(t^2 - 1)}{t})$.
The focus is $S(a, 0)$.
The slope of $SP$ is $m_1 = \frac{2at - 0}{at^2 - a} = \frac{2t}{t^2 - 1}$.
The slope of $SQ$ is $m_2 = \frac{\frac{a(t^2 - 1)}{t} - 0}{-a - a} = \frac{a(t^2 - 1)}{t(-2a)} = -\frac{t^2 - 1}{2t}$.
Since $m_1 \times m_2 = \left(\frac{2t}{t^2 - 1}\right) \times \left(-\frac{t^2 - 1}{2t}\right) = -1$,the lines $SP$ and $SQ$ are perpendicular.
Therefore,$\theta = \frac{\pi}{2}$.

(D) The focus is $S(0, 1)$ and the directrix is $x + 2 = 0$.
The definition of a parabola is $SP = PM$,where $P(x, y)$ is a point on the parabola.
$SP^2 = PM^2 \implies (x - 0)^2 + (y - 1)^2 = (x + 2)^2$
$x^2 + (y - 1)^2 = x^2 + 4x + 4$
$(y - 1)^2 = 4(x + 1)$
Let $Y = y - 1$ and $X = x + 1$. Then the equation becomes $Y^2 = 4X$,which is of the form $Y^2 = 4aX$ with $a = 1$.
The parametric coordinates for $Y^2 = 4aX$ are $(at^2, 2at)$.
Here,$X = t^2$ and $Y = 2t$.
Substituting back for $x$ and $y$:
$x + 1 = t^2 \implies x = t^2 - 1$
$y - 1 = 2t \implies y = 2t + 1$
Thus,the parametric coordinates are $(t^2 - 1, 2t + 1)$.

(D) The axis of the parabola is perpendicular to the directrix $x + y - 2 = 0$.
Thus,the equation of the axis is $x - y + k = 0$.
Since it passes through the focus $S(3, -4)$,we have $3 - (-4) + k = 0$,which gives $k = -7$.
The equation of the axis is $x - y - 7 = 0$.
Let $Z$ be the intersection of the axis and the directrix. Solving $x + y - 2 = 0$ and $x - y - 7 = 0$,we get $Z = (9/2, -5/2)$.
The vertex $A$ is the midpoint of $SZ$.
$A = \left( \frac{3 + 9/2}{2}, \frac{-4 - 5/2}{2} \right) = \left( \frac{15/2}{2}, \frac{-13/2}{2} \right) = \left( \frac{15}{4}, -\frac{13}{4} \right)$.

(C) Points are collinear if the area of the triangle formed by them is zero.
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$.
Substituting the points $(a, 0)$,$(at_1^2, 2at_1)$,and $(at_2^2, 2at_2)$:
$\frac{1}{2} |a(2at_1 - 2at_2) + at_1^2(2at_2 - 0) + at_2^2(0 - 2at_1)| = 0$.
Dividing by $2a^2$ (assuming $a \neq 0$):
$|(t_1 - t_2) + t_1^2 t_2 - t_2^2 t_1| = 0$.
$t_1 - t_2 + t_1 t_2 (t_1 - t_2) = 0$.
$(t_1 - t_2)(1 + t_1 t_2) = 0$.
Since $t_1 \neq t_2$ for distinct points,we must have $1 + t_1 t_2 = 0$,which implies $t_1 t_2 = -1$.

(B) Let the coordinates of $P, Q, R$ be $(at_i^2, 2at_i)$ for $i = 1, 2, 3$.
Since the ordinates $2at_1, 2at_2, 2at_3$ are in geometric progression,$t_1, t_2, t_3$ are also in geometric progression.
Thus,$t_1 t_3 = t_2^2$.
The equations of the tangents at $P$ and $R$ are $t_1y = x + at_1^2$ and $t_3y = x + at_3^2$.
Solving these,the intersection point $(x, y)$ satisfies $x = at_1t_3 = at_2^2$.
Since the $x$-coordinate of the intersection is $at_2^2$,which is the $x$-coordinate of $Q$,the intersection point lies on a line passing through $Q$ parallel to the $y$-axis.

