Parabola Questions in English

(B) Given the equation $\frac{l}{r} = 2 \sin^2 \frac{\theta}{2}$.
Using the trigonometric identity $2 \sin^2 \frac{\theta}{2} = 1 - \cos \theta$,we get $\frac{l}{r} = 1 - \cos \theta$.
Rearranging,$l = r(1 - \cos \theta) = r - r \cos \theta$.
Since $x = r \cos \theta$ and $r = \sqrt{x^2 + y^2}$,we have $l = \sqrt{x^2 + y^2} - x$.
Thus,$\sqrt{x^2 + y^2} = x + l$.
Squaring both sides,$x^2 + y^2 = (x + l)^2 = x^2 + 2lx + l^2$.
Simplifying,$y^2 = 2lx + l^2 = 2l(x + \frac{l}{2})$.
This is the standard form of a parabola $Y^2 = 4AX$ with vertex at $(-\frac{l}{2}, 0)$.

(B) Let the coordinates of point $Q$ be $(x, y)$ on the parabola $y^2=4x$.
The distance $PQ$ is given by $PQ = \sqrt{(x-2)^2 + (y-0)^2}$.
Substituting $y^2 = 4x$,we get $PQ = \sqrt{(x-2)^2 + 4x} = \sqrt{x^2 - 4x + 4 + 4x} = \sqrt{x^2 + 4}$.
To find the minimum distance,let $f(x) = x^2 + 4$.
Differentiating with respect to $x$,$f'(x) = 2x$.
Setting $f'(x) = 0$,we get $x = 0$.
Since $f''(x) = 2 > 0$,the function has a minimum at $x = 0$.
The minimum distance is $PQ = \sqrt{0^2 + 4} = \sqrt{4} = 2$.

(B) Given the angle $\theta = \sin^{-1}(\frac{3}{5})$,we have $\tan \theta = \frac{3}{4}$. The slope of the line is $m = \frac{3}{4}$.
The equation of the line passing through $(-1, 1)$ with slope $m = \frac{3}{4}$ is $y - 1 = \frac{3}{4}(x + 1)$,which simplifies to $4y - 4 = 3x + 3$,or $4y = 3x + 7$.
Substitute $4y = 3x + 7$ into the curve equation $x^2 = 4y - 9$:
$x^2 = (3x + 7) - 9$
$x^2 - 3x + 2 = 0$
$(x - 1)(x - 2) = 0$
So,$x = 1$ and $x = 2$.
For $x = 1$,$4y = 3(1) + 7 = 10 \Rightarrow y = \frac{5}{2}$. Point $A = (1, \frac{5}{2})$.
For $x = 2$,$4y = 3(2) + 7 = 13 \Rightarrow y = \frac{13}{4}$. Point $B = (2, \frac{13}{4})$.
The distance $|AB| = \sqrt{(2 - 1)^2 + (\frac{13}{4} - \frac{5}{2})^2} = \sqrt{1^2 + (\frac{13-10}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$ units.

(A) The given equation is $x^{2}+2 x y+y^{2}-5 x+5 y-5=0$.
This can be rewritten as $(x+y)^{2} = 5x - 5y + 5$.
Let $X = x+y$ and $Y = x-y+1$.
The equation is of the form $X^{2} = 5Y$,which represents a parabola.
The axis of a parabola of the form $(ax+by+c)^2 = k(bx-ay+d)$ is given by $ax+by+c=0$.
Here,$x+y=0$ is the axis of the parabola.

(C) Given equation of the conic is $x^{2}-6x+4y+1=0$.
Completing the square for the $x$ terms:
$(x^{2}-6x+9)-9+4y+1=0$
$(x-3)^{2}+4y-8=0$
$(x-3)^{2}=-4(y-2)$.
This is a parabola of the form $(x-h)^{2}=-4a(y-k)$,where $(h,k)=(3,2)$ and $4a=4$,so $a=1$.
The focus of this parabola is given by $(h, k-a)$.
Substituting the values,we get $(3, 2-1) = (3,1)$.

(D) Given equation: $y^{2}+4x+4y+k=0$
Complete the square for $y$:
$y^{2}+4y+4-4+4x+k=0$
$(y+2)^{2} = -4x+4-k$
$(y+2)^{2} = -4(x - \frac{4-k}{4})$
Comparing this with the standard form of a parabola $(y-k')^{2} = -4a(x-h')$,we get $4a = 4$.
Therefore,the length of the latus rectum is $4$ units.

(B) Given the equation of the parabola: $y^2 + 6x - 2y + 13 = 0$.
Rearrange the terms to group $y$ variables: $y^2 - 2y = -6x - 13$.
Complete the square for the $y$ terms: $(y^2 - 2y + 1) = -6x - 13 + 1$.
This simplifies to: $(y - 1)^2 = -6x - 12$.
Factor out $-6$ on the right side: $(y - 1)^2 = -6(x + 2)$.
Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex:
Here,$h = -2$ and $k = 1$.
Therefore,the vertex is $(-2, 1)$.

(C) Given the coordinates of the point $P$ are $x = 2t^2 + 4$ and $y = 4t + 6$.
From the second equation,we have $4t = y - 6$,which implies $t = \frac{y - 6}{4}$.
Substituting this value of $t$ into the equation for $x$:
$x = 2\left(\frac{y - 6}{4}\right)^2 + 4$
$x - 4 = 2 \cdot \frac{(y - 6)^2}{16}$
$x - 4 = \frac{(y - 6)^2}{8}$
$(y - 6)^2 = 8(x - 4)$
This equation is of the form $(y - k)^2 = 4a(x - h)$,which represents a parabola.

