Parabola Questions in English

(D) Let the vertex of the parabola $y^2 = 4x$ be $O(0, 0)$.
Let the other end of the chord passing through the vertex be $P(h, k)$.
Since $P(h, k)$ lies on the parabola,we have $k^2 = 4h$.
The midpoint $(a, b)$ of the chord $OP$ is given by $a = \frac{0+h}{2} = \frac{h}{2}$ and $b = \frac{0+k}{2} = \frac{k}{2}$.
This implies $h = 2a$ and $k = 2b$.
Substituting these into the parabola equation $k^2 = 4h$,we get $(2b)^2 = 4(2a)$.
$4b^2 = 8a$,which simplifies to $b^2 = 2a$ or $2a = b^2$.

(A) For the parabola $y^2 = 4ax$,we have $4a = 2/3$,which implies $a = 1/6$.
Given the system of parallel chords $y + 2x + 1 = 0$,the slope $m$ of these chords is $-2$.
The equation of the diameter corresponding to a system of parallel chords with slope $m$ for the parabola $y^2 = 4ax$ is given by $y = \frac{2a}{m}$.
Substituting the values $a = 1/6$ and $m = -2$:
$y = \frac{2(1/6)}{-2} = \frac{1/3}{-2} = -1/6$.
Thus,the equation of the diameter is $y = -1/6$.

(B) The given line is $x + y - 1 = 0$,which can be written as $y = -x + 1$. Comparing this with the slope-intercept form $y = mx + c$,we have $m = -1$ and $c = 1$.
The condition for the line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$ is $c = \frac{a}{m}$.
Here,the parabola is $y^2 = kx$,so $4a = k$,which implies $a = \frac{k}{4}$.
Substituting the values into the condition $c = \frac{a}{m}$:
$1 = \frac{k/4}{-1}$
$1 = -\frac{k}{4}$
$k = -4$.

(D) The given tangent is $y = x + 2$,which is of the form $y = mx + c$ where $m = 1$ and $c = 2$.
For a parabola $y^2 = 4ax$,the condition for $y = mx + c$ to be a tangent is $c = a/m$.
Here,$4a = 8$,so $a = 2$. Checking the condition: $c = 2/1 = 2$,which is satisfied.
The locus of the intersection of two perpendicular tangents to a parabola is its directrix.
The directrix of the parabola $y^2 = 8x$ is $x = -a$,which is $x = -2$.
We are looking for a point on the line $y = x + 2$ that also lies on the directrix $x = -2$.
Substituting $x = -2$ into the line equation: $y = -2 + 2 = 0$.
Thus,the required point is $(-2, 0)$.

(C) The equation of the normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.
Here,$a = 1$ and the normal passes through the point $(3, 0)$.
Substituting these values into the equation: $0 = 3m - 2(1)m - (1)m^3$.
This simplifies to $0 = m - m^3$,or $m(1 - m^2) = 0$.
Thus,$m = 0$ or $m = \pm 1$.
For $m = 0$,the equation is $y = 0$.
For $m = 1$,the equation is $y = x - 2(1) - 1(1)^3 = x - 3$.
For $m = -1$,the equation is $y = -x - 2(1)(-1) - 1(-1)^3 = -x + 2 + 1 = -x + 3$.
Both $y = x - 3$ and $y = -x + 3$ are valid normals. Given the options,$y = -x + 3$ is provided as option $C$.

(D) The point $(a, 2a)$ on the parabola $y^2 = 4ax$ corresponds to the parameter $t'$ where $2at' = 2a$,which gives $t' = 1$.
The relation between the parameter $t'$ of the point where the normal is drawn and the parameter $t$ of the point where it meets the parabola again is given by the formula $t = -t' - \frac{2}{t'}$.
Substituting $t' = 1$ into the formula:
$t = -1 - \frac{2}{1} = -1 - 2 = -3$.

(B) For any parabola,the latus rectum is a line parallel to the directrix.
Since the directrix is $2x - 3y + 1 = 0$,the equation of the latus rectum is of the form $2x - 3y + k = 0$.
Since the latus rectum passes through the focus $(2, 1)$,we substitute these coordinates into the equation:
$2(2) - 3(1) + k = 0$
$4 - 3 + k = 0$
$1 + k = 0$
$k = -1$
Therefore,the equation of the latus rectum is $2x - 3y - 1 = 0$.