(B) Let the parabola be $y^2 = 4ax$ with vertex $A(0, 0)$.
Let the double ordinate be at $x = x_1$,so the points are $P(x_1, 2\sqrt{ax_1})$ and $Q(x_1, -2\sqrt{ax_1})$.
For the latus rectum (a specific double ordinate),$x_1 = a$,so $P(a, 2a)$ and $Q(a, -2a)$.
The slope of $AP$ is $m_1 = \frac{2a - 0}{a - 0} = 2$.
However,the question implies the angle subtended by the latus rectum at the vertex.
Slope of $AP = \frac{2a}{a} = 2$ and slope of $AQ = \frac{-2a}{a} = -2$.
The angle $\theta$ between $AP$ and $AQ$ is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{2 - (-2)}{1 + (2)(-2)}| = |\frac{4}{1 - 4}| = |\frac{4}{-3}| = \frac{4}{3}$.
Wait,if the double ordinate is such that the angle is $90^\circ$,then $m_1 m_2 = -1$. Given $m_1 = \frac{y_1}{x_1}$ and $m_2 = -\frac{y_1}{x_1}$,then $-\frac{y_1^2}{x_1^2} = -1 \implies y_1^2 = x_1^2$. Since $y_1^2 = 4ax_1$,we have $x_1^2 = 4ax_1 \implies x_1 = 4a$.
Thus,for the angle to be $90^\circ$,the double ordinate must be at $x = 4a$.

(A) The equation of any tangent to the parabola $y^2 = 4x$ is given by $y = mx + \frac{1}{m}$.
Since the tangent passes through the point $(-2, -1)$,we have:
$-1 = -2m + \frac{1}{m}$
$-m = -2m^2 + 1$
$2m^2 - m - 1 = 0$
Let the slopes of the two tangents be $m_1$ and $m_2$. From the quadratic equation,we have:
$m_1 + m_2 = \frac{1}{2}$
$m_1 m_2 = -\frac{1}{2}$
The angle $\alpha$ between the two tangents is given by:
$\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$:
$(m_1 - m_2)^2 = (\frac{1}{2})^2 - 4(-\frac{1}{2}) = \frac{1}{4} + 2 = \frac{9}{4}$
$|m_1 - m_2| = \frac{3}{2}$
Substituting the values:
$\tan \alpha = \left| \frac{3/2}{1 - 1/2} \right| = \left| \frac{3/2}{1/2} \right| = 3$

(C) The vertex is $V(0, 4)$ and the focus is $S(0, 2)$.
Since both points lie on the $y$-axis,the axis of the parabola is the $y$-axis.
The distance between the vertex and the focus is $a = |4 - 2| = 2$.
Since the focus is below the vertex,the parabola opens downwards.
The standard equation for a parabola with vertex $(h, k)$ opening downwards is $(x - h)^2 = -4a(y - k)$.
Substituting $h = 0, k = 4$,and $a = 2$,we get:
$(x - 0)^2 = -4(2)(y - 4)$
$x^2 = -8(y - 4)$
$x^2 = -8y + 32$
$x^2 + 8y = 32$.

(D) Given equation: $y^2 - 2y = x - 2$.
Completing the square: $(y - 1)^2 - 1 = x - 2$,which simplifies to $(y - 1)^2 = x - 1$.
Let $Y = y - 1$ and $X = x - 1$. The equation becomes $Y^2 = X$.
This is a standard parabola of the form $Y^2 = 4aX$,where $4a = 1$,so $a = 1/4$.
The focus in the $(X, Y)$ coordinate system is $(a, 0) = (1/4, 0)$.
To find the focus in the original $(x, y)$ coordinate system,we use $x = X + 1$ and $y = Y + 1$.
Thus,$x = 1/4 + 1 = 5/4$ and $y = 0 + 1 = 1$.
The focus is $(5/4, 1)$.

(D) The given equation is $y = \frac{a^3x^2}{3} + \frac{a^2x}{2} - 2a$.
To find the vertex,we complete the square for $x$:
$y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a}x \right) - 2a$
$y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a}x + \left(\frac{3}{4a}\right)^2 - \left(\frac{3}{4a}\right)^2 \right) - 2a$
$y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{a^3}{3} \cdot \frac{9}{16a^2} - 2a$
$y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{3a}{16} - 2a$
$y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{35a}{16}$
The vertex $(x, y)$ is given by $x = -\frac{3}{4a}$ and $y = -\frac{35a}{16}$.
To find the locus,we multiply $x$ and $y$:
$xy = \left( -\frac{3}{4a} \right) \left( -\frac{35a}{16} \right) = \frac{105}{64}$.
Thus,the locus is $xy = \frac{105}{64}$.