(A) The equation of the parabola is $y^{2}=4ax$.
Let the coordinates of point $P$ on the parabola be $(at^{2}, 2at)$.
The focus $F$ is $(a, 0)$ and the directrix is $x = -a$.
The foot of the perpendicular $Q$ from $P(at^{2}, 2at)$ onto the directrix $x = -a$ is $(-a, 2at)$.
By the definition of a parabola,the distance $PF$ is equal to the distance $PQ$.
$PQ = \sqrt{(at^{2} - (-a))^{2} + (2at - 2at)^{2}} = \sqrt{(a(t^{2}+1))^{2}} = a(t^{2}+1)$.
$PF = \sqrt{(at^{2}-a)^{2} + (2at-0)^{2}} = \sqrt{a^{2}(t^{2}-1)^{2} + 4a^{2}t^{2}} = \sqrt{a^{2}(t^{4}-2t^{2}+1+4t^{2})} = \sqrt{a^{2}(t^{2}+1)^{2}} = a(t^{2}+1)$.
Since $PQ = PF$,the triangle $\triangle PQF$ is an isosceles triangle with $PQ = PF$.
In an isosceles triangle,the angles opposite to the equal sides are equal,so $\angle PQF = \angle PFQ$.
Therefore,$\tan \angle PQF = \tan \angle PFQ$.
Hence,$\frac{\tan \angle PQF}{\tan \angle PFQ} = 1$.

(D) Let the coordinates of point $B$ be $(at^2, 2at)$. The vertex $A$ is at $(0, 0)$.
The slope of $AB$ is $m_{AB} = \frac{2at - 0}{at^2 - 0} = \frac{2}{t}$.
Since $BC \perp AB$,the slope of $BC$ is $m_{BC} = -\frac{1}{m_{AB}} = -\frac{t}{2}$.
The equation of line $BC$ passing through $B(at^2, 2at)$ with slope $-\frac{t}{2}$ is:
$y - 2at = -\frac{t}{2}(x - at^2)$
To find the point $C$ on the axis of the parabola $(y = 0)$:
$0 - 2at = -\frac{t}{2}(x - at^2)$
$4at = t(x - at^2)$
$4a = x - at^2$
$x = 4a + at^2$
So,the coordinates of $C$ are $(4a + at^2, 0)$.
The projection of $BC$ on the axis is the distance $DC$,where $D$ is the projection of $B$ on the axis,i.e.,$D(at^2, 0)$.
$DC = |x_C - x_D| = |(4a + at^2) - at^2| = 4a$ units.

(A) Let the coordinates of the points be $P_i(at_i^2, 2at_i)$ for $i = 1, 2, 3, 4$.
Since $P_1 P_2$ and $P_3 P_4$ are focal chords,we have $t_1 t_2 = -1$ and $t_3 t_4 = -1$.
The equation of the chord joining $P_i$ and $P_j$ is $(t_i + t_j)y = 2x + 2at_i t_j$.
For $P_1 P_3$,the equation is $(t_1 + t_3)y = 2x + 2at_1 t_3$ ... $(1)$.
For $P_2 P_4$,the equation is $(t_2 + t_4)y = 2x + 2at_2 t_4$ ... $(2)$.
Since $t_2 = -1/t_1$ and $t_4 = -1/t_3$,substituting these into $(2)$ gives $( -1/t_1 - 1/t_3 )y = 2x + 2a(-1/t_1)(-1/t_3)$,which simplifies to $-(t_1 + t_3)y = 2xt_1 t_3 + 2a$.
Solving the system of equations $(1)$ and $(2)$ shows that the intersection point lies on the line $x = -a$,which is the directrix of the parabola.

(B) The vertex of the parabola $y^{2}=4ax$ is at the origin $O(0,0)$.
The line passing through the vertex $O$ making an angle $\alpha$ with the $x$-axis is $y = x \tan \alpha$.
To find the intersection point $P$ of this line and the parabola,substitute $y = x \tan \alpha$ into $y^{2}=4ax$:
$(x \tan \alpha)^{2} = 4ax$
$x^{2} \tan^{2} \alpha = 4ax$
Since $P$ is not the origin,$x \neq 0$,so $x \tan^{2} \alpha = 4a$,which gives $x = 4a \cot^{2} \alpha$.
Then $y = (4a \cot^{2} \alpha) \tan \alpha = 4a \cot \alpha$.
Thus,the coordinates of $P$ are $(4a \cot^{2} \alpha, 4a \cot \alpha)$.
The length of the chord $OP$ is the distance from $(0,0)$ to $(4a \cot^{2} \alpha, 4a \cot \alpha)$:
$OP = \sqrt{(4a \cot^{2} \alpha)^{2} + (4a \cot \alpha)^{2}}$
$OP = \sqrt{16a^{2} \cot^{4} \alpha + 16a^{2} \cot^{2} \alpha}$
$OP = 4a \cot \alpha \sqrt{\cot^{2} \alpha + 1}$
$OP = 4a \cot \alpha \operatorname{cosec} \alpha$ (since $0 < \alpha < 90^{\circ}$,$\cot \alpha > 0$ and $\operatorname{cosec} \alpha > 0$).

(D) Since $PQ$ is a focal chord with $P(at^{2}, 2at)$,the coordinates of $Q$ are $(\frac{a}{t^{2}}, \frac{-2a}{t})$.
Slope of $QR = \frac{2ar - (-2a/t)}{ar^{2} - a/t^{2}} = \frac{2a(r + 1/t)}{a(r - 1/t)(r + 1/t)} = \frac{2}{r - 1/t} = \frac{2t}{rt - 1}$.
Slope of $PK = \frac{2at - 0}{at^{2} - 2a} = \frac{2at}{a(t^{2} - 2)} = \frac{2t}{t^{2} - 2}$.
Since $PK \parallel QR$,their slopes are equal:
$\frac{2t}{rt - 1} = \frac{2t}{t^{2} - 2}$.
Assuming $t \neq 0$,we have $rt - 1 = t^{2} - 2$.
$rt = t^{2} - 1$.
$r = \frac{t^{2} - 1}{t}$.

(C) The area of $\Delta PQR$ is maximized when the distance from $R$ to the line $PQ$ is maximum.
Let $R$ be $(t^{2}, 2t)$. The line $PQ$ passes through $P(4, -4)$ and $Q(9, 6)$.
The slope of $PQ$ is $m = \frac{6 - (-4)}{9 - 4} = \frac{10}{5} = 2$.
The equation of line $PQ$ is $y - 6 = 2(x - 9) \Rightarrow 2x - y - 12 = 0$.
The perpendicular distance from $R(t^{2}, 2t)$ to $2x - y - 12 = 0$ is $d = \frac{|2t^{2} - 2t - 12|}{\sqrt{2^{2} + (-1)^{2}}} = \frac{|2(t^{2} - t - 6)|}{\sqrt{5}} = \frac{2|t - 3||t + 2|}{\sqrt{5}}$.
For $R$ to be on the arc between $P$ and $Q$,the parameter $t$ must lie between $-2$ and $3$.
Let $f(t) = t^{2} - t - 6$. To maximize the distance,we find the critical point of $f(t)$ by setting $f'(t) = 2t - 1 = 0$,which gives $t = \frac{1}{2}$.
At $t = \frac{1}{2}$,the coordinates of $R$ are $\left(\left(\frac{1}{2}\right)^{2}, 2\left(\frac{1}{2}\right)\right) = \left(\frac{1}{4}, 1\right)$.

