Parabola Questions in English

(C) For the parabola $y^2 = 4ax$,the equation of a normal at point $t$ is $y + tx = 2at + at^3$. Here $a = 1$,so the normal is $y + tx = 2t + t^3$.
If a chord joining $t_1$ and $t_2$ subtends a right angle at the vertex $(0,0)$,then the product of the slopes of the lines joining the vertex to the points is $-1$. The slope of the line joining $(0,0)$ and $(at_1^2, 2at_1)$ is $m_1 = \frac{2}{t_1}$ and $m_2 = \frac{2}{t_2}$.
Thus,$\frac{2}{t_1} \times \frac{2}{t_2} = -1 \Rightarrow t_1t_2 = -4$.
Since the chord is a normal,$t_2 = -t_1 - \frac{2}{t_1}$.
Substituting $t_2 = -\frac{4}{t_1}$,we get $-\frac{4}{t_1} = -t_1 - \frac{2}{t_1}$ $\Rightarrow t_1^2 = 2$ $\Rightarrow t_1 = \sqrt{2}$ (for acute angle).
The slope of the normal is $m = -t_1 = -\sqrt{2}$.
The angle $\theta$ with the $x$-axis is given by $\tan(\pi - \theta) = |m| = \sqrt{2}$.
Therefore,$\tan(\theta) = \frac{1}{\sqrt{2}}$,which implies $\theta = \tan^{-1}(\frac{1}{\sqrt{2}}) = \cot^{-1}(\sqrt{2})$.

(A) Let the chord be $y = mx + c$. The intersection points of the line and the parabola $y^2 = 4ax$ are given by $(mx + c)^2 = 4ax$,which simplifies to $m^2x^2 + (2mc - 4a)x + c^2 = 0$.
Let the points be $A(x_1, y_1)$ and $B(x_2, y_2)$. Then $x_1x_2 = \frac{c^2}{m^2}$ and $y_1y_2 = \frac{4ac}{m}$.
Since the chord subtends a right angle at the vertex $(0,0)$,the product of the slopes of $OA$ and $OB$ is $-1$,so $\frac{y_1}{x_1} \times \frac{y_2}{x_2} = -1$,which implies $y_1y_2 = -x_1x_2$.
Substituting the values,$\frac{4ac}{m} = -\frac{c^2}{m^2}$,which gives $c = -4am$.
The equation of the chord is $y = mx - 4am$.
Let $(h, k)$ be the foot of the perpendicular from the origin to this line. The slope of the perpendicular line is $-\frac{1}{m}$,so $\frac{k}{h} = -\frac{1}{m}$,which gives $m = -\frac{h}{k}$.
Since $(h, k)$ lies on the line $y = mx - 4am$,we have $k = mh - 4am = m(h - 4a)$.
Substituting $m = -\frac{h}{k}$,we get $k = -\frac{h}{k}(h - 4a)$,which simplifies to $k^2 = -h^2 + 4ah$.
Thus,$h^2 + k^2 - 4ah = 0$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - 4ax = 0$.

(C) Let the point of contact be $(x_1, y_1)$.
For the first parabola $y^2 = 4a(x - l_1)$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y_1}$.
For the second parabola $x^2 = 4a(y - l_2)$,differentiating with respect to $x$ gives $2x = 4a \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x_1}{2a}$.
Since the parabolas touch at $(x_1, y_1)$,their slopes must be equal at this point.
Therefore,$\frac{2a}{y_1} = \frac{x_1}{2a}$.
This implies $x_1 y_1 = 4a^2$.
Thus,the locus of the point of contact is $xy = 4a^2$.

(C) The vertex of the parabola is $(h, k) = (2, 2)$.
Since the extremities of the latus rectum are $(-2, 0)$ and $(6, 0)$,the axis of the parabola is the vertical line $x = 2$.
The parabola opens downwards,so its equation is of the form $(x - h)^2 = -4a(y - k)$.
Substituting the vertex $(2, 2)$,we get $(x - 2)^2 = -4a(y - 2)$.
The length of the latus rectum is the distance between $(-2, 0)$ and $(6, 0)$,which is $|6 - (-2)| = 8$.
Thus,$4a = 8$,so $a = 2$.
The equation becomes $(x - 2)^2 = -8(y - 2)$.
Expanding this,we get $x^2 - 4x + 4 = -8y + 16$,which simplifies to $x^2 - 4x + 8y - 12 = 0$.

(D) The equation of the chord of the parabola $y^2 = 4ax$ bisected at $(x_1, y_1)$ is given by $T = S_1$.
Here,$y^2 = x$,so $4a = 1 \Rightarrow a = 1/4$.
The equation of the chord is $y y_1 - 2a(x + x_1) = y_1^2 - 4ax_1$.
Substituting $(x_1, y_1) = (2, 1)$ and $a = 1/4$:
$y(1) - 2(1/4)(x + 2) = 1^2 - 4(1/4)(2)$
$y - 1/2(x + 2) = 1 - 2$
$y - 1/2x - 1 = -1$
$y = 1/2x \Rightarrow x = 2y$.
Substitute $x = 2y$ into $y^2 = x$:
$y^2 = 2y \Rightarrow y(y - 2) = 0$.
So,$y = 0$ or $y = 2$.
If $y = 0$,$x = 0$. Point $A = (0, 0)$.
If $y = 2$,$x = 4$. Point $B = (4, 2)$.
The length of the chord $AB = \sqrt{(4 - 0)^2 + (2 - 0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.

(C) Let the extremities of the focal chord be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$. Since $PQ$ is a focal chord,$t_1t_2 = -1$.
The intersection of tangents at $P$ and $Q$ is $(x_1, y_1) = (at_1t_2, a(t_1 + t_2))$.
Substituting $t_1t_2 = -1$,we get $x_1 = -a$ and $y_1 = a(t_1 + t_2)$.
The intersection of normals at $P$ and $Q$ is $(x_2, y_2) = (a(t_1^2 + t_2^2 + t_1t_2 + 2), -at_1t_2(t_1 + t_2))$.
Substituting $t_1t_2 = -1$,we get $x_2 = a(t_1^2 + t_2^2 + 1)$ and $y_2 = a(t_1 + t_2)$.
Comparing the coordinates,we see that $y_1 = y_2$.