(C) The given equation of the parabola is $4y^2 + 12x - 20y + 67 = 0$.
Rearranging the terms,we get $4y^2 - 20y = -12x - 67$.
Dividing by $4$,we have $y^2 - 5y = -3x - 67/4$.
Completing the square on the left side: $(y - 5/2)^2 = -3x - 67/4 + 25/4$.
$(y - 5/2)^2 = -3x - 42/4 = -3x - 21/2$.
$(y - 5/2)^2 = -3(x + 7/2)$.
Comparing this with the standard form $Y^2 = 4aX$,where $Y = y - 5/2$,$X = x + 7/2$,and $4a = -3$ (so $a = -3/4$).
The focus of $Y^2 = 4aX$ is $(a, 0) = (-3/4, 0)$.
Thus,$x + 7/2 = -3/4$ and $y - 5/2 = 0$.
$x = -3/4 - 14/4 = -17/4$ and $y = 5/2$.
The focus is $(-17/4, 5/2)$.

(A) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$. Here $a = 1$,so the equation is $y = mx - 2m - m^3$.
Since the normal passes through $(15, 12)$,we have $12 = 15m - 2m - m^3$,which simplifies to $m^3 - 13m + 12 = 0$.
Let the slopes of the three normals be $m_1, m_2, m_3$. From the cubic equation,the sum of the roots is $m_1 + m_2 + m_3 = 0$.
Given the two normals $4x + y = 72$ (slope $m_1 = -4$) and $3x - y = 33$ (slope $m_2 = 3$),we have $-4 + 3 + m_3 = 0$,which gives $m_3 = 1$.
The equation of the third normal with slope $m = 1$ passing through $(15, 12)$ is $y - 12 = 1(x - 15)$,which simplifies to $y = x - 3$.

(D) The equation of any tangent to the parabola $y^{2} = 8ax$ is $y = mx + \frac{2a}{m}$.
For this line to be a tangent to the circle $x^{2} + y^{2} = 2a^{2}$,the perpendicular distance from the center $(0, 0)$ to the line $mx - y + \frac{2a}{m} = 0$ must be equal to the radius $r = a\sqrt{2}$.
Using the distance formula: $\frac{|\frac{2a}{m}|}{\sqrt{m^{2} + 1}} = a\sqrt{2}$.
Squaring both sides: $\frac{4a^{2}}{m^{2}(m^{2} + 1)} = 2a^{2}$.
$\frac{2}{m^{2}(m^{2} + 1)} = 1 \Rightarrow m^{4} + m^{2} - 2 = 0$.
$(m^{2} + 2)(m^{2} - 1) = 0$. Since $m$ must be real,$m^{2} = 1$,so $m = \pm 1$.
Substituting $m = 1$ into the tangent equation: $y = x + \frac{2a}{1} = x + 2a$.
Thus,$y = x + 2a$ is a common tangent.

(A) Let the vertex be $A(4, -1)$ and the focus be $S(4, -3)$.
The axis of the parabola is the line passing through the focus and the vertex,which is $x = 4$.
Since the vertex $A(4, -1)$ is at a distance $a = |-1 - (-3)| = 2$ from the focus $S(4, -3)$,and the parabola opens downwards (as the focus is below the vertex),the equation of the parabola is of the form $(x - h)^{2} = -4a(y - k)$,where $(h, k)$ is the vertex.
Here,$(h, k) = (4, -1)$ and $a = 2$.
Substituting these values,we get:
$(x - 4)^{2} = -4(2)(y - (-1))$
$(x - 4)^{2} = -8(y + 1)$
$x^{2} - 8x + 16 = -8y - 8$
$x^{2} - 8x + 8y + 24 = 0$.

(B) The equation of the parabola is $y^2 = 64x$,so $4a = 64$,which gives $a = 16$.
The equation of the line is $4y = 3x - 48$,which can be written as $y = \frac{3}{4}x - 12$. Here,$m = \frac{3}{4}$ and $c = -12$.
The length of the chord intercepted by the line $y = mx + c$ on the parabola $y^2 = 4ax$ is given by the formula $L = \frac{4}{m^2} \sqrt{a(1+m^2)(a-mc)}$.
Substituting the values: $L = \frac{4}{(3/4)^2} \sqrt{16(1 + (3/4)^2)(16 - (3/4)(-12))}$.
$L = \frac{4}{9/16} \sqrt{16(1 + 9/16)(16 + 9)}$.
$L = \frac{64}{9} \sqrt{16(\frac{25}{16})(25)}$.
$L = \frac{64}{9} \sqrt{25 \times 25} = \frac{64}{9} \times 25 = \frac{1600}{9}$.