(D) The length of the latus rectum of a parabola is given by $2 \times d$,where $d$ is the perpendicular distance from the focus to the directrix.
The focus is $(x_1, y_1) = (2, 3)$ and the directrix is $Ax + By + C = 0$,where $x - 4y + 3 = 0$.
The perpendicular distance $d$ is given by the formula:
$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
Substituting the values:
$d = \frac{|1(2) - 4(3) + 3|}{\sqrt{1^2 + (-4)^2}} = \frac{|2 - 12 + 3|}{\sqrt{1 + 16}} = \frac{|-7|}{\sqrt{17}} = \frac{7}{\sqrt{17}}$
The length of the latus rectum is $2d = 2 \times \frac{7}{\sqrt{17}} = \frac{14}{\sqrt{17}}$.

(C) Given the equation of the parabola: $x^2 + 4x + 2y = 0$.
Completing the square for $x$: $(x^2 + 4x + 4) = -2y + 4$.
$(x + 2)^2 = -2(y - 2)$.
Comparing this with the standard form $(x - h)^2 = -4a(y - k)$,we get $4a = 2$,so $a = 1/2$.
The vertex is $(h, k) = (-2, 2)$.
The latus rectum is a horizontal line at $y = k - a$.
$y = 2 - 1/2 = 3/2$.
Therefore,$2y = 3$.

(C) Let the midpoint of the focal chord be $(h, k)$.
The equation of a chord of the parabola $y^2 = 4ax$ with midpoint $(h, k)$ is given by $T = S_1$,where $T = yk - 2a(x + h)$ and $S_1 = k^2 - 4ah$.
So,the equation is $yk - 2a(x + h) = k^2 - 4ah$.
Since this chord passes through the focus $(a, 0)$,we substitute $x = a$ and $y = 0$ into the equation:
$0(k) - 2a(a + h) = k^2 - 4ah$
$-2a^2 - 2ah = k^2 - 4ah$
$k^2 = 2ah - 2a^2$
$k^2 = 2a(h - a)$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2a(x - a)$.

(A) For the parabola $y^2 = 4ax$,the point $(2, 4)$ lies on it,so $4^2 = 4a(2) \implies 16 = 8a \implies a = 2$.
Comparing $(2, 4)$ with $(at^2, 2at)$,we get $2t = 4 \implies t_1 = 2$.
The normal at $t_1$ meets the parabola at $t_2 = -t_1 - \frac{2}{t_1}$.
Substituting $t_1 = 2$,we get $t_2 = -2 - \frac{2}{2} = -2 - 1 = -3$.
The coordinates of the point are $(at_2^2, 2at_2) = (2(-3)^2, 2(2)(-3)) = (2 \times 9, -12) = (18, -12)$.

(A) The equation of a chord of the parabola $y^2 = 4ax$ with midpoint $(x_1, y_1)$ is given by $T = S_1$.
Here,$y^2 = 6x$,so $4a = 6$,which implies $a = 3/2$.
The midpoint $(x_1, y_1) = (-1, 1)$.
The equation $T = S_1$ is $yy_1 - 2a(x + x_1) = y_1^2 - 4ax_1$.
Substituting the values: $y(1) - 2(3/2)(x - 1) = (1)^2 - 6(-1)$.
$y - 3(x - 1) = 1 + 6$.
$y - 3x + 3 = 7$.
$y - 3x = 4$.

(A) For the parabola $y^2 = 4x$,we have $a = 1$.
The ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(1, 2)$ and $(1, -2)$.
For $y^2 = 4ax$,the parameter $T$ at $(at^2, 2at)$ is $t$. Here $t^2 = 1$,so $T_1 = \pm 1$.
The relation for the normal at $T_1$ meeting the parabola again at $T_2$ is $T_2 = -T_1 - \frac{2}{T_1}$.
For $T_1 = 1$,$T_2 = -1 - 2 = -3$. The point is $(aT_2^2, 2aT_2) = (9, -6)$.
For $T_1 = -1$,$T_2 = 1 + 2 = 3$. The point is $(aT_2^2, 2aT_2) = (9, 6)$.
The distance $PP' = \sqrt{(9-9)^2 + (6 - (-6))^2} = \sqrt{0^2 + 12^2} = 12$ units.
Both $A$ and $R$ are true,and $R$ explains $A$.