(B) Given equations are $x+y=4$ $(i)$ and $x-y=2$ $(ii)$.
Solving $(i)$ and $(ii)$,we get $x=3$ and $y=1$.
The line passing through $(3, 1)$ with slope $m = \tan(\tan^{-1}(3/4)) = 3/4$ is given by:
$(y-1) = \frac{3}{4}(x-3) \Rightarrow y = \frac{3x-5}{4}$.
Substituting this into the parabola equation $y^{2}=4(x-3)$:
$\left(\frac{3x-5}{4}\right)^{2} = 4(x-3)$
$\frac{9x^{2}-30x+25}{16} = 4x-12$
$9x^{2}-30x+25 = 64x-192$
$9x^{2}-94x+217 = 0$.
For this quadratic equation,$x_{1}+x_{2} = \frac{94}{9}$ and $x_{1}x_{2} = \frac{217}{9}$.
Then $|x_{1}-x_{2}| = \sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}$
$= \sqrt{\left(\frac{94}{9}\right)^{2} - 4\left(\frac{217}{9}\right)}$
$= \sqrt{\frac{8836}{81} - \frac{868}{9}} = \sqrt{\frac{8836-7812}{81}} = \sqrt{\frac{1024}{81}} = \frac{32}{9}$.

(B) The vertex of the parabola $y^{2}=8x$ is $M(0,0)$.
Let the other point on the parabola be $N(2t^{2}, 4t)$,where $t$ is a parameter.
Let $P(x, y)$ be the mid-point of the chord $MN$.
Using the mid-point formula,we have:
$x = \frac{0 + 2t^{2}}{2} = t^{2}$
$y = \frac{0 + 4t}{2} = 2t$
From the second equation,$t = \frac{y}{2}$.
Substituting this into the first equation:
$x = (\frac{y}{2})^{2}$
$x = \frac{y^{2}}{4}$
$y^{2} = 4x$
Thus,the locus of $P$ is $y^{2}=4x$.

(C) For a parabola $y^2 = 4ax$,the coordinates of any point on it can be represented as $(at^2, 2at)$.
Let the two end points of a focal chord be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.
The slope of the chord $PQ$ is $m = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$.
Since the chord passes through the focus $(a, 0)$,the slope is also $m = \frac{2at_1 - 0}{at_1^2 - a} = \frac{2at_1}{a(t_1^2 - 1)} = \frac{2t_1}{t_1^2 - 1}$.
Equating the slopes: $\frac{2}{t_1 + t_2} = \frac{2t_1}{t_1^2 - 1}$.
$t_1^2 - 1 = t_1(t_1 + t_2) = t_1^2 + t_1 t_2$.
$-1 = t_1 t_2$.
Therefore,$t_1 t_2 = -1$.

(A) The line equation is $y = \sqrt{3}x - 3$. The slope $m = \sqrt{3}$,so $\tan \theta = \sqrt{3} \Rightarrow \theta = 60^\circ$.
Any point on the line at a distance $r$ from $X(\sqrt{3}, 0)$ is given by $x = \sqrt{3} + r \cos 60^\circ = \sqrt{3} + \frac{r}{2}$ and $y = 0 + r \sin 60^\circ = \frac{r\sqrt{3}}{2}$.
Substituting these into the parabola equation $y^2 = x + 2$:
$(\frac{r\sqrt{3}}{2})^2 = (\sqrt{3} + \frac{r}{2}) + 2$
$\frac{3r^2}{4} = \sqrt{3} + 2 + \frac{r}{2}$
$3r^2 - 2r - 4(\sqrt{3} + 2) = 0$.
Let $r_1$ and $r_2$ be the roots of this quadratic equation,which represent the directed distances $XP$ and $XQ$.
The product of the roots $r_1 r_2 = \frac{-4(\sqrt{3} + 2)}{3}$.
Since $XP \cdot XQ = |r_1 r_2|$,we have $XP \cdot XQ = |\frac{-4(\sqrt{3} + 2)}{3}| = \frac{4(2 + \sqrt{3})}{3}$.

(B) The parabola is $y^2 = 12x$,so $4a = 12$,which implies $a = 3$. The focus is $(3, 0)$.
The incident ray passes through the focus $(3, 0)$ with slope $m = \tan(\tan^{-1} \frac{3}{4}) = \frac{3}{4}$.
The equation of the incident ray is $y - 0 = \frac{3}{4}(x - 3)$,or $x = \frac{4y}{3} + 3$.
Substituting this into the parabola equation $y^2 = 12x$:
$y^2 = 12(\frac{4y}{3} + 3) = 16y + 36$.
$y^2 - 16y - 36 = 0$.
$(y - 18)(y + 2) = 0$.
Since the ray is directed into the first quadrant,we take $y = 18$.
$A$ property of the parabola states that any ray passing through the focus reflects parallel to the axis of the parabola.
The axis of the parabola $y^2 = 12x$ is the $x$-axis $(y = 0)$.
Thus,the reflected ray is a horizontal line passing through the point $P$ with $y$-coordinate $18$.
The equation of the reflected ray is $y = 18$.

(D) The distance between the latus rectum and the tangent at the vertex of a parabola is equal to $a$,where $4a$ is the length of the latus rectum.
Given the equations $x+y-8=0$ and $x+y-12=0$.
The distance $a$ between these two parallel lines is given by the formula $d = \frac{|c_1 - c_2|}{\sqrt{A^2 + B^2}}$.
$a = \frac{|-8 - (-12)|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$.
The length of the latus rectum is $4a$.
Length $= 4 \times (2 \sqrt{2}) = 8 \sqrt{2} \text{ units}$.