(A) For a parabola,the semi-latus rectum $l$ is the harmonic mean of the focal segments $SP$ and $SQ$.
Thus,$\frac{1}{SP} + \frac{1}{SQ} = \frac{2}{l}$.
Given $SP = 3$ and $SQ = 2$,we have $\frac{1}{3} + \frac{1}{2} = \frac{2}{l}$.
$\frac{2+3}{6} = \frac{2}{l} \implies \frac{5}{6} = \frac{2}{l}$.
$l = \frac{12}{5}$.
The latus rectum $L = 2l = 2 \times \frac{12}{5} = \frac{24}{5}$.

(B) Let the two normals to the parabola $y^2 = 4ax$ be at points $t_1$ and $t_2$. The equation of a normal at point $t$ is $y + tx = 2at + at^3$.
If the normals intersect at $(h, k)$,then $at^3 + (2a - h)t + k = 0$. Let the roots be $t_1, t_2, t_3$.
Given that the normals are at right angles,the product of their slopes $m_1 m_2 = -1$. Since $m = -t$,we have $(-t_1)(-t_2) = -1$,so $t_1 t_2 = -1$.
The equation of the chord joining $t_1$ and $t_2$ is $y(t_1 + t_2) = 2x + 2a t_1 t_2$.
Substituting $t_1 t_2 = -1$,we get $y(t_1 + t_2) = 2x - 2a$,or $2x - 2a - y(t_1 + t_2) = 0$.
For this line to pass through a fixed point independent of $t_1$ and $t_2$,the coefficient of $(t_1 + t_2)$ must be zero,so $y = 0$.
Substituting $y = 0$ into the equation,we get $2x - 2a = 0$,which gives $x = a$.
Thus,the fixed point is $(a, 0)$.

(D) The given equation is $y^2 + 4y - 6x - 2 = 0$.
Completing the square for $y$,we get $(y^2 + 4y + 4) - 4 - 6x - 2 = 0$,which simplifies to $(y + 2)^2 = 6x + 6 = 6(x + 1)$.
Let $Y = y + 2$ and $X = x + 1$. The equation becomes $Y^2 = 6X$.
This is a parabola of the form $Y^2 = 4aX$,where $4a = 6$,so $a = 1.5$.
The locus of the point of intersection of perpendicular tangents to a parabola is its directrix.
The directrix of the parabola $Y^2 = 4aX$ is $X = -a$.
Substituting back,we get $x + 1 = -1.5$,which implies $x + 1 = -3/2$.
Multiplying by $2$,we get $2x + 2 = -3$,or $2x + 5 = 0$.

(C) Let the focal chord make an angle $\theta$ with the $x$-axis. The length of the focal chord is given by $L = 4a \csc^2 \theta$.
The distance $p$ of the focal chord from the vertex $(0,0)$ is the perpendicular distance from the origin to the line passing through the focus $(a,0)$ with inclination $\theta$.
The equation of the focal chord is $(x - a) \sin \theta - y \cos \theta = 0$.
The distance $p$ from the origin $(0,0)$ is $p = |\frac{-a \sin \theta}{\sqrt{\sin^2 \theta + \cos^2 \theta}}| = |a \sin \theta|$.
Thus,$\sin \theta = \frac{p}{a}$,which implies $\csc \theta = \frac{a}{p}$.
Substituting this into the length formula: $L = 4a (\frac{a}{p})^2 = \frac{4a^3}{p^2}$.

(A) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
If this tangent passes through the point $(h, k)$,then $k = mh + \frac{a}{m}$,which simplifies to the quadratic equation $m^2h - km + a = 0$.
Let the slopes of the two tangents be $m$ and $2m$.
From the properties of quadratic equations,the sum of the roots is $m + 2m = 3m = \frac{k}{h}$ and the product of the roots is $m \times 2m = 2m^2 = \frac{a}{h}$.
From the first equation,$m = \frac{k}{3h}$.
Substituting this into the second equation: $2(\frac{k}{3h})^2 = \frac{a}{h}$.
$2(\frac{k^2}{9h^2}) = \frac{a}{h}$.
$2k^2 = 9ah$.
Replacing $(h, k)$ with $(x, y)$,the locus is $2y^2 = 9ax$,or $y^2 = \frac{9}{2}ax$.

(A) Let the parabola be $y^2 = 4ax$. The equation of a normal to the parabola is $y = mx - 2am - am^3$.
If the normal passes through $P(h, k)$,then $k = mh - 2am - am^3$,which simplifies to $am^3 + (2a - h)m + k = 0$.
Let the roots of this cubic equation be $m_1, m_2, m_3$. These represent the slopes of the three normals.
The angles these normals make with the axis of the parabola (the $x$-axis) are $\theta_1, \theta_2, \theta_3$,where $\tan \theta_i = m_i$.
The sum of the angles is $\theta_1 + \theta_2 + \theta_3 = C$ (constant).
Taking the tangent on both sides,$\tan(\theta_1 + \theta_2 + \theta_3) = \tan C$.
Using the identity $\tan(\theta_1 + \theta_2 + \theta_3) = \frac{S_1 - S_3}{1 - S_2}$,where $S_1 = \sum m_i = 0$,$S_2 = \sum m_i m_j = \frac{2a - h}{a}$,and $S_3 = m_1 m_2 m_3 = -\frac{k}{a}$.
Substituting these,$\frac{0 - (-k/a)}{1 - (2a - h)/a} = \tan C$.
This simplifies to $\frac{k/a}{(a - 2a + h)/a} = \tan C$,so $\frac{k}{h - a} = \tan C$.
Thus,$k = (h - a) \tan C$,which represents a straight line passing through $(a, 0)$.