(B) Let $S$ be the focus of the parabola $y^2 = 4ax$.
Let $T$ be the point of intersection of the tangents at $P$ and $Q$.
For a parabola,the distance of the intersection of tangents at two points $P$ and $Q$ from the focus $S$ is given by the geometric mean of the focal distances of $P$ and $Q$.
Thus,$ST^2 = SP \times SQ$.
Given $SP = 4$ and $SQ = 9$.
$ST^2 = 4 \times 9 = 36$.
Therefore,$ST = \sqrt{36} = 6$.

(A) The given equation is $(y - 1)^2 = 4(x + 1)$.
Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get the vertex $(h, k) = (-1, 1)$ and $4a = 4$,so $a = 1$.
The latus rectum of a parabola $Y^2 = 4aX$ has endpoints at $(a, 2a)$ and $(a, -2a)$.
Here,$X = x - h = x + 1$ and $Y = y - k = y - 1$.
Substituting $a = 1$,the endpoints in terms of $(X, Y)$ are $(1, 2)$ and $(1, -2)$.
Now,converting back to $(x, y)$ coordinates:
For $(1, 2)$: $x + 1 = 1 \implies x = 0$ and $y - 1 = 2 \implies y = 3$.
For $(1, -2)$: $x + 1 = 1 \implies x = 0$ and $y - 1 = -2 \implies y = -1$.
Thus,the endpoints are $(0, 3)$ and $(0, -1)$.

(A) The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is $y + tx = 2at + at^3$.
Here,$4a = 4$,so $a = 1$.
Thus,the normal at $(t^2, 2t)$ is $y + tx = 2t + t^3$.
Comparing this with the given line $y = -tx + (2t + t^3)$,we have the slope $-t = 2$,which implies $t = -2$.
Substituting $t = -2$ into the equation $y = 2x + k$,we get $y = -2x + (2(-2) + (-2)^3) = -2x + (-4 - 8) = -2x - 12$.
Wait,the given line is $y = 2x + k$.
Comparing $y = -tx + (2t + t^3)$ with $y = 2x + k$,we get $-t = 2 \implies t = -2$.
Then $k = 2t + t^3 = 2(-2) + (-2)^3 = -4 - 8 = -12$.
Therefore,$k = -12$ and $t = -2$.

(B) The equation of any tangent to the parabola $x^2 = 16y$ is of the form $y = mx - 4m^2$,where $a = 4$ in $x^2 = 4ay$.
If this line is also a tangent to the parabola $y^2 = 2x$ (where $a = 1/2$),the condition for tangency $c = a/m$ must be satisfied.
Here,$c = -4m^2$ and $a = 1/2$,so $-4m^2 = (1/2)/m$.
This simplifies to $m^3 = -1/8$,which gives $m = -1/2$.
Substituting $m = -1/2$ into the tangent equation: $y = (-1/2)x - 4(-1/2)^2$.
$y = -x/2 - 1$.
Multiplying by $2$,we get $2y = -x - 2$,which rearranges to $x + 2y + 2 = 0$.

(B) The equation of a normal to the parabola $y^2 = 4ax$ at point $t$ is given by $y + tx = 2at + at^3$.
Given $y^2 = 12x$,we have $4a = 12$,so $a = 3$.
The equation of the normal becomes $y + tx = 2(3)t + 3t^3$,which simplifies to $y + tx = 6t + 3t^3$.
Comparing this with the given normal $x + y = k$,we have the coefficients of $x$ and $y$ proportional:
$\frac{t}{1} = \frac{1}{1} \implies t = 1$.
Substituting $t = 1$ into the constant term:
$k = 6(1) + 3(1)^3 = 6 + 3 = 9$.