(D) Let the coordinates of points $A$ and $B$ be $(t_{1}^{2}, 2t_{1})$ and $(t_{2}^{2}, 2t_{2})$ respectively.
The center of the circle with diameter $AB$ is given by $(\frac{t_{1}^{2}+t_{2}^{2}}{2}, t_{1}+t_{2})$.
The axis of the parabola $y^{2}=4x$ is the $x$-axis,which has the equation $y=0$.
Since the circle touches the $x$-axis,the radius of the circle is equal to the absolute value of the $y$-coordinate of the center.
Thus,$r = |t_{1}+t_{2}|$,which implies $t_{1}+t_{2} = \pm r$.
The slope of the line $AB$ is given by $m = \frac{2t_{2}-2t_{1}}{t_{2}^{2}-t_{1}^{2}} = \frac{2(t_{2}-t_{1})}{(t_{2}-t_{1})(t_{2}+t_{1})} = \frac{2}{t_{1}+t_{2}}$.
Substituting $t_{1}+t_{2} = \pm r$,we get the slope $m = \frac{2}{\pm r} = \pm \frac{2}{r}$.
Since $\pm \frac{2}{r}$ is not explicitly listed in the options $A, B,$ or $C$,the correct choice is $D$.

(A, B) Given,the equation of the parabola is $x^{2}=ay$,which implies $y=\frac{x^{2}}{a}$.
The equation of the line is $y=2x+1$.
Substituting $y$ in the parabola equation: $\frac{x^{2}}{a}=2x+1 \Rightarrow x^{2}-2ax-a=0$.
Let the roots be $x_{1}$ and $x_{2}$. Then $x_{1}+x_{2}=2a$ and $x_{1}x_{2}=-a$.
The difference $|x_{1}-x_{2}| = \sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = \sqrt{4a^{2}+4a} = 2\sqrt{a^{2}+a}$.
Since the points lie on $y=2x+1$,the distance $d$ between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.
Since $y_{2}-y_{1} = 2(x_{2}-x_{1})$,we have $d = \sqrt{(x_{2}-x_{1})^{2}+4(x_{2}-x_{1})^{2}} = |x_{2}-x_{1}|\sqrt{5}$.
Given $d=\sqrt{40}$,so $\sqrt{40} = 2\sqrt{a^{2}+a} \cdot \sqrt{5} \Rightarrow \sqrt{40} = \sqrt{20(a^{2}+a)}$.
Squaring both sides: $40 = 20(a^{2}+a) \Rightarrow a^{2}+a-2=0$.
$(a+2)(a-1)=0$,so $a=1$ or $a=-2$.

(A) Since $\triangle OAB$ is an equilateral triangle and $O$ is the vertex $(0,0)$,the axis of the parabola (the $x$-axis) bisects the angle $\angle AOB$.
Thus,the angle that $OA$ makes with the $x$-axis is $30^{\circ}$.
The slope of $OA$ is $m = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
The equation of line $OA$ is $y = \frac{1}{\sqrt{3}}x$.
Substituting $y = \frac{x}{\sqrt{3}}$ into the parabola equation $y^2 = 4ax$:
$\left(\frac{x}{\sqrt{3}}\right)^2 = 4ax \Rightarrow \frac{x^2}{3} = 4ax$.
Since $x \neq 0$ for point $A$,we have $x = 12a$.
Then $y = \frac{12a}{\sqrt{3}} = 4\sqrt{3}a$.
Point $A$ is $(12a, 4\sqrt{3}a)$.
Since $AB$ is perpendicular to the $x$-axis,the length $AB = 2y_A = 2(4\sqrt{3}a) = 8\sqrt{3}a$.
Since $\triangle OAB$ is equilateral,the side length is $8\sqrt{3}a$.

(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $yt = x + at^2$.
Here,$4a = 9$,so $a = \frac{9}{4}$.
The tangent passes through $(4, 10)$,so $10t = 4 + \frac{9}{4}t^2$.
Multiplying by $4$,we get $40t = 16 + 9t^2$,or $9t^2 - 40t + 16 = 0$.
Factoring the quadratic: $(9t - 4)(t - 4) = 0$,which gives $t = 4$ or $t = \frac{4}{9}$.
The slope of the tangent is $m = \frac{1}{t} = \tan \theta$.
Given $\tan \theta > 2$,we have $\frac{1}{t} > 2$,which implies $t < \frac{1}{2}$.
Thus,$t = \frac{4}{9}$ is the valid parameter.
The point of contact is $(at^2, 2at) = \left(\frac{9}{4} \times \left(\frac{4}{9}\right)^2, 2 \times \frac{9}{4} \times \frac{4}{9}\right) = \left(\frac{4}{9}, 2\right)$.

(A) The coordinates of point $P$ are $(at^2, 2at)$.
The equation of the tangent at $P$ is $ty = x + at^2$. Setting $y=0$,we get $x = -at^2$,so $T = (-at^2, 0)$.
The equation of the normal at $P$ is $y = -tx + 2at + at^3$. Setting $y=0$,we get $x = 2a + at^2$,so $G = (2a + at^2, 0)$.
Since the tangent and normal are perpendicular,$\angle PTG = 90^\circ$,which implies that $TG$ is the diameter of the circle passing through $P, T,$ and $G$.
The length of the diameter $TG = |(2a + at^2) - (-at^2)| = |2a + 2at^2| = 2a(1+t^2)$.
Therefore,the radius of the circle is $\frac{1}{2} TG = a(1+t^2)$.

(C) The equation of the parabola is $y = ax^2 + bx + c$.
Since the point $(1,1)$ lies on the parabola,we have $1 = a(1)^2 + b(1) + c$,which gives $a + b + c = 1$ ...$(1)$.
Since the point $(-1,0)$ lies on the parabola,we have $0 = a(-1)^2 + b(-1) + c$,which gives $a - b + c = 0$ ...$(2)$.
Subtracting $(2)$ from $(1)$,we get $2b = 1$,so $b = \frac{1}{2}$.
Substituting $b = \frac{1}{2}$ into $(1)$,we get $a + c = \frac{1}{2}$ ...$(3)$.
The slope of the tangent to the parabola at any point $(x,y)$ is given by $\frac{dy}{dx} = 2ax + b$.
At the point $(1,1)$,the slope of the tangent is $2a(1) + b = 2a + b$.
Since the line $y=x$ is tangent at $(1,1)$,its slope is $1$. Thus,$2a + b = 1$.
Substituting $b = \frac{1}{2}$,we get $2a + \frac{1}{2} = 1$,which implies $2a = \frac{1}{2}$,so $a = \frac{1}{4}$.
From $(3)$,$c = \frac{1}{2} - a = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.
Thus,$a = \frac{1}{4}, b = \frac{1}{2}, c = \frac{1}{4}$.