(B) The equation of the normal in parametric form is $y = -tx + 2at + at^3$.
Rearranging,we get $at^3 + (2a - x)t - y = 0$.
This is a cubic equation in $t$,so the sum of the roots $t_1 + t_2 + t_3 = 0$,which implies $t_1 + t_3 = -t_2$ $(i)$.
At $x = 2a$,the equation becomes $at^3 + (2a - 2a)t - y = 0$,so $y = at^3$.
The ordinates at $x = 2a$ are $at_1^3, at_2^3, at_3^3$.
Given these are in $A$.$P$.,we have $2at_2^3 = at_1^3 + at_3^3$,or $2t_2^3 = t_1^3 + t_3^3$.
Using the identity $t_1^3 + t_3^3 = (t_1 + t_3)(t_1^2 + t_3^2 - t_1t_3)$,we substitute $t_1 + t_3 = -t_2$:
$2t_2^3 = -t_2(t_1^2 + t_3^2 - t_1t_3)$.
Dividing by $-t_2$ (assuming $t_2 \neq 0$),we get $-2t_2^2 = t_1^2 + t_3^2 - t_1t_3$.
Since $t_1^2 + t_3^2 = (t_1 + t_3)^2 - 2t_1t_3 = (-t_2)^2 - 2t_1t_3 = t_2^2 - 2t_1t_3$,we have:
$-2t_2^2 = t_2^2 - 2t_1t_3 - t_1t_3 = t_2^2 - 3t_1t_3$.
$-3t_2^2 = -3t_1t_3$,so $t_2^2 = t_1t_3$.
The slope of the normal is $m = -t$,so $\tan \phi = -t$.
Thus,$(-\tan \phi_2)^2 = (-\tan \phi_1)(-\tan \phi_3)$,which simplifies to $\tan^2 \phi_2 = \tan \phi_1 \tan \phi_3$.
Therefore,$\tan \phi_1, \tan \phi_2, \tan \phi_3$ are in $G.P.$

(A) The condition for the line $y = mx + c$ to be a tangent to the parabola $y^2 = 4a(x - h)$ is $c = mh + \frac{a}{m}$.
Here,$m = 2$,$c = -3$,and $h = \frac{1}{3}$.
Substituting these values into the condition:
$-3 = 2(\frac{1}{3}) + \frac{a}{2}$
$-3 = \frac{2}{3} + \frac{a}{2}$
$-3 - \frac{2}{3} = \frac{a}{2}$
$-\frac{11}{3} = \frac{a}{2}$
$a = -\frac{22}{3}$.

(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to the parabola $S: y^2 - 4x = 0$ is given by $SS_1 = T^2$.
Here,$S = y^2 - 4x$,$S_1 = (2)^2 - 4(-1) = 4 + 4 = 8$,and $T = y(2) - 2(x - 1) = 2y - 2x + 2$.
Substituting these into the equation:
$(y^2 - 4x)(8) = (2y - 2x + 2)^2$
$8(y^2 - 4x) = 4(y - x + 1)^2$
$2(y^2 - 4x) = (y - x + 1)^2$
To find the intercept on the line $x = 2$,substitute $x = 2$ into the equation:
$2(y^2 - 8) = (y - 2 + 1)^2$
$2y^2 - 16 = (y - 1)^2$
$2y^2 - 16 = y^2 - 2y + 1$
$y^2 + 2y - 17 = 0$
Let the roots be $y_1$ and $y_2$. Then $y_1 + y_2 = -2$ and $y_1 y_2 = -17$.
The length of the intercept is $|y_1 - y_2| = \sqrt{(y_1 + y_2)^2 - 4y_1 y_2}$.
$|y_1 - y_2| = \sqrt{(-2)^2 - 4(-17)} = \sqrt{4 + 68} = \sqrt{72} = 6 \sqrt{2}$.

(D) Given $f(x) = ax^2 + bx + c$. Since the absolute term is $3$,$c = 3$. Since the leading coefficient is unity,$a = 1$. Thus,$f(x) = x^2 + bx + 3$.
Since the parabola is symmetric about $x = 1$,the vertex is at $x = 1$. The derivative $f'(x) = 2x + b$. Setting $f'(1) = 0$,we get $2(1) + b = 0$,so $b = -2$.
Therefore,the polynomial is $f(x) = x^2 - 2x + 3$.
We can rewrite this as $y = (x - 1)^2 + 2$,or $(x - 1)^2 = 1(y - 2)$.
Comparing this with the standard form of a parabola $(x - h)^2 = 4a(y - k)$,we have $h = 1$,$k = 2$,and $4a = 1$,which means $a = 1/4$.
The focus of the parabola $(x - h)^2 = 4a(y - k)$ is given by $(h, k + a)$.
Substituting the values,the focus is $(1, 2 + 1/4) = (1, 9/4)$.

(D) Let the variable point on the parabola $y^2 = 4ax$ be $P(at^2, 2at)$.
The focus of the parabola is $S(a, 0)$.
The focal radius is the segment $SP$. The midpoint $M(x, y)$ of $SP$ is given by:
$x = \frac{at^2 + a}{2}, y = \frac{2at + 0}{2} = at$.
From $y = at$,we get $t = \frac{y}{a}$.
Substituting $t$ into the equation for $x$:
$x = \frac{a(\frac{y}{a})^2 + a}{2} = \frac{\frac{y^2}{a} + a}{2} = \frac{y^2 + a^2}{2a}$.
$2ax = y^2 + a^2 \implies y^2 = 2ax - a^2 = 2a(x - \frac{a}{2})$.
This is a parabola with vertex at $(\frac{a}{2}, 0)$.
The latus rectum of this new parabola is $4A = 2a$,which is half the latus rectum of the original parabola $(4a)$.
The directrix is $x - \frac{a}{2} = -\frac{a}{2} \implies x = 0$,which is the $y$-axis.
Thus,all the given statements are correct.

(D) Let the point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.
The vertex $A$ is $(0, 0)$.
The equation of the line $PA$ passing through $(0, 0)$ and $(at^2, 2at)$ is $y = \frac{2}{t}x$.
The directrix of the parabola is $x = -a$.
To find $D$,substitute $x = -a$ into the line equation: $y = \frac{2}{t}(-a) = -\frac{2a}{t}$. So,$D = (-a, -\frac{2a}{t})$.
$M$ is the foot of the perpendicular from $P(at^2, 2at)$ to the directrix $x = -a$,so $M = (-a, 2at)$.
The circle with $MD$ as diameter has the equation $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Here,$(x_1, y_1) = (-a, 2at)$ and $(x_2, y_2) = (-a, -\frac{2a}{t})$.
$(x + a)(x + a) + (y - 2at)(y + \frac{2a}{t}) = 0$.
To find the intersection with the $x$-axis,set $y = 0$:
$(x + a)^2 + (-2at)(\frac{2a}{t}) = 0$.
$(x + a)^2 - 4a^2 = 0$.
$(x + a)^2 = 4a^2$.
$x + a = \pm 2a$.
$x = a$ or $x = -3a$.
Thus,the coordinates are $(a, 0)$ and $(-3a, 0)$.