(A) The equation of a normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.
If the normal passes through a point on the axis at a distance of $4a$ from the vertex $(0, 0)$,the point is $(4a, 0)$.
Substituting $(4a, 0)$ into the equation: $0 = m(4a) - 2am - am^3$.
$0 = 2am - am^3$.
Since $a \neq 0$,we have $2m - m^3 = 0$,which implies $m(2 - m^2) = 0$.
Thus,the possible slopes are $m = 0$ and $m^2 = 2$,so $m = 0, \sqrt{2}, -\sqrt{2}$.
The slopes are $-\sqrt{2}, 0, \sqrt{2}$,which form an $A.P.$ with a common difference of $\sqrt{2}$.

(B) The equation of the line is $y = -x + 1$,which is in the form $y = mx + c$,where $m = -1$ and $c = 1$.
For a parabola $y^2 = 4ax$,the condition for the line $y = mx + c$ to be a normal is $c = -2am - am^3$.
Comparing $y^2 = kx$ with $y^2 = 4ax$,we get $4a = k$,so $a = k/4$.
Substituting $m = -1$ and $c = 1$ into the condition:
$1 = -2(k/4)(-1) - (k/4)(-1)^3$
$1 = (k/2) + (k/4)$
$1 = (2k + k) / 4$
$1 = 3k / 4$
$k = 4/3$.

(C) The distance from the vertex to the directrix is $a$.
Given the vertex $(x_1, y_1) = (0, 0)$ and the directrix $Ax + By + C = 0$,the distance $a$ is given by:
$a = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
$a = \frac{|3(0) - 4(0) + 2|}{\sqrt{3^2 + (-4)^2}} = \frac{2}{\sqrt{9 + 16}} = \frac{2}{5}$
The length of the latus rectum is $4a$.
Length $= 4 \times \frac{2}{5} = \frac{8}{5}$.

(A) For the parabola $y^2 = 4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At the point $(16, 8)$,the slope $m_1 = \frac{2}{8} = \frac{1}{4}$.
For the parabola $x^2 = 32y$,differentiating with respect to $x$ gives $2x = 32 \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x}{16}$. At the point $(16, 8)$,the slope $m_2 = \frac{16}{16} = 1$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.
Substituting the values,$\tan \theta = |\frac{1 - 1/4}{1 + (1)(1/4)}| = |\frac{3/4}{5/4}| = \frac{3}{5}$.
Thus,$\theta = \tan^{-1}(3/5)$.

(A) For the parabola $y^2 = 4ax$,the equation of the tangent with slope $m$ is $y = mx + a/m$.
Here,$y^2 = -4x$,so $4a = -4$,which implies $a = -1$.
Substituting $a = -1$ into the tangent equation,we get $y = mx - 1/m$.
Thus,Statement $- 1$ is true.
For the parabola $y^2 = -4x$,the tangent at point $(x_1, y_1)$ is $yy_1 = -2(x + x_1)$.
The $x$-intercept is found by setting $y = 0$,which gives $0 = -2(x + x_1)$,so $x = -x_1$.
Since the parabola $y^2 = -4x$ lies in the region $x \leq 0$,the $x$-coordinate of any point on the parabola is non-positive $(x_1 \leq 0)$.
Therefore,the $x$-intercept $x = -x_1$ is non-negative $(x \geq 0)$.
Statement $- 2$ is true and it explains why the tangent behaves as described.

(D) The equation of the parabola is $y^2 = 4x$.
Given the $y$-coordinates of the vertices are $y_1 = 1$,$y_2 = 2$,and $y_3 = 4$.
Since the points lie on the parabola,their $x$-coordinates are $x = y^2/4$.
Thus,the vertices are $A(1/4, 1)$,$B(1, 2)$,and $C(4, 4)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the values:
$\text{Area} = \frac{1}{2} |\frac{1}{4}(2 - 4) + 1(4 - 1) + 4(1 - 2)|$
$\text{Area} = \frac{1}{2} |\frac{1}{4}(-2) + 1(3) + 4(-1)|$
$\text{Area} = \frac{1}{2} |-\frac{1}{2} + 3 - 4| = \frac{1}{2} |-\frac{3}{2}| = \frac{3}{4}$
Note: The provided options in the prompt were incorrect; the calculated area is $3/4$.