(B) The given equation is $y^{2}-4y=4x-4a$.
Completing the square for $y$,we get $(y-2)^{2}-4=4x-4a$.
This simplifies to $(y-2)^{2}=4x-4a+4$,or $(y-2)^{2}=4(x-(a-1))$.
Comparing this with the standard form $(y-k)^{2}=4A(x-h)$,the vertex is $(h, k) = (a-1, 2)$.
The vertex lies between the lines $L_1: x+y-3=0$ and $L_2: 2x+2y-1=0$.
For a point $(x_0, y_0)$ to lie between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$,the expressions $(ax_0+by_0+c_1)$ and $(ax_0+by_0+c_2)$ must have opposite signs,i.e.,$(ax_0+by_0+c_1)(ax_0+by_0+c_2) < 0$.
Substituting the vertex $(a-1, 2)$ into the lines:
$L_1(a-1, 2) = (a-1)+2-3 = a-2$.
$L_2(a-1, 2) = 2(a-1)+2(2)-1 = 2a-2+4-1 = 2a+1$.
Thus,$(a-2)(2a+1) < 0$.
Solving this inequality,we find $a \in \left(-\frac{1}{2}, 2\right)$.

(B) The equation of the parabola is $y^{2}=12x$. Comparing with $y^{2}=4ax$,we get $4a=12$,so $a=3$.
The slope of the given line $y=4x+3$ is $m=4$.
The slope of the normal to the parabola $y^{2}=4ax$ at point $(x_{1}, y_{1})$ is given by $m_{n} = -\frac{y_{1}}{2a}$.
Since the normal is parallel to the line $y=4x+3$,its slope must be $4$. Thus,$-\frac{y_{1}}{2(3)} = 4$,which gives $y_{1} = -24$.
Substituting $y_{1} = -24$ into the parabola equation $y^{2}=12x$,we get $(-24)^{2} = 12x$,so $576 = 12x$,which gives $x_{1} = 48$.
The point on the parabola is $(48, -24)$.
The equation of the normal at $(48, -24)$ is $y - (-24) = 4(x - 48)$,which simplifies to $y + 24 = 4x - 192$,or $4x - y - 216 = 0$.
The given line is $4x - y + 3 = 0$.
The distance between two parallel lines $Ax + By + C_{1} = 0$ and $Ax + By + C_{2} = 0$ is given by $d = \frac{|C_{1} - C_{2}|}{\sqrt{A^{2} + B^{2}}}$.
Here,$d = \frac{|3 - (-216)|}{\sqrt{4^{2} + (-1)^{2}}} = \frac{|3 + 216|}{\sqrt{16 + 1}} = \frac{219}{\sqrt{17}}$.

(A) The given equation of the parabola is $y^{2} = 64x$ $(i)$.
The point on the parabola nearest to the line $4x + 3y + 35 = 0$ is the point where the tangent is parallel to the given line.
The slope of the line $4x + 3y + 35 = 0$ is $m = -\frac{4}{3}$.
Differentiating $(i)$ with respect to $x$,we get $2y \frac{dy}{dx} = 64$,which implies $\frac{dy}{dx} = \frac{32}{y}$.
Since the tangent is parallel to the line,their slopes must be equal:
$\frac{32}{y} = -\frac{4}{3} \Rightarrow y = -24$.
Substituting $y = -24$ into $(i)$:
$(-24)^{2} = 64x$ $\Rightarrow 576 = 64x$ $\Rightarrow x = 9$.
Thus,the required point is $(9, -24)$.

(A) Given that $\alpha$ and $\beta$ are the roots of $f(x) = x^{2} + bx + c = 0$.
By the properties of quadratic equations,the sum of roots is $\alpha + \beta = -b$.
The vertex of the parabola $y = f(x)$ occurs at $x = -\frac{b}{2} = \frac{\alpha + \beta}{2}$.
Now,find the derivative of $f(x)$:
$f'(x) = \frac{dy}{dx} = 2x + b$.
At the point $x = \frac{\alpha + \beta}{2} = -\frac{b}{2}$,the slope of the tangent is:
$f'\left(-\frac{b}{2}\right) = 2\left(-\frac{b}{2}\right) + b = -b + b = 0$.
$A$ slope of $0$ indicates that the tangent line is horizontal,meaning it is parallel to the $x$-axis.
Therefore,the angle between the tangent and the positive direction of the $x$-axis is $0^{\circ}$.

(A) Given parabolas are $y = x^{2}$ and $y = -(x-2)^{2}$.
Let the tangent to $y = x^{2}$ be $y = mx - \frac{m^{2}}{4}$.
This line is also a tangent to $y = -(x-2)^{2}$,which can be written as $(y-0) = -1(x-2)^{2}$.
The condition for a line $y = mx + c$ to be a tangent to $y = a(x-h)^{2} + k$ is $c = k - \frac{m^{2}}{4a}$.
Here,$a = -1, h = 2, k = 0$. So,$c = 0 - \frac{m^{2}}{4(-1)} = \frac{m^{2}}{4}$.
Equating the two expressions for $c$:
$-\frac{m^{2}}{4} = \frac{m^{2}}{4}$ $\Rightarrow \frac{m^{2}}{2} = 0$ $\Rightarrow m = 0$.
For $m = 0$,the tangent is $y = 0$.
Since there is only one value of $m$,there is only $1$ common tangent.

(C) The equations of the parabolas are $y^2=4ax$ and $x^2=4by$.
The equation of the normal to $y^2=4ax$ with slope $m$ is $y=mx-2am-am^3$.
The equation of the normal to $x^2=4by$ with slope $m$ is $y=mx+2b+\frac{b}{m^2}$.
For a common normal,the equations must be identical,so $-2am-am^3 = 2b+\frac{b}{m^2}$.
Multiplying by $m^2$,we get $-2am^3-am^5 = 2bm^2+b$.
Rearranging the terms,we obtain $am^5+2am^3+2bm^2+b=0$.
Since this is a polynomial equation of degree $5$ in $m$,there are at most $5$ real roots for $m$.
Therefore,the maximum number of common normals is $5$.