(B) Let the vertex be $A(0,0)$. Let $P$ be a point on the parabola $x^2 = 4ay$ such that the chord $AP$ has a slope of $\tan \alpha$. The coordinates of $P$ are $(2at, at^2)$.
The slope of $AP$ is given by $\frac{at^2 - 0}{2at - 0} = \frac{t}{2}$.
Given the slope is $\tan \alpha$,we have $\frac{t}{2} = \tan \alpha$,which implies $t = 2 \tan \alpha$.
The length of the chord $AP$ is $\sqrt{(2at - 0)^2 + (at^2 - 0)^2} = \sqrt{4a^2t^2 + a^2t^4} = at \sqrt{4 + t^2}$.
Substituting $t = 2 \tan \alpha$:
$AP = a(2 \tan \alpha) \sqrt{4 + (2 \tan \alpha)^2} = 2a \tan \alpha \sqrt{4(1 + \tan^2 \alpha)}$.
Since $1 + \tan^2 \alpha = \sec^2 \alpha$,we get:
$AP = 2a \tan \alpha \sqrt{4 \sec^2 \alpha} = 2a \tan \alpha (2 \sec \alpha) = 4a \tan \alpha \sec \alpha$.

(A) Let $P\ (x, y)$ be any point on the parabola with focus $S\ (-1, -2)$ and directrix $x - 2y + 3 = 0$.
By the definition of a parabola,the distance from $P$ to the focus $S$ is equal to the perpendicular distance from $P$ to the directrix.
$SP = PM$
$SP^2 = PM^2$
$(x + 1)^2 + (y + 2)^2 = \left( \frac{|x - 2y + 3|}{\sqrt{1^2 + (-2)^2}} \right)^2$
$(x^2 + 2x + 1) + (y^2 + 4y + 4) = \frac{(x - 2y + 3)^2}{5}$
$5(x^2 + y^2 + 2x + 4y + 5) = x^2 + 4y^2 + 9 - 4xy + 6x - 12y$
$5x^2 + 5y^2 + 10x + 20y + 25 = x^2 + 4y^2 - 4xy + 6x - 12y + 9$
$4x^2 + y^2 + 4xy + 4x + 32y + 16 = 0$

(B) The vertex of the parabola is $V(0, a)$ and the focus is $S(0, 0)$.
Since the focus and vertex lie on the $y$-axis,the axis of the parabola is the $y$-axis.
The distance between the vertex and the focus is $a = |a - 0| = |a|$.
Since the focus is below the vertex,the parabola opens downwards.
The standard form for a downward-opening parabola with vertex $(h, k)$ is $(x - h)^2 = -4a(y - k)$.
Here,$(h, k) = (0, a)$ and the distance from vertex to focus is $a$.
Substituting these values,we get $(x - 0)^2 = -4a(y - a)$.
$x^2 = -4a(y - a)$.
$x^2 = 4a(a - y)$.

(C) Given equation: $4y^2 - 4y = 6x + 5$
Completing the square for $y$: $4(y^2 - y) = 6x + 5$
$4(y^2 - y + \frac{1}{4}) = 6x + 5 + 1$
$4(y - \frac{1}{2})^2 = 6(x + 1)$
$(y - \frac{1}{2})^2 = \frac{6}{4}(x + 1) = \frac{3}{2}(x + 1)$
Comparing with $(Y)^2 = 4aX$,where $Y = y - \frac{1}{2}$,$X = x + 1$,and $4a = \frac{3}{2} \Rightarrow a = \frac{3}{8}$.
The axis is $Y = 0$ $\Rightarrow y - \frac{1}{2} = 0$ $\Rightarrow 2y - 1 = 0$.
The directrix is $X + a = 0$ $\Rightarrow x + 1 + \frac{3}{8} = 0$ $\Rightarrow x + \frac{11}{8} = 0$ $\Rightarrow 8x + 11 = 0$.