(D) For the parabola $y^2 = 4ax$,the coordinates of a point on the parabola are given by $(at^2, 2at)$.
Comparing $(2a, 2\sqrt{2}a)$ with $(at^2, 2at)$,we get $t^2 = 2$,so $t = \sqrt{2}$.
The length of the normal chord at parameter $t$ is given by the formula $L = 2a|t| (t^2 + 2)$.
Substituting $t = \sqrt{2}$:
$L = 2a(\sqrt{2})((\sqrt{2})^2 + 2)$
$L = 2a(\sqrt{2})(2 + 2)$
$L = 2a(\sqrt{2})(4) = 8\sqrt{2}a$.
Wait,re-evaluating the standard formula for the length of a normal chord $L = 4a \frac{(1+m^2)^{3/2}}{m^2}$ where $m = -t$ is the slope of the normal.
For $t = \sqrt{2}$,the slope of the normal $m = -t = -\sqrt{2}$.
$L = 4a \frac{(1 + (-\sqrt{2})^2)^{3/2}}{(-\sqrt{2})^2} = 4a \frac{(1+2)^{3/2}}{2} = 2a(3)^{3/2} = 2a(3\sqrt{3}) = 6\sqrt{3}a$.

(A) The equation of the parabola is $y^2 = 4ax$. Comparing this with $y^2 = 4x$,we get $a = 1$.
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,$m = 1$.

(C) The axis of the parabola is perpendicular to the directrix $x + y = 5$. Thus,the equation of the axis is of the form $x - y = k$.
Since the axis passes through the focus $(1, 0)$,we have $1 - 0 = k$,so $k = 1$. The equation of the axis is $x - y = 1$.
The intersection of the axis $x - y = 1$ and the directrix $x + y = 5$ gives the point $Z$. Adding the two equations: $2x = 6 \implies x = 3$. Substituting $x=3$ into $x - y = 1$,we get $y = 2$. So,$Z = (3, 2)$.
The vertex $V$ is the midpoint of the focus $S(1, 0)$ and the point $Z(3, 2)$.
$V = \left( \frac{1 + 3}{2}, \frac{0 + 2}{2} \right) = (2, 1)$.

(B) Let the point of contact be $(x_1, y_1)$.
The equation of the tangent to the parabola $y^2 = 4ax$ at $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
Here,$4a = 4$,so $a = 1$. The tangent equation is $yy_1 = 2(x + x_1)$,which simplifies to $2x - yy_1 + 2x_1 = 0$.
The given line is $4x - 7y + 10 = 0$.
Comparing the coefficients of the two equations:
$\frac{2}{4} = \frac{-y_1}{-7} = \frac{2x_1}{10}$
From $\frac{2}{4} = \frac{y_1}{7}$,we get $y_1 = \frac{14}{4} = \frac{7}{2}$.
From $\frac{2}{4} = \frac{2x_1}{10}$,we get $x_1 = \frac{20}{8} = \frac{5}{2}$.
Thus,the point of contact is $\left( \frac{5}{2}, \frac{7}{2} \right)$.

(C) The given equation of the parabola is $x^2 = -12y$.
Comparing this with the standard form $x^2 = -4ay$,we get $4a = 12$,which implies $a = 3$.
The focus of this parabola is $(0, -a) = (0, -3)$.
The latus rectum is a line passing through the focus and perpendicular to the axis of the parabola.
Since the axis of the parabola is the $y$-axis,the latus rectum is a horizontal line given by $y = -a$.
Therefore,the equation of the latus rectum is $y = -3$.

(B) The given equation of the parabola is $x^2 - x - 8y + 19 = 0$.
Completing the square for $x$:
$x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 8y + 19 = 0$
$\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - 8y + 19 = 0$
$\left(x - \frac{1}{2}\right)^2 = 8y - 19 + \frac{1}{4}$
$\left(x - \frac{1}{2}\right)^2 = 8y - \frac{75}{4}$
$\left(x - \frac{1}{2}\right)^2 = 8\left(y - \frac{75}{32}\right)$
Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,the vertex $(h, k)$ is $\left(\frac{1}{2}, \frac{75}{32}\right)$.