(D) Let the chord be $y = mx + c$. Since it passes through the parabola $y^2 = 4ax$,the combined equation of the lines joining the vertex $(0, 0)$ to the points of intersection is $y^2 = 4ax \left(\frac{y-mx}{c}\right)$.
Since the chord subtends a right angle at the vertex,the sum of the coefficients of $x^2$ and $y^2$ is $0$.
$1 + \frac{4am}{c} = 0 \implies c = -4am$.
Thus,the chord is $y = mx - 4am$,which always passes through the fixed point $(4a, 0)$.
The locus of the feet of the perpendiculars from the vertex to the chord $y = mx - 4am$ is $x^2 + y^2 - 4ax = 0$,which is a circle.
The locus of the midpoints of the chords is $y^2 = 2a(x - 4a)$,which is a parabola.
Therefore,all the given statements are correct.

(B) The slope of the chord joining the points $(3, 0)$ and $(4, 1)$ is given by $m = \frac{1 - 0}{4 - 3} = 1$.
Since the tangent is parallel to this chord,its slope must also be $1$.
Given the parabola $y = (x - 3)^2$,we differentiate with respect to $x$ to find the slope of the tangent at any point $(x, y)$:
$\frac{dy}{dx} = 2(x - 3)$.
Setting the slope equal to $1$:
$2(x - 3) = 1$ $\Rightarrow x - 3 = \frac{1}{2}$ $\Rightarrow x = 3.5 = \frac{7}{2}$.
Substituting $x = \frac{7}{2}$ into the parabola equation to find the $y$-coordinate:
$y = (\frac{7}{2} - 3)^2 = (\frac{1}{2})^2 = \frac{1}{4}$.
The point of tangency is $(\frac{7}{2}, \frac{1}{4})$.
The equation of the tangent line with slope $m = 1$ passing through $(\frac{7}{2}, \frac{1}{4})$ is:
$y - \frac{1}{4} = 1(x - \frac{7}{2})$
$y - \frac{1}{4} = x - \frac{14}{4}$
$x - y = \frac{14}{4} - \frac{1}{4} = \frac{13}{4}$
$4x - 4y = 13$.

(D) Given parabola: $y^2 - 2y - 4x - 7 = 0$
$(y - 1)^2 = 4x + 8 = 4(x + 2)$
Vertex $A = (-2, 1)$ and length of latus rectum $L = 4$.
For the new parabola,the vertex is $A(-2, 1)$,the length of latus rectum is $2L = 8$,and the axis is perpendicular to the axis of the given parabola $(y = 1)$.
Thus,the axis of the new parabola is $x = -2$.
The equation of the new parabola is $(x + 2)^2 = \pm 4(2)(y - 1)$.
Case $1$: $(x + 2)^2 = 8(y - 1) \implies x^2 + 4x + 4 = 8y - 8 \implies x^2 + 4x - 8y + 12 = 0$.
Case $2$: $(x + 2)^2 = -8(y - 1) \implies x^2 + 4x + 4 = -8y + 8 \implies x^2 + 4x + 8y - 4 = 0$.
Both equations are valid,so the correct option is $(D)$.

(D) line touches a parabola if the quadratic equation formed by substituting the line equation into the parabola equation has a discriminant $D = 0$.
For $(A) x^2 + 4y = 0$:
Substitute $y = 1 - x$:
$x^2 + 4(1 - x) = 0 \Rightarrow x^2 - 4x + 4 = 0$.
$D = (-4)^2 - 4(1)(4) = 16 - 16 = 0$. (Touches)
For $(B) x^2 - x + y = 0$:
Substitute $y = 1 - x$:
$x^2 - x + 1 - x = 0 \Rightarrow x^2 - 2x + 1 = 0$.
$D = (-2)^2 - 4(1)(1) = 4 - 4 = 0$. (Touches)
For $(C) 4x^2 - 3x + y = 0$:
Substitute $y = 1 - x$:
$4x^2 - 3x + 1 - x = 0 \Rightarrow 4x^2 - 4x + 1 = 0$.
$D = (-4)^2 - 4(4)(1) = 16 - 16 = 0$. (Touches)
Since all options satisfy the condition,the correct answer is $(D)$.

(A) The equation of a line passing through $(1/2, 2)$ with slope $m$ is $y - 2 = m(x - 1/2)$,which simplifies to $y = mx - m/2 + 2$.
For this line to be tangent to the parabola $y = -x^2/2 + 2$,we substitute $y$ into the parabola equation:
$mx - m/2 + 2 = -x^2/2 + 2$
$x^2/2 + mx - m/2 = 0$
$x^2 + 2mx - m = 0$
For tangency,the discriminant $D = 0$:
$D = (2m)^2 - 4(1)(-m) = 0$
$4m^2 + 4m = 0$
$4m(m + 1) = 0$
So,$m = 0$ or $m = -1$.
If $m = 0$,the line is $y = 2$,which is a tangent to the parabola at $(0, 2)$ but is not a secant to the curve $y = \sqrt{4 - x^2}$ (it is a tangent to the circle at $(0, 2)$).
If $m = -1$,the line is $y - 2 = -1(x - 1/2)$,which simplifies to $y = -x + 5/2$,or $2x + 2y - 5 = 0$. This line intersects the curve $y = \sqrt{4 - x^2}$ at two points,thus it is a secant line.

(D) Given the line $y = 2x - 3$. At $x = 1$,$y = 2(1) - 3 = -1$.
Since the parabola $y = x^2 + px + q$ passes through $(1, -1)$,we have $-1 = 1^2 + p(1) + q$,which simplifies to $p + q = -2$,or $q = -2 - p$.
The vertex of the parabola $y = x^2 + px + q$ is at $x = -p/2$.
The $y$-coordinate of the vertex is $y_v = (-p/2)^2 + p(-p/2) + q = p^2/4 - p^2/2 + q = q - p^2/4$.
The distance of the vertex from the $x$-axis is $|y_v| = |q - p^2/4|$.
Substituting $q = -2 - p$,we get $d(p) = |(-2 - p) - p^2/4| = |- (p^2/4 + p + 2)|$.
To minimize the distance,we analyze $f(p) = p^2/4 + p + 2$.
Setting $f'(p) = p/2 + 1 = 0$,we get $p = -2$.
For $p = -2$,$q = -2 - (-2) = 0$.
The distance is $|0 - (-2)^2/4| = |-1| = 1$.
Thus,both $(B)$ and $(C)$ are correct.