(A) The equation of the chord of a parabola $S = 0$ with midpoint $(h, k)$ is given by $T = S_1$.
For the parabola $x^2 + 4y = 0$,the equation of the chord with midpoint $(h, k)$ is $xh + 2(y + k) = h^2 + 4k$.
Since the chord passes through the focus $(0, -1)$,we substitute these coordinates into the equation:
$x(0) + 2(-1 + k) = h^2 + 4k$
$-2 + 2k = h^2 + 4k$
$h^2 + 2k + 2 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + 2y + 2 = 0$.

(B) The equation of the chord passing through the vertex $(0, 0)$ and making an angle $\theta$ with the $x-$axis is $y = x \tan \theta$.
Substituting $y = x \tan \theta$ into the parabola equation $y^2 = 4ax$:
$(x \tan \theta)^2 = 4ax$
$x^2 \tan^2 \theta = 4ax$
$x(x \tan^2 \theta - 4a) = 0$
So,$x = 0$ or $x = \frac{4a}{\tan^2 \theta}$.
For $x = \frac{4a}{\tan^2 \theta}$,$y = \frac{4a}{\tan^2 \theta} \cdot \tan \theta = \frac{4a}{\tan \theta}$.
The intersection points are $(0, 0)$ and $(\frac{4a}{\tan^2 \theta}, \frac{4a}{\tan \theta})$.
The length of the chord $L$ is given by the distance formula:
$L = \sqrt{(\frac{4a}{\tan^2 \theta} - 0)^2 + (\frac{4a}{\tan \theta} - 0)^2}$
$L = \sqrt{\frac{16a^2}{\tan^4 \theta} + \frac{16a^2}{\tan^2 \theta}}$
$L = \frac{4a}{\tan \theta} \sqrt{\frac{1}{\tan^2 \theta} + 1}$
$L = \frac{4a}{\tan \theta} \sqrt{\csc^2 \theta}$
$L = \frac{4a}{\tan \theta} \cdot \csc \theta = 4a \cdot \cot \theta \cdot \csc \theta = 4a \cdot \frac{\cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta} = 4a \cos \theta \csc^2 \theta$.

(C) Step $1$: Identify the point on the parabola.
The parabola is $y^2 = 4ax$. The end of the latus rectum is $P(a, 2a)$.
Step $2$: Find the equations of the tangent and normal.
The slope of the tangent at $P(a, 2a)$ is given by $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2a} = 1$.
The equation of the tangent at $P(a, 2a)$ is $y - 2a = 1(x - a) \Rightarrow y = x + a$.
This tangent meets the $x$-axis $(y=0)$ at $T(-a, 0)$.
The slope of the normal at $P(a, 2a)$ is $-1$ (since the tangent slope is $1$).
The equation of the normal at $P(a, 2a)$ is $y - 2a = -1(x - a) \Rightarrow y = -x + 3a$.
This normal meets the $x$-axis $(y=0)$ at $N(3a, 0)$.
Step $3$: Calculate the area of the triangle $PTN$.
The vertices of the triangle are $P(a, 2a)$,$T(-a, 0)$,and $N(3a, 0)$.
The base $TN$ lies on the $x$-axis,with length $TN = |3a - (-a)| = 4a$.
The height of the triangle is the $y$-coordinate of $P$,which is $2a$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4a \times 2a = 4a^2$.

(B) Given the parabola $y^2 = 4x$.
For the line $y = x$,substituting $y$ in the parabola equation: $x^2 = 4x \implies x(x - 4) = 0$.
Since $x \neq 0$,we have $x = 4$,so $y = 4$. Thus,point $A = (4, 4)$.
For the line $y = -x$,substituting $y$ in the parabola equation: $(-x)^2 = 4x \implies x^2 = 4x \implies x(x - 4) = 0$.
Since $x \neq 0$,we have $x = 4$,so $y = -4$. Thus,point $B = (4, -4)$.
The length of $AB$ is the distance between $(4, 4)$ and $(4, -4)$.
$AB = \sqrt{(4 - 4)^2 + (4 - (-4))^2} = \sqrt{0^2 + 8^2} = 8$.