(B) For the parabola $y^2 = 4ax$,where $a = 1$,and the external point $(x_1, y_1) = (-2, -1)$.
The length of the chord of contact is given by the formula $L = \frac{1}{a} \sqrt{(y_1^2 - 4ax_1)(y_1^2 + 4a^2)}$.
Substituting the values: $L = \frac{1}{1} \sqrt{((-1)^2 - 4(1)(-2))((-1)^2 + 4(1)^2)}$.
$L = \sqrt{(1 + 8)(1 + 4)} = \sqrt{9 \times 5} = 3\sqrt{5}$.

(B) The area of the triangle formed by the tangents from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ and their chord of contact is given by the formula: $\text{Area} = \frac{(y_1^2 - 4ax_1)^{3/2}}{2a}$.
Given the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.
The point is $(x_1, y_1) = (4, 6)$.
Substituting these values into the formula:
$\text{Area} = \frac{(6^2 - 4(2)(4))^{3/2}}{2(2)}$
$\text{Area} = \frac{(36 - 32)^{3/2}}{4}$
$\text{Area} = \frac{(4)^{3/2}}{4}$
$\text{Area} = \frac{8}{4} = 2$.

(B) The equation of any tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$.
Since the tangent passes through $(1, 4)$,we have $4 = m(1) + \frac{1}{m}$.
Multiplying by $m$,we get $m^2 - 4m + 1 = 0$.
Let the slopes of the two tangents be $m_1$ and $m_2$. Then $m_1 + m_2 = 4$ and $m_1m_2 = 1$.
The difference between the slopes is $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1m_2} = \sqrt{4^2 - 4(1)} = \sqrt{12} = 2\sqrt{3}$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{2\sqrt{3}}{1 + 1} \right| = \sqrt{3}$.
Therefore,$\theta = \frac{\pi}{3}$.

(A) Let the line passing through the origin be $y = mx$.
If this line is a tangent to the parabola $y^2 = 4a(x - a)$,then 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 discriminant of this quadratic equation must be zero:
$D = (-4a)^2 - 4(m^2)(4a^2) = 0$
$16a^2 - 16a^2m^2 = 0$
$16a^2(1 - m^2) = 0$
Since $a \neq 0$,we have $m^2 = 1$,which implies $m = 1$ 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 them is $90^o$.

(C) The given equation of the parabola is $y^2 = kx - 8$,which can be rewritten as $y^2 = k(x - 8/k)$.
Comparing this with the standard form $Y^2 = 4AX$,where $Y = y$ and $X = x - 8/k$,we have $4A = k$,so $A = k/4$.
The directrix of the parabola $Y^2 = 4AX$ is $X + A = 0$.
Substituting the values of $X$ and $A$,we get $(x - 8/k) + k/4 = 0$,which simplifies to $x = 8/k - k/4$.
We are given that the directrix is $x - 1 = 0$,or $x = 1$.
Equating the two expressions for $x$,we have $8/k - k/4 = 1$.
Multiplying by $4k$,we get $32 - k^2 = 4k$,which leads to the quadratic equation $k^2 + 4k - 32 = 0$.
Factoring the quadratic,we get $(k + 8)(k - 4) = 0$.
Thus,the possible values for $k$ are $k = -8$ or $k = 4$.

(B) The equation of the parabola is $y^2 = 4ax$,where $a = 1$.
The chord passes through the vertex $(0, 0)$ and makes an angle $\theta = 45^{\circ}$ with the $x$-axis.
The slope of the chord is $m = \tan(45^{\circ}) = 1$.
The equation of the line passing through $(0, 0)$ with slope $m = 1$ is $y = x$.
Substitute $y = x$ into the parabola equation $y^2 = 4x$:
$x^2 = 4x$
$x^2 - 4x = 0$
$x(x - 4) = 0$
So,$x = 0$ or $x = 4$.
If $x = 0$,then $y = 0$. The point is $(0, 0)$.
If $x = 4$,then $y = 4$. The point is $(4, 4)$.
The length of the chord is the distance between $(0, 0)$ and $(4, 4)$:
$L = \sqrt{(4 - 0)^2 + (4 - 0)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.