(C) The parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$. The general point on the parabola is $P(2t^2, 4t)$.
The slope of the tangent at $P$ is given by $\frac{dy}{dx} = \frac{4}{2t} = \frac{2}{t}$.
For the minimum distance,the tangent at $P$ must be parallel to the line $x + y + 4 = 0$,which has a slope of $-1$.
Setting $\frac{2}{t} = -1$,we get $t = -2$.
The point $P$ is $(2(-2)^2, 4(-2)) = (8, -8)$.
The distance from point $P(8, -8)$ to the line $x + y + 4 = 0$ is $d = \frac{|8 - 8 + 4|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
The minimum distance between the parabola and its image is twice the distance from the point $P$ to the line of reflection,which is $2 \times (2\sqrt{2}) = 4\sqrt{2}$.

(B) The focus is $(2, 0)$ and the directrix is $x = -2$. The vertex is the midpoint of the focus and the intersection of the axis with the directrix,which is $(0, 0)$. The distance $a$ from the vertex to the focus is $2$. The equation of the parabola is $y^2 = 4ax = 8x$.
The incident beam travels along $y = -4$. Substituting $y = -4$ into $y^2 = 8x$,we get $(-4)^2 = 8x$,so $16 = 8x$,which gives $x = 2$. Thus,point $P$ is $(2, -4)$.
According to the reflection property of a parabola,any light ray traveling parallel to the axis of the parabola will reflect off the parabola and pass through its focus. The axis of this parabola is the $x$-axis $(y = 0)$. The incident ray $y = -4$ is parallel to the $x$-axis.
Therefore,the reflected ray passes through the focus $S(2, 0)$. The reflected ray is a vertical line passing through $P(2, -4)$ and $S(2, 0)$,which is the line $x = 2$.
To find the point $Q$ where the reflected ray hits the parabola again,we substitute $x = 2$ into the parabola equation $y^2 = 8x$:
$y^2 = 8(2) = 16$
$y = \pm 4$.
Since $P$ is $(2, -4)$,the other point $Q$ is $(2, 4)$.

(B) Let the vertex of the parabola $x^2 = 4ay$ be $A(0, 0)$.
Let the other end of the chord be $P(x_1, y_1) = (2at, at^2)$.
The slope of the chord $AP$ is given by $m = \frac{at^2 - 0}{2at - 0} = \frac{t}{2}$.
Given that 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 $t = 2\tan\alpha$:
$AP = a(2\tan\alpha)\sqrt{4 + (2\tan\alpha)^2} = 2a\tan\alpha \sqrt{4(1 + \tan^2\alpha)} = 2a\tan\alpha \sqrt{4\sec^2\alpha} = 2a\tan\alpha (2\sec\alpha) = 4a\tan\alpha \sec\alpha$.

(C) The line equation is $y = \sqrt{3}x - 3$,which can be written as $\frac{y - 0}{x - \sqrt{3}} = \sqrt{3} = \tan(60^{\circ})$.
Using the parametric form of the line passing through $P(\sqrt{3}, 0)$ with angle $\theta = 60^{\circ}$:
$x = \sqrt{3} + r \cos(60^{\circ}) = \sqrt{3} + \frac{r}{2}$
$y = 0 + r \sin(60^{\circ}) = \frac{r\sqrt{3}}{2}$
Substitute these into the parabola equation $2y^2 = 2x + 3$:
$2(\frac{r\sqrt{3}}{2})^2 = 2(\sqrt{3} + \frac{r}{2}) + 3$
$\frac{3r^2}{2} = 2\sqrt{3} + r + 3$
$3r^2 - 2r - (6 + 4\sqrt{3}) = 0$
Let $r_1 = PA$ and $r_2 = -PB$ be the roots of this quadratic equation (since $P$ lies between $A$ and $B$ or on one side,we consider the directed distances).
The sum of roots $r_1 + r_2 = \frac{2}{3}$ and the product $r_1 r_2 = -\frac{6 + 4\sqrt{3}}{3}$.
We need $|PA - PB| = |r_1 - r_2|$.
Using $|r_1 - r_2| = \sqrt{(r_1 + r_2)^2 - 4r_1 r_2}$:
$|r_1 - r_2| = \sqrt{(\frac{2}{3})^2 - 4(-\frac{6 + 4\sqrt{3}}{3})} = \sqrt{\frac{4}{9} + \frac{24 + 16\sqrt{3}}{3}} = \sqrt{\frac{4 + 72 + 48\sqrt{3}}{9}} = \frac{\sqrt{76 + 48\sqrt{3}}}{3}$.

(C) The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$.
The equation of the line is $x + y = 6$,which can be written as $y = -x + 6$. Comparing this with $y = mx + c$,we get $m = -1$.
The condition for the line $y = mx + c$ to be a normal to the parabola $y^2 = 4ax$ is $c = -2am - am^3$.
Substituting the values: $c = -2(2)(-1) - 2(-1)^3 = 4 + 2 = 6$. This confirms the line is a normal.
The point of contact for a normal with slope $m$ is $(am^2, -2am)$.
Substituting $a = 2$ and $m = -1$,we get the point $(2(-1)^2, -2(2)(-1)) = (2, 4)$.

(B) The equation of the parabola is $y^2 = 12x$. Comparing this with $y^2 = 4ax$,we get $4a = 12$,so $a = 3$.
For any point $P(x, y)$ on the parabola,the focal distance is given by $x + a$.
Given the focal distance is $12$,we have $x + 3 = 12$,which implies $x = 9$.
Substituting $x = 9$ into the parabola equation $y^2 = 12x$,we get $y^2 = 12(9) = 108$.
Thus,$y = \pm \sqrt{108} = \pm 6\sqrt{3}$.
Therefore,the point is $(9, 6\sqrt{3})$ or $(9, -6\sqrt{3})$.
Comparing this with the given options,the correct point is $(9, 6\sqrt{3})$.