(A) To find the length of the chord,we first find the points of intersection of the parabola $y = x^2 + 3x$ and the line $y = 5 - x$.
Equating the two expressions for $y$:
$x^2 + 3x = 5 - x$
$x^2 + 4x - 5 = 0$
$(x + 5)(x - 1) = 0$
So,$x = 1$ and $x = -5$.
For $x = 1$,$y = 5 - 1 = 4$. Point is $(1, 4)$.
For $x = -5$,$y = 5 - (-5) = 10$. Point is $(-5, 10)$.
The length of the chord is the distance between $(1, 4)$ and $(-5, 10)$:
$L = \sqrt{(-5 - 1)^2 + (10 - 4)^2}$
$L = \sqrt{(-6)^2 + (6)^2}$
$L = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$.

(D) Given equation: $y^2 + 4y + 4x + 2 = 0$
Completing the square for $y$: $(y^2 + 4y + 4) - 4 + 4x + 2 = 0$
$(y + 2)^2 = -4x + 2$
$(y + 2)^2 = -4(x - \frac{1}{2})$
Comparing this with the standard form $(Y)^2 = -4a(X)$,where $Y = y + 2$,$X = x - \frac{1}{2}$,and $4a = 4 \Rightarrow a = 1$.
The vertex is $(\frac{1}{2}, -2)$.
The equation of the directrix for $Y^2 = -4aX$ is $X = a$.
Substituting the values: $x - \frac{1}{2} = 1$
$x = 1 + \frac{1}{2} = \frac{3}{2}$.

(A) The given equation is $y = x \tan \alpha - \frac{g x^2}{2 u^2 \cos^2 \alpha}$.
Rearranging the terms to the standard form of a parabola $x^2 = -4ay$:
$\frac{g x^2}{2 u^2 \cos^2 \alpha} = x \tan \alpha - y$
$x^2 = \frac{2 u^2 \cos^2 \alpha}{g} (x \tan \alpha - y)$.
This is of the form $(x - h)^2 = -4a(y - k)$.
Comparing the coefficient of $y$,we have $4a = \frac{2 u^2 \cos^2 \alpha}{g}$.
The length of the latus rectum is $4a = \frac{2 u^2 \cos^2 \alpha}{g}$.

(D) The equation of the parabola is $y^2 = \frac{25}{7}x$. Comparing this with $y^2 = 4ax$,we get $4a = \frac{25}{7}$,so $a = \frac{25}{28}$.
Given the system of parallel chords is $4x - y + \lambda = 0$,the slope of these chords is $m = 4$.
The equation of the diameter of a parabola $y^2 = 4ax$ corresponding to chords of slope $m$ is given by $y = \frac{2a}{m}$.
Substituting the values,$y = \frac{2 \times (25/28)}{4} = \frac{50/28}{4} = \frac{50}{112} = \frac{25}{56}$.
Thus,$56y = 25$.

(A) The standard equation of a parabola is given by $x^2 = 4ay$ or $x^2 = -4ay$.
Comparing the given equation $x^2 = -y$ with $x^2 = -4ay$,we get:
$4a = 1$
$a = 1/4$
The length of the latus rectum of a parabola is defined as $|4a|$.
Therefore,the length of the latus rectum $= |4 \times (1/4)| = 1$.

(C) For a parabola $y^2 = 4ax$,the equation of a tangent with slope $m$ is $y = mx + \frac{a}{m}$.
For $y^2 = 4x$,we have $a = 1$,so the tangent is $y = mx + \frac{1}{m}$.
If the point $P(h, k)$ lies on the tangent,then $k = mh + \frac{1}{m}$,which simplifies to $m^2h - mk + 1 = 0$.
Let $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$ be the slopes of the two tangents passing through $P(h, k)$.
Then $m_1$ and $m_2$ are roots of the quadratic equation $m^2h - km + 1 = 0$.
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: $\frac{k/h}{1 - 1/h} = 1$.
$\frac{k/h}{(h-1)/h} = 1 \Rightarrow \frac{k}{h-1} = 1$.
Thus,$k = h - 1$,or $h - k - 1 = 0$.
The locus of $P(h, k)$ is $x - y - 1 = 0$.