(C) Let the equation of the parabola be $S(x, y) = y^2 - 8x = 0$.
To find the position of the point $(2, 5)$,we calculate the value of $S(2, 5)$:
$S(2, 5) = (5)^2 - 8(2) = 25 - 16 = 9$.
Since $S(2, 5) > 0$,the point $(2, 5)$ lies outside the parabola $y^2 = 8x$.

(C) The given parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,so $a = 3$.
The vertex of the parabola is at $(0, 0)$.
The endpoints of the latus rectum are $(2a, a)$ and $(-2a, a)$,which are $(6, 3)$ and $(-6, 3)$.
The area of the triangle formed by the vertex $(0, 0)$ and the points $(6, 3)$ and $(-6, 3)$ is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Here,the base is the length of the latus rectum,which is $|6 - (-6)| = 12$.
The height is the $y$-coordinate of the latus rectum,which is $3$.
Therefore,$\text{Area} = \frac{1}{2} \times 12 \times 3 = 18$ square units.

(D) Given the line equation is $2x - 3y = k$,which can be written as $x = \frac{3y + k}{2}$.
Substituting this into the parabola equation $y^2 = 6x$:
$y^2 = 6 \left( \frac{3y + k}{2} \right)$
$y^2 = 3(3y + k)$
$y^2 - 9y - 3k = 0$.
Since the line is tangent to the parabola,the quadratic equation must have equal roots,meaning the discriminant $D = 0$.
$D = (-9)^2 - 4(1)(-3k) = 0$
$81 + 12k = 0$
$12k = -81$
$k = -\frac{81}{12} = -\frac{27}{4}$.

(D) The point of intersection of the latus rectum and the axis of a parabola is its focus.
Given the equation of the parabola: $y^2 - 4y - 2x - 8 = 0$.
Rewrite the equation by completing the square for $y$:
$(y^2 - 4y + 4) - 4 - 2x - 8 = 0$
$(y - 2)^2 = 2x + 12$
$(y - 2)^2 = 2(x + 6)$
Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get:
$h = -6$,$k = 2$,and $4a = 2 \implies a = 1/2$.
The focus of the parabola is given by $(h + a, k)$.
Focus $= (-6 + 1/2, 2) = (-11/2, 2)$.

(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.
Let the coordinates of the ends of the focal chord be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$.
For a focal chord,the relation between the parameters is $t_1 t_2 = -1$.
Given one end is $(1, -4)$,we have $2at_1 = -4$.
Substituting $a = 4$,we get $2(4)t_1 = -4$,which gives $t_1 = -1/2$.
Since $t_1 t_2 = -1$,we have $t_2 = -1 / t_1 = -1 / (-1/2) = 2$.
The other end is $(at_2^2, 2at_2) = (4(2)^2, 2(4)(2)) = (4 \times 4, 16) = (16, 16)$.

(B) For the parabola $y^2 = 4ax$,we have $a = 1$. The point $(4, 4)$ corresponds to $at^2 = 4$ and $2at = 4$,which gives $t_1 = 2$.
The normal at $t_1$ meets the parabola again at $t_2$,where $t_2 = -t_1 - \frac{2}{t_1}$.
Substituting $t_1 = 2$,we get $t_2 = -2 - \frac{2}{2} = -2 - 1 = -3$.
The coordinates of the point are $(at_2^2, 2at_2) = (1 \times (-3)^2, 2 \times 1 \times (-3)) = (9, -6)$.

(A) The axis of the parabola is the $x$-axis,as it passes through the vertex $(-3, 0)$ and is perpendicular to the directrix $x + 5 = 0$.
Let the focus of the parabola be $S(a, 0)$. The vertex $(-3, 0)$ is the midpoint of the segment joining the focus $S(a, 0)$ and the point $Z(-5, 0)$ on the directrix.
$-3 = \frac{a - 5}{2}$
$-6 = a - 5$
$a = -1$
Thus,the focus is $S(-1, 0)$.
Using the definition of a parabola,the distance from any point $P(x, y)$ on the parabola to the focus equals the distance to the directrix:
$\sqrt{(x + 1)^2 + (y - 0)^2} = |x + 5|$
$(x + 1)^2 + y^2 = (x + 5)^2$
$x^2 + 2x + 1 + y^2 = x^2 + 10x + 25$
$y^2 = 8x + 24$
$y^2 = 8(x + 3)$