(B) The equation of the parabola is $y^{2} = x$,which is of the form $y^{2} = 4ax$,where $4a = 1$,so $a = \frac{1}{4}$.
For a parabola $y^{2} = 4ax$,the condition for three distinct normals to be drawn from a point $(h, k)$ is that $h > 2a$.
Here,the point is $(d, 0)$,so $h = d$.
Substituting the values,we get $d > 2 \times \frac{1}{4}$.
Therefore,$d > \frac{1}{2}$.

(D) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.
The directrix of the parabola $y^2 = 4x$ is given by $x = -a$,which is $x = -1$.
The given point is $(-1, -6)$.
Since the $x$-coordinate of the point is $-1$,the point $(-1, -6)$ lies on the directrix of the parabola.
It is a standard property of parabolas that the tangents drawn from any point on the directrix to the parabola are perpendicular to each other.
Therefore,the angle between the two tangents is $\pi / 2$.

(C) Let the coordinates of points $P$ and $Q$ on the parabola $y^{2}=4x$ be $(t^{2}, 2t)$ and $(m^{2}, 2m)$ respectively. The vertex of the parabola is $X(0, 0)$.
Since $PQ$ subtends a right angle at the vertex $X$,the product of the slopes of $XP$ and $XQ$ is $-1$.
Slope of $XP = \frac{2t-0}{t^{2}-0} = \frac{2}{t}$.
Slope of $XQ = \frac{2m-0}{m^{2}-0} = \frac{2}{m}$.
Given,$(\frac{2}{t}) \times (\frac{2}{m}) = -1 \Rightarrow tm = -4$.
The equation of the line $PQ$ passing through $(t^{2}, 2t)$ and $(m^{2}, 2m)$ is:
$y - 2t = \frac{2m-2t}{m^{2}-t^{2}}(x - t^{2})$
$y - 2t = \frac{2(m-t)}{(m-t)(m+t)}(x - t^{2})$
$y - 2t = \frac{2}{m+t}(x - t^{2})$.
Since $PQ$ intersects the axis of the parabola ($x$-axis) at $R(\alpha, 0)$,we substitute $y=0$ and $x=\alpha$:
$0 - 2t = \frac{2}{m+t}(\alpha - t^{2})$
$-t(m+t) = \alpha - t^{2}$
$-tm - t^{2} = \alpha - t^{2}$
$\alpha = -tm$.
Since $tm = -4$,we have $\alpha = -(-4) = 4$.
Thus,the distance of the vertex $X(0, 0)$ from $R(4, 0)$ is $4$.

(D) The vertex of the parabola $x^2=8y$ is $O(0, 0)$.
Let the coordinates of point $Q$ be $(x_1, y_1)$. Since $Q$ lies on the parabola,$x_1^2 = 8y_1$.
Let the coordinates of point $P$ be $(h, k)$.
Since $P$ divides $OQ$ in the ratio $1:3$,by the section formula:
$h = \frac{1 \cdot x_1 + 3 \cdot 0}{1+3} = \frac{x_1}{4} \Rightarrow x_1 = 4h$
$k = \frac{1 \cdot y_1 + 3 \cdot 0}{1+3} = \frac{y_1}{4} \Rightarrow y_1 = 4k$
Substituting these into the parabola equation $x_1^2 = 8y_1$:
$(4h)^2 = 8(4k)$
$16h^2 = 32k$
$h^2 = 2k$
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $x^2 = 2y$.

(D) Let the coordinates of point $Q$ on the parabola $(y - 6)^2 = 2(x - 4)$ be $(4 + \frac{t^2}{2}, 6 + t)$.
Given $P = (2, 0)$.
Let the mid-point of $PQ$ be $R(h, k)$.
Then $h = \frac{2 + 4 + \frac{t^2}{2}}{2} = 3 + \frac{t^2}{4}$ and $k = \frac{0 + 6 + t}{2} = 3 + \frac{t}{2}$.
From the second equation,$\frac{t}{2} = k - 3$,so $t = 2(k - 3)$.
Substituting $t$ into the equation for $h$:
$h = 3 + \frac{(2(k - 3))^2}{4} = 3 + \frac{4(k - 3)^2}{4} = 3 + (k - 3)^2$.
$h - 3 = (k - 3)^2$.
Replacing $(h, k)$ with $(x, y)$,we get $(y - 3)^2 = x - 3$.
$y^2 - 6y + 9 = x - 3$.
$y^2 - 6y - x + 12 = 0$.

(A) The given equation is $6y = 2a^3x^2 + 3a^2x - 12a$.
Dividing by $2a^3$ (assuming $a \neq 0$),we get $x^2 + \frac{3}{2a}x = \frac{6y + 12a}{2a^3}$.
Completing the square for $x$: $(x + \frac{3}{4a})^2 = \frac{6y + 12a}{2a^3} + \frac{9}{16a^2} = \frac{48y + 96a + 9a}{16a^3} = \frac{48y + 105a}{16a^3}$.
Thus,$(x + \frac{3}{4a})^2 = \frac{3}{a^2}(y + \frac{35a}{16})$.
The vertex $(h, k)$ is given by $h = -\frac{3}{4a}$ and $k = -\frac{35a}{16}$.
From $h = -\frac{3}{4a}$,we have $a = -\frac{3}{4h}$.
Substituting $a$ into $k$: $k = -\frac{35}{16} \times (-\frac{3}{4h}) = \frac{105}{64h}$.
Therefore,$hk = \frac{105}{64}$.
The locus is $xy = \frac{105}{64}$.