(B) Let the two points on the parabola $y^2 = 4ax$ be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.
The equation of the normal at point $t$ is $y = -tx + 2at + at^3$.
If the normals at $t_1$ and $t_2$ meet at a point $(x, y)$ on the parabola,then the point of intersection of the normals is given by $x = 2a + a(t_1^2 + t_1t_2 + t_2^2)$ and $y = -at_1t_2(t_1 + t_2)$.
Since the intersection point lies on the parabola $y^2 = 4ax$,we have $(-at_1t_2(t_1 + t_2))^2 = 4a(2a + a(t_1^2 + t_1t_2 + t_2^2))$.
This simplifies to $a^2t_1^2t_2^2(t_1 + t_2)^2 = 4a^2(2 + t_1^2 + t_1t_2 + t_2^2)$.
For the normals to meet on the parabola,the condition is $t_1t_2 = 2$.
The ordinates of the points are $y_1 = 2at_1$ and $y_2 = 2at_2$.
The product of the ordinates is $y_1y_2 = (2at_1)(2at_2) = 4a^2(t_1t_2)$.
Substituting $t_1t_2 = 2$,we get $y_1y_2 = 4a^2(2) = 8a^2$.

(C) The given parabola is $y^2 = x$. Comparing this with $y^2 = 4ax$,we get $4a = 1$,so $a = 1/4$.
Let the point on the parabola be $P(x_1, y_1)$. The focal distance of a point $P(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $x_1 + a$.
Given $x_1 + 1/4 = 17/4$,we get $x_1 = 16/4 = 4$.
Since $y^2 = x$,$y^2 = 4$,so $y = \pm 2$. Since the point is in the first quadrant,$y_1 = 2$.
Thus,the point is $P(4, 2)$.
The slope of the tangent at $(x_1, y_1)$ is $m_t = \frac{1}{2y_1} = \frac{1}{2(2)} = 1/4$.
The slope of the normal $m_n$ is $-1/m_t = -4$.
The equation of the normal at $(4, 2)$ is $y - 2 = -4(x - 4)$.
$y - 2 = -4x + 16$.
$4x + y = 18$.

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

(D) The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.
Here,$a = 1$,so the normal at $(t_1^2, 2t_1)$ is $y = -t_1x + 2t_1 + t_1^3$.
Since this normal passes through $(t_2^2, 2t_2)$,we substitute these coordinates into the equation:
$2t_2 = -t_1(t_2^2) + 2t_1 + t_1^3$.
Rearranging the terms:
$2(t_2 - t_1) = -t_1(t_2^2 - t_1^2)$.
$2(t_2 - t_1) = -t_1(t_2 - t_1)(t_2 + t_1)$.
Since $t_1 \neq t_2$ for the normal to intersect at a different point,we can divide by $(t_2 - t_1)$:
$2 = -t_1(t_2 + t_1)$.
$2 = -t_1t_2 - t_1^2$.
$t_1t_2 = -(t_1^2 + 2)$.

(A) The given equation of the parabola is $y^2 = 12x$.
Comparing this with the standard form $y^2 = 4ax$,we get $4a = 12$,which implies $a = 3$.
The focus of the parabola $y^2 = 4ax$ is $(a, 0)$,so the focus here is $(3, 0)$.
The focal distance of any point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by the formula $|x_1 + a|$.
Substituting the value of $a = 3$,the focal distance is $|x_1 + 3|$.
Since the point $(x_1, y_1)$ lies on the parabola $y^2 = 12x$,$x_1$ must be greater than or equal to $0$,so $x_1 + 3$ is always positive.
Thus,the focal distance is $x_1 + 3$.

(A) The vertex of the parabola is $V(0, 1)$ and the focus is $F(0, 0)$.
Since the $x$-coordinates are the same,the axis of the parabola is the $y$-axis $(x = 0)$.
The parabola opens downwards because the focus is below the vertex.
The distance $a$ between the vertex and the focus is $|1 - 0| = 1$.
The standard equation of a downward-opening parabola with vertex $(h, k)$ is $(x - h)^2 = -4a(y - k)$.
Substituting $h = 0, k = 1$,and $a = 1$:
$(x - 0)^2 = -4(1)(y - 1)$
$x^2 = -4y + 4$
$x^2 + 4y - 4 = 0$.

(A) Given the line $2x + y - 1 = 0$,we can express $y$ as $y = 1 - 2x$.
Substitute this into the parabola equation $y^2 = 4x$:
$(1 - 2x)^2 = 4x$
$1 - 4x + 4x^2 = 4x$
$4x^2 - 8x + 1 = 0$
Now,calculate the discriminant $D = b^2 - 4ac$ for this quadratic equation:
$D = (-8)^2 - 4(4)(1) = 64 - 16 = 48$.
Since $D > 0$,the quadratic equation has two distinct real roots.
Therefore,the line intersects the parabola at two real and distinct points.