(A) The slope of the tangent $m = \tan(45^{\circ}) = 1$.
For the parabola $y^2 = 4Ax$,the slope of the tangent at point $(At^2, 2At)$ is $m = 1/t$.
Comparing $y^2 = ax$ with $y^2 = 4Ax$,we get $4A = a$,so $A = a/4$.
Since $m = 1/t = 1$,we have $t = 1$.
The point of contact is $(At^2, 2At) = (\frac{a}{4}(1)^2, 2(\frac{a}{4})(1)) = (\frac{a}{4}, \frac{a}{2})$.

(D) For the parabola $y^2 = 4ax$,the coordinates of any point on it are given by $(at^2, 2at)$.
If $t_1$ and $t_2$ are the parameters of the endpoints of a focal chord,then the condition is $t_1 t_2 = -1$.
Given $t_1 = t$,the parameter of the other endpoint is $t_2 = -\frac{1}{t}$.
Substituting $t_2 = -\frac{1}{t}$ into the parametric coordinates $(at_2^2, 2at_2)$,we get:
$x = a\left(-\frac{1}{t}\right)^2 = \frac{a}{t^2}$
$y = 2a\left(-\frac{1}{t}\right) = -\frac{2a}{t}$
Thus,the coordinates of the other end are $\left( \frac{a}{t^2}, -\frac{2a}{t} \right)$.

(A) The given equation of the parabola is $(y - 2)^2 = 12(x - 4)$.
Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get $h = 4$,$k = 2$,and $4a = 12$,which implies $a = 3$.
The parametric equations for a parabola of the form $(y - k)^2 = 4a(x - h)$ are given by $x = h + at^2$ and $y = k + 2at$.
Substituting the values $h = 4$,$k = 2$,and $a = 3$:
$x = 4 + 3t^2$
$y = 2 + 2(3)t = 2 + 6t$
Thus,the parametric equations are $x = 4 + 3t^2$ and $y = 2 + 6t$.

(D) The given equation of the parabola is $y^2 + 2y + x = 0$.
Completing the square for $y$,we get $(y^2 + 2y + 1) - 1 + x = 0$.
This simplifies to $(y + 1)^2 = -(x - 1)$.
Comparing this with the standard form $(y - k)^2 = -4a(x - h)$,the vertex $(h, k)$ is $(1, -1)$.
Since the $x$-coordinate is positive and the $y$-coordinate is negative,the vertex $(1, -1)$ lies in the $IV$ quadrant.

(B) The equation of a tangent with slope $m$ to the parabola $y^2 = 4ax$ is $y = mx + \frac{a}{m}$.
Here,$4a = 9$,so $a = \frac{9}{4}$.
The equation of the tangent is $y = mx + \frac{9}{4m}$.
Since the tangent passes through the point $(4, 10)$,we have:
$10 = 4m + \frac{9}{4m}$
Multiplying by $4m$,we get:
$40m = 16m^2 + 9$
$16m^2 - 40m + 9 = 0$
$(4m - 1)(4m - 9) = 0$
Thus,$m = \frac{1}{4}$ or $m = \frac{9}{4}$.

(A) Given the equation of the parabola: $9y^2 - 12y - 16x - 57 = 0$.
Rearranging the terms: $9y^2 - 12y = 16x + 57$.
Completing the square for the $y$ terms: $(3y)^2 - 2(3y)(2) + 2^2 = 16x + 57 + 4$.
$(3y - 2)^2 = 16x + 61$.
$9(y - 2/3)^2 = 16(x + 61/16)$.
$(y - 2/3)^2 = \frac{16}{9}(x + 61/16)$.
The axis of the parabola is given by the linear term inside the square: $3y - 2 = 0$,which simplifies to $3y = 2$.

(D) The equation of the chord of contact from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.
Given the parabola $y^2 = 4x$,we have $a = 1$.
The point is $(x_1, y_1) = (2, 4)$.
Substituting these values into the formula:
$y(4) = 2(1)(x + 2)$
$4y = 2(x + 2)$
Dividing both sides by $2$,we get:
$2y = x + 2$.