(C) The equation of the line is $ax + by + c = 0$,which can be rewritten as $y = -\frac{a}{b}x - \frac{c}{b}$.
For a line $y = mx + k$ to be tangent to the parabola $y^2 = 4Ax$ (where $A=a$),the condition is $k = \frac{A}{m}$.
Here,$m = -\frac{a}{b}$ and $k = -\frac{c}{b}$.
Substituting these into the condition: $-\frac{c}{b} = \frac{a}{-\frac{a}{b}}$.
$-\frac{c}{b} = -\frac{ab}{a} = -b$.
Therefore,$c = b^2$ is incorrect based on standard derivation; the correct condition is $c = ab^2/a = b^2$ is not standard. Let us re-evaluate: $k = A/m$ $\Rightarrow -c/b = a/(-a/b)$ $\Rightarrow -c/b = -b$ $\Rightarrow c = b^2$. Wait,the standard form $y^2=4ax$ with line $y=mx+c'$ gives $c'=a/m$. Here $m=-a/b$ and $c'=-c/b$. So $-c/b = a/(-a/b) = -b$. Thus $c = b^2$. The correct option is $c = b^2$.

(A) $1$. The parabola opens downwards,so the coefficient of $x^2$ must be negative. Thus,$a < 0$.
$2$. The $y$-intercept is the value of $y$ when $x = 0$. From the graph,the parabola intersects the $y$-axis below the origin,so $c < 0$.
$3$. The $x$-coordinate of the vertex is given by $x = -(-b) / (2a) = b / (2a)$.
$4$. From the graph,the vertex lies to the right of the $y$-axis,so the $x$-coordinate of the vertex is positive,i.e.,$b / (2a) > 0$.
$5$. Since $a < 0$,for the ratio $b / (2a)$ to be positive,$b$ must also be negative. Thus,$b < 0$.
$6$. Therefore,$a < 0, b < 0, c < 0$.

(C) Let the point $P$ be $(h, k)$. $A$ point $(x, y)$ on the parabola $y^2 = 16x$ is equidistant from $P$ if it lies on a circle centered at $P$ with radius $r$.
Thus,the distance condition is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting $x = \frac{y^2}{16}$ into the circle equation,we get $(\frac{y^2}{16} - h)^2 + (y - k)^2 = r^2$.
This simplifies to a quartic equation in $y$: $\frac{y^4}{256} - \frac{hy^2}{8} + h^2 + y^2 - 2ky + k^2 = r^2$,which is $\frac{y^4}{256} + (1 - \frac{h}{8})y^2 - 2ky + (h^2 + k^2 - r^2) = 0$.
$A$ quartic equation can have at most $4$ real roots.
Therefore,a circle can intersect a parabola at a maximum of $4$ points.

(B) Let $P(h, k)$ be the point from which two tangents are drawn to $y^2 = 4x$.
Any tangent to the parabola $y^2 = 4x$ is given by $y = mx + \frac{1}{m}$.
If it passes through $P(h, k)$,then $k = mh + \frac{1}{m}$,which simplifies to $m^2h - mk + 1 = 0$.
Let $m_1$ and $m_2$ be the roots of this quadratic equation.
From the properties of roots,we have $m_1 + m_2 = \frac{k}{h}$ and $m_1 m_2 = \frac{1}{h}$.
Given that $m_1 = 2m_2$,we substitute this into the equations:
$2m_2 + m_2 = 3m_2 = \frac{k}{h} \Rightarrow m_2 = \frac{k}{3h}$.
$2m_2 \times m_2 = 2m_2^2 = \frac{1}{h} \Rightarrow m_2^2 = \frac{1}{2h}$.
Substituting $m_2$ from the first into the second:
$2(\frac{k}{3h})^2 = \frac{1}{h}$ $\Rightarrow 2(\frac{k^2}{9h^2}) = \frac{1}{h}$ $\Rightarrow \frac{2k^2}{9h} = 1$ $\Rightarrow 2k^2 = 9h$.
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $2y^2 = 9x$.

(D) For the parabola $y^2 = 16x$,we have $4a = 16$,so $a = 4$.
Point $Q(36, -24)$ is on the parabola,so $(36, -24) = (at_2^2, 2at_2)$.
$2at_2 = -24$ $\Rightarrow 2(4)t_2 = -24$ $\Rightarrow t_2 = -3$.
The relation between the parameters of the normal at $t_1$ meeting the parabola at $t_2$ is $t_2 = -t_1 - \frac{2}{t_1}$.
Substituting $t_2 = -3$,we get $-3 = -t_1 - \frac{2}{t_1} \Rightarrow t_1^2 - 3t_1 + 2 = 0$.
Solving for $t_1$,we get $(t_1 - 1)(t_1 - 2) = 0$,so $t_1 = 1$ or $t_1 = 2$.
The focal distance of point $P(at_1^2, 2at_1)$ is $a(1 + t_1^2)$.
For $t_1 = 1$,focal distance $= 4(1 + 1^2) = 8$.
For $t_1 = 2$,focal distance $= 4(1 + 2^2) = 20$.
The maximum focal distance is $20$.

(A) The equation of any normal to the parabola $y^{2}=4ax$ is $y=mx-2am-am^{3}$.
If it passes through $(h, k)$,then $k=mh-2am-am^{3}$,or $am^{3}+(2a-h)m+k=0$.
This is a cubic equation in $m$. Let its roots be $m_{1}, m_{2}, m_{3}$.
Then $m_{1}+m_{2}+m_{3}=0$ and $m_{1}m_{2}+m_{2}m_{3}+m_{3}m_{1}=\frac{2a-h}{a}$.
The coordinates of $P, Q, R$,the feet of the three normals,are $(am_{1}^{2}, -2am_{1}), (am_{2}^{2}, -2am_{2}), (am_{3}^{2}, -2am_{3})$.
Let $G(\bar{x}, \bar{y})$ be the centroid of $\Delta PQR$.
Then $\bar{x} = \frac{a(m_{1}^{2}+m_{2}^{2}+m_{3}^{2})}{3} = \frac{a}{3}[(m_{1}+m_{2}+m_{3})^{2}-2(m_{1}m_{2}+m_{2}m_{3}+m_{3}m_{1})] = \frac{a}{3}[0 - 2(\frac{2a-h}{a})] = \frac{2}{3}(h-2a)$.
And $\bar{y} = \frac{-2a(m_{1}+m_{2}+m_{3})}{3} = -\frac{2a}{3}(0) = 0$.
Thus,the centroid of $\Delta PQR$ is $G(\frac{2}{3}(h-2a), 0)$,which lies on the line $y=0$.