(D) Let the chord pass through the vertex $V(0, 0)$ and intersect the parabola $y^{2}=4ax$ at $P(at^{2}, 2at)$.
Let $(h, k)$ be the mid-point of the chord $VP$.
Then,$h = \frac{at^{2}+0}{2} = \frac{at^{2}}{2}$ and $k = \frac{2at+0}{2} = at$.
From $k = at$,we have $t = \frac{k}{a}$.
Substituting $t$ in the expression for $h$: $h = \frac{a}{2} \left(\frac{k}{a}\right)^{2} = \frac{k^{2}}{2a}$.
Thus,$k^{2} = 2ah$.
The locus of the mid-point $(h, k)$ is $y^{2} = 2ax$.
This can be rewritten as $(y-0)^{2} = 4\left(\frac{a}{2}\right)(x-0)$.
For a parabola $Y^{2} = 4AX$,the directrix is $X = -A$.
Here,$A = \frac{a}{2}$,so the directrix is $x = -\frac{a}{2}$.

(B) The given equation is $(7x+5)^{2}+(7y+3)^{2}=\lambda^{2}(4x+3y-24)^{2}$.
This is of the form $SP^{2} = \lambda^{2} PM^{2}$,where $S$ is the focus,$P$ is a point on the curve,and $PM$ is the perpendicular distance from $P$ to the directrix.
For the curve to be a parabola,the eccentricity $e$ must be equal to $1$.
Here,the distance from the point $(x, y)$ to the focus $S$ is $\sqrt{(x - x_s)^2 + (y - y_s)^2}$.
Rewriting the equation: $\frac{(7x+5)^{2}+(7y+3)^{2}}{49} = \lambda^{2} \frac{(4x+3y-24)^{2}}{49}$.
$\left(\sqrt{(x + 5/7)^2 + (y + 3/7)^2}\right)^2 = \lambda^{2} \left(\frac{4x+3y-24}{7}\right)^2$.
Comparing this with the definition of a conic $SP = e PM$,we have $e = \lambda \cdot \frac{\sqrt{4^2+3^2}}{7} = \lambda \cdot \frac{5}{7}$.
For a parabola,$e = 1$,so $\lambda \cdot \frac{5}{7} = 1$.
Thus,$\lambda = \pm \frac{7}{5}$.

(C) Let the midpoint of a chord passing through the vertex $(0, 0)$ be $(h, k)$.
Since the chord passes through the vertex $(0, 0)$ and the midpoint $(h, k)$,the equation of the chord is $y - 0 = \frac{k - 0}{h - 0}(x - 0)$,which simplifies to $y = \frac{k}{h}x$.
Substituting this into the parabola equation $y^2 = 4ax$,we get $(\frac{k}{h}x)^2 = 4ax$.
$\frac{k^2}{h^2}x^2 = 4ax$.
Since $x=0$ is the vertex,for the chord,we consider $x \neq 0$,so $\frac{k^2}{h^2}x = 4a$,which gives $x = \frac{4ah^2}{k^2}$.
However,the midpoint $(h, k)$ must satisfy the chord equation $k = \frac{k}{h}h$,which is trivial.
The condition for $(h, k)$ to be the midpoint of a chord of $y^2 = 4ax$ is $k^2 = 2a(h - x_1)$ where $x_1$ is the other end. For a chord through the vertex,the equation of the chord with midpoint $(h, k)$ is $T = S_1$,i.e.,$yk - 2a(x + h) = k^2 - 4ah$.
Since it passes through $(0, 0)$,we have $0 - 2ah = k^2 - 4ah$,which simplifies to $k^2 = 2ah$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2ax$,which is a parabola.

(B) The parabola is $y^{2}=8x$,so $4a=8 \Rightarrow a=2$. The focus $A$ is $(2,0)$.
Let the points $B$ and $C$ be $(2t_{1}^{2}, 4t_{1})$ and $(2t_{2}^{2}, 4t_{2})$ respectively.
The centroid of $\triangle ABC$ is given by $(\frac{2t_{1}^{2}+2t_{2}^{2}+2}{3}, \frac{4t_{1}+4t_{2}+0}{3}) = (\frac{7}{3}, \frac{4}{3})$.
Equating the coordinates:
$2(t_{1}^{2}+t_{2}^{2}+1) = 7 \Rightarrow t_{1}^{2}+t_{2}^{2} = \frac{5}{2}$.
$4(t_{1}+t_{2}) = 4 \Rightarrow t_{1}+t_{2} = 1$.
Now,$(t_{1}-t_{2})^{2} = (t_{1}+t_{2})^{2} - 4t_{1}t_{2}$.
Since $t_{1}^{2}+t_{2}^{2} = (t_{1}+t_{2})^{2} - 2t_{1}t_{2} = \frac{5}{2}$,we have $1 - 2t_{1}t_{2} = \frac{5}{2}$ $\Rightarrow 2t_{1}t_{2} = -\frac{3}{2}$ $\Rightarrow t_{1}t_{2} = -\frac{3}{4}$.
Thus,$(t_{1}-t_{2})^{2} = 1 - 4(-\frac{3}{4}) = 1+3 = 4$.
$(BC)^{2} = (2t_{1}^{2}-2t_{2}^{2})^{2} + (4t_{1}-4t_{2})^{2} = 4(t_{1}^{2}-t_{2}^{2})^{2} + 16(t_{1}-t_{2})^{2}$.
$(BC)^{2} = 4(t_{1}+t_{2})^{2}(t_{1}-t_{2})^{2} + 16(t_{1}-t_{2})^{2} = 4(1)^{2}(4) + 16(4) = 16 + 64 = 80$.

(C) The parametric point $P$ on the parabola $x^2 = 4y$ is $(2t, t^2)$.
Let the mirror image of $P$ in the line $x - y - 1 = 0$ be $Q(h, k)$.
Using the reflection formula $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$,we get:
$\frac{h - 2t}{1} = \frac{k - t^2}{-1} = -2 \frac{2t - t^2 - 1}{1^2 + (-1)^2} = -(2t - t^2 - 1) = t^2 - 2t + 1$.
Thus,$h = 2t + t^2 - 2t + 1 = t^2 + 1$ and $k = t^2 - (t^2 - 2t + 1) = 2t - 1$.
From $k = 2t - 1$,we have $t = \frac{k + 1}{2}$.
Substituting $t$ into $h = t^2 + 1$,we get $h = (\frac{k + 1}{2})^2 + 1$.
$h - 1 = \frac{(k + 1)^2}{4} \implies (k + 1)^2 = 4(h - 1)$.
Replacing $(h, k)$ with $(x, y)$,the equation of the image parabola is $(y + 1)^2 = 4(x - 1)$.
Comparing this with $(y + a)^2 = b(x - c)$,we get $a = 1, b = 4, c = 1$.
Therefore,$a + b + c = 1 + 4 + 1 = 6$.