(C) The given equation of the parabola is $y^2=32x$.
Comparing this with the standard form $y^2=4ax$,we get $4a=32$,which implies $a=8$.
The length of a focal chord of a parabola $y^2=4ax$ is given by $L = a(t + \frac{1}{t})^2$,where $t$ is the parameter of one end of the chord.
The minimum length of a focal chord is the length of the latus rectum,which is $4a$.
Therefore,the minimum length $= 4 \times 8 = 32 \text{ units}$.
Thus,the length of $PQ$ can never be less than $32$ units.

(A) By the definition of a parabola,the distance from any point $P(x, y)$ on the parabola to the focus $S(3, -4)$ is equal to the perpendicular distance from $P$ to the directrix $y - 4 = 0$.
$SP = PM$
$\sqrt{(x - 3)^2 + (y + 4)^2} = |y - 4|$
Squaring both sides:
$(x - 3)^2 + (y + 4)^2 = (y - 4)^2$
$(x - 3)^2 + y^2 + 8y + 16 = y^2 - 8y + 16$
$(x - 3)^2 = -8y - 8y$
$(x - 3)^2 = -16y$

(B) The given equation of the parabola is $3y^2 + 4y - 6x + 8 = 0$.
Rearranging the terms: $3(y^2 + \frac{4}{3}y) = 6x - 8$.
Completing the square: $3(y^2 + \frac{4}{3}y + \frac{4}{9}) = 6x - 8 + \frac{4}{3}$.
$3(y + \frac{2}{3})^2 = 6x - \frac{20}{3}$.
$(y + \frac{2}{3})^2 = 2(x - \frac{10}{9})$.
This is of the form $Y^2 = 4AX$,where $Y = y + \frac{2}{3}$,$X = x - \frac{10}{9}$,and $4A = 2 \implies A = \frac{1}{2}$.
The axis of the parabola is $Y = 0$,which implies $y = -\frac{2}{3}$.
Any point on the axis is $(a, -\frac{2}{3})$.
For a point $(X_0, 0)$ on the axis of $Y^2 = 4AX$ to have $3$ real normals,we must have $X_0 > 2A$.
Substituting the values: $x - \frac{10}{9} > 2(\frac{1}{2}) = 1$.
$x > 1 + \frac{10}{9} = \frac{19}{9}$.
Thus,the point is $(a, -\frac{2}{3})$ where $a > \frac{19}{9}$.

(A) The line equation is $y - \sqrt{3}x + 3 = 0$. We can write it in parametric form as $\frac{x - \sqrt{3}}{\cos \theta} = \frac{y - 0}{\sin \theta} = r$.
Comparing with the line equation,the slope is $\tan \theta = \sqrt{3}$,so $\theta = 60^\circ$.
Thus,$\cos \theta = \frac{1}{2}$ and $\sin \theta = \frac{\sqrt{3}}{2}$.
The parametric coordinates are $x = \sqrt{3} + \frac{r}{2}$ and $y = \frac{\sqrt{3}r}{2}$.
Substituting these into the parabola equation $y^2 = -x - 2$:
$(\frac{\sqrt{3}r}{2})^2 = -(\sqrt{3} + \frac{r}{2}) - 2$
$\frac{3r^2}{4} = -\sqrt{3} - \frac{r}{2} - 2$
$3r^2 + 2r + 4(\sqrt{3} + 2) = 0$.
Since $PA \cdot PB = |r_1 r_2|$,by the product of roots formula,$|r_1 r_2| = |\frac{c}{a}| = |\frac{4(\sqrt{3} + 2)}{3}| = \frac{4(\sqrt{3} + 2)}{3}$.

(C) The given equation of the parabola is $(y - 2)^2 = 4(x + 1)$.
Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get the vertex $(h, k) = (-1, 2)$ and $4a = 4$,which implies $a = 1$.
The axis of the parabola is $y - 2 = 0$,which is parallel to the $x$-axis.
According to the optical property of a parabola,any ray of light moving parallel to the axis of the parabola passes through its focus after reflection.
The focus of the parabola is given by $(h + a, k) = (-1 + 1, 2) = (0, 2)$.
Therefore,the reflected ray must pass through the point $(0, 2)$.

(D) The given equation is $2\{(x - 1)^2 + (y - 3)^2\} = (x + y - 1)^2$.
Dividing by $2$,we get $(x - 1)^2 + (y - 3)^2 = \frac{1}{2}(x + y - 1)^2$.
This is in the form $SP^2 = e^2 PM^2$,where $S(1, 3)$ is the focus and $x + y - 1 = 0$ is the directrix.
Here,$e^2 = \frac{1}{2}$,so $e = \frac{1}{\sqrt{2}}$.
The distance from the focus $S(1, 3)$ to the directrix $L: x + y - 1 = 0$ is $d = \frac{|1 + 3 - 1|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}}$.
The length of the latus rectum is $2e^2d$ or $2 \times \text{distance from focus to directrix} \times e^2$ is not correct here; rather,for a parabola $SP = PM$,the definition is $SP^2 = PM^2$.
Wait,the standard form is $SP^2 = PM^2$. Here,$SP^2 = \frac{1}{2} PM^2$. This implies $e^2 = 1/2$,which is an ellipse. However,the question states it is a parabola. Let's re-examine: $2\{(x-1)^2 + (y-3)^2\} = (x+y-1)^2 \implies (x-1)^2 + (y-3)^2 = \frac{(x+y-1)^2}{2} = \left(\frac{x+y-1}{\sqrt{2}}\right)^2$.
This is $SP^2 = PM^2$ where $PM = \frac{|x+y-1|}{\sqrt{2}}$. This is a parabola with focus $S(1, 3)$ and directrix $x+y-1=0$.
The distance $d$ from $S$ to the directrix is $\frac{|1+3-1|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$.
The length of the latus rectum of a parabola is $2d = 2 \times \frac{3}{\sqrt{2}} = 3\sqrt{2}$.

(C) Let the coordinates of point $T$ be $(h, k)$.
The equation of the chord of contact $PQ$ for the parabola $y^2 = 4ax$ from point $(h, k)$ is given by $ky = 2a(x + h)$.
For the given parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.
Thus,the equation of the chord of contact is $ky = 2(2)(x + h)$,which simplifies to $ky = 4(x + h)$.
Since the chord $PQ$ passes through the point $(-2, 3)$,we substitute $x = -2$ and $y = 3$ into the equation:
$3k = 4(-2 + h)$
$3k = 4h - 8$
Replacing $(h, k)$ with $(x, y)$ to find the locus of $T$:
$3y = 4x - 8$
$y = \frac{4}{3}x - \frac{8}{3}$
Comparing this with $y = mx + c$,we get $m = \frac{4}{3}$ and $c = -\frac{8}{3}$.
Therefore,$m + c = \frac{4}{3} - \frac{8}{3} = -\frac{4}{3}$.

(A) The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$.
The focal distance of a point $P(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $x_1 + a$.
Given that the focal distance is $4$,we have $x_1 + 2 = 4$,which implies $x_1 = 2$.
Substituting $x_1 = 2$ into the parabola equation $y^2 = 8x$,we get $y^2 = 8(2) = 16$.
Therefore,$y = \pm 4$.
Thus,the coordinates of the point are $(2, 4)$ and $(2, -4)$.

(D) The ends 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 given by $yy_1 = 2a(x + x_1)$.
For the point $(a, 2a)$:
$y(2a) = 2a(x + a)$
$y = x + a$
$x - y + a = 0$
For the point $(a, -2a)$:
$y(-2a) = 2a(x + a)$
$-y = x + a$
$x + y + a = 0$
Thus,the equations of the tangents are $x - y + a = 0$ and $x + y + a = 0$.

(C) Let the focus of the parabola be $F = (3, 2)$.
The tangents at the endpoints of the latus rectum of a parabola intersect at the point where the axis of the parabola meets the directrix.
This intersection point lies on the line $x + y = 2$.
Since the axis of the parabola passes through the focus $(3, 2)$ and is perpendicular to the directrix,and the intersection point of the tangents lies on the axis,the axis must pass through the point of intersection of the tangents.
However,the problem states the tangents intersect on the line $x + y = 2$. The intersection point of the tangents at the ends of the latus rectum is the point where the axis meets the directrix.
Let the intersection point be $P(h, k)$. Since $P$ lies on $x + y = 2$,we have $h + k = 2$.
The axis passes through the focus $(3, 2)$ and the point $P(h, k)$.
The slope of the axis is $m = \frac{2 - k}{3 - h}$.
Since the directrix is perpendicular to the axis,and the intersection point $P$ lies on the directrix,the line $x + y = 2$ is not necessarily the directrix,but the intersection point lies on it.
Given the standard properties,the axis of the parabola is the line passing through the focus $(3, 2)$ and perpendicular to the directrix.
By testing the options,the line $x - y = 1$ passes through $(3, 2)$ since $3 - 2 = 1$. Thus,$x - y = 1$ is the correct axis.

(B) The equation of the normal to the parabola $y^2 = 4ax$ (where $a=1$) is $y = mx - 2am - am^3$. Substituting $a=1$,we get $y = mx - 2m - m^3$.
Since the normal passes through $(3, 0)$,we have $0 = 3m - 2m - m^3$,which simplifies to $m^3 - m = 0$.
Solving for $m$,we get $m(m-1)(m+1) = 0$,so $m = 0, 1, -1$.
The coordinates of the points are given by $(am^2, -2am)$.
For $m=0$,$P = (0, 0)$.
For $m=1$,$Q = (1, -2)$.
For $m=-1$,$R = (1, 2)$.
Now,calculate the side lengths:
$PQ = \sqrt{(1-0)^2 + (-2-0)^2} = \sqrt{1+4} = \sqrt{5}$.
$PR = \sqrt{(1-0)^2 + (2-0)^2} = \sqrt{1+4} = \sqrt{5}$.
$QR = \sqrt{(1-1)^2 + (2-(-2))^2} = \sqrt{0+16} = 4$.
Since $(QR)^2 = 16$ and $(PQ)^2 + (PR)^2 = 5 + 5 = 10$,we have $(QR)^2 > (PQ)^2 + (PR)^2$.
Therefore,$\Delta PQR$ is an obtuse-angled triangle.

(A) Given points on $y = x^2$ are $(x_i, x_i^2)$.
$x_1 = -1$,so $y_1 = 1$.
$y_3 = 9 \Rightarrow x_3^2 = 9$. Since $x_1, x_2, x_3$ is an increasing $AP$,$x_3$ must be $3$.
Common difference $d = \frac{x_3 - x_1}{3 - 1} = \frac{3 - (-1)}{2} = 2$.
Thus,$x_2 = x_1 + d = -1 + 2 = 1$,so $y_2 = 1^2 = 1$.
The points are $A(-1, 1)$,$B(1, 1)$,and $C(3, 9)$.
The tangents to $y = x^2$ at $(x_i, x_i^2)$ are given by $y - x_i^2 = 2x_i(x - x_i)$,or $y = 2x_ix - x_i^2$.
Tangent at $A(-1, 1): y = -2x - 1$.
Tangent at $B(1, 1): y = 2x - 1$.
Tangent at $C(3, 9): y = 6x - 9$.
Intersection of $A$ and $B$: $-2x - 1 = 2x - 1$ $\Rightarrow 4x = 0$ $\Rightarrow x = 0, y = -1$. Point $P(0, -1)$.
Intersection of $B$ and $C$: $2x - 1 = 6x - 9$ $\Rightarrow 4x = 8$ $\Rightarrow x = 2, y = 3$. Point $Q(2, 3)$.
Intersection of $A$ and $C$: $-2x - 1 = 6x - 9$ $\Rightarrow 8x = 8$ $\Rightarrow x = 1, y = -3$. Point $R(1, -3)$.
Area $\Delta = \frac{1}{2} |0(3 - (-3)) + 2(-3 - (-1)) + 1(-1 - 3)| = \frac{1}{2} |0 - 4 - 4| = \frac{1}{2} |-8| = 4$.