(B) The equation of the parabola is $y^{2}=16x$,which is of the form $y^{2}=4ax$ with $a=4$.
Let the coordinates of point $A$ be $(at_{1}^{2}, 2at_{1}) = (16, 16)$.
Thus,$2(4)t_{1} = 16 \Rightarrow t_{1}=2$.
Since $AB$ is a focal chord,the product of the parameters of its endpoints is $t_{1}t_{2}=-1$.
Therefore,$2t_{2}=-1 \Rightarrow t_{2}=-\frac{1}{2}$.
The coordinates of point $B$ are $(at_{2}^{2}, 2at_{2}) = (4(-\frac{1}{2})^{2}, 2(4)(-\frac{1}{2})) = (1, -4)$.
Point $P(\alpha, \beta)$ divides the chord $AB$ internally in the ratio $5:2$.
Case $1$: $P$ divides $AB$ in ratio $5:2$ (from $A$ to $B$):
$\alpha = \frac{5(1) + 2(16)}{5+2} = \frac{37}{7}$,$\beta = \frac{5(-4) + 2(16)}{5+2} = \frac{12}{7}$.
$\alpha+\beta = \frac{37+12}{7} = \frac{49}{7} = 7$.
Case $2$: $P$ divides $BA$ in ratio $5:2$ (from $B$ to $A$):
$\alpha = \frac{5(16) + 2(1)}{5+2} = \frac{82}{7}$,$\beta = \frac{5(16) + 2(-4)}{5+2} = \frac{72}{7}$.
$\alpha+\beta = \frac{82+72}{7} = \frac{154}{7} = 22$.
The minimum value of $\alpha+\beta$ is $7$.

(C) The parabola is $y^{2}=12x$,so $4a=12 \Rightarrow a=3$. Any point $P$ on the parabola can be represented as $(3t^{2}, 6t)$.
The slope of $OP$ is $m_{OP} = \frac{6t-0}{3t^{2}-0} = \frac{2}{t}$.
Since $\angle OPA=90^{\circ}$,$OP \perp PA$,so the slope of $PA$ is $m_{PA} = -\frac{t}{2}$.
The equation of line $PA$ is $y-6t = -\frac{t}{2}(x-3t^{2})$.
For point $A$ on the $x$-axis,set $y=0$: $-6t = -\frac{t}{2}(x-3t^{2})$ $\Rightarrow 12 = x-3t^{2}$ $\Rightarrow x = 3t^{2}+12$.
Thus,$A = (3t^{2}+12, 0)$.
Let the centroid of $\triangle OPA$ be $G(h, k)$.
$h = \frac{0+3t^{2}+3t^{2}+12}{3} = \frac{6t^{2}+12}{3} = 2t^{2}+4$.
$k = \frac{0+6t+0}{3} = 2t$.
From $k=2t$,we have $t = \frac{k}{2}$.
Substituting into $h$: $h = 2(\frac{k}{2})^{2}+4 = \frac{k^{2}}{2}+4$.
$2h = k^{2}+8 \Rightarrow k^{2} = 2h-8$.
The locus is $y^{2} = 2x-8$,which can be written as $y^{2}-2x+8=0$.

(B) Let the parabola be $y^2=4x$. Any point on the parabola is $P(t^2, 2t)$.
The chord passes through the origin $O(0,0)$ and $P(t^2, 2t)$.
The mid-point $M(h, k)$ of the chord $OP$ is given by $h = \frac{t^2+0}{2} = \frac{t^2}{2}$ and $k = \frac{2t+0}{2} = t$.
Substituting $t=k$ into $h = \frac{t^2}{2}$,we get $h = \frac{k^2}{2}$,so $k^2=2h$.
The locus $S$ is $y^2=2x$.
Let $P(x_0, y_0)$ be a point on $S$,so $y_0^2=2x_0$. Since $P$ is on $S$,we can write $P$ as $(2t^2, 2t)$ because $(2t)^2 = 2(2t^2)$.
Let $R(x, y)$ be the point that divides $OP$ internally in the ratio $3:1$. Using the section formula:
$x = \frac{3(2t^2) + 1(0)}{3+1} = \frac{6t^2}{4} = \frac{3t^2}{2}$
$y = \frac{3(2t) + 1(0)}{3+1} = \frac{6t}{4} = \frac{3t}{2}$
From $y = \frac{3t}{2}$,we get $t = \frac{2y}{3}$.
Substituting $t$ into $x = \frac{3t^2}{2}$:
$x = \frac{3}{2} \left(\frac{2y}{3}\right)^2 = \frac{3}{2} \cdot \frac{4y^2}{9} = \frac{2y^2}{3}$
Thus,$3x = 2y^2$,or $2y^2=3x$.

(A) For the parabola $y^{2} = 4ax$,where $4a = 12$,we have $a = 3$. The points on the parabola are $P_{1}(3t_{1}^{2}, 6t_{1})$ and $P_{2}(3t_{2}^{2}, 6t_{2})$.
Since the chord $P_{1}P_{2}$ subtends a right angle at the vertex $(0, 0)$,the product of the slopes of $OP_{1}$ and $OP_{2}$ is $-1$.
Slope $m_{1} = \frac{6t_{1}}{3t_{1}^{2}} = \frac{2}{t_{1}}$ and $m_{2} = \frac{6t_{2}}{3t_{2}^{2}} = \frac{2}{t_{2}}$.
Thus,$(\frac{2}{t_{1}})(\frac{2}{t_{2}}) = -1 \implies t_{1}t_{2} = -4$.
Now,$x_{1}x_{2} = (3t_{1}^{2})(3t_{2}^{2}) = 9(t_{1}t_{2})^{2} = 9(-4)^{2} = 9(16) = 144$.
And $y_{1}y_{2} = (6t_{1})(6t_{2}) = 36(t_{1}t_{2}) = 36(-4) = -144$.
Therefore,$x_{1}x_{2} - y_{1}y_{2} = 144 - (-144) = 144 + 144 = 288$.