Parabola

• Focus $A$ : The focus is $(a, 0)$, which is $(2,0)$.

1. Define Parametric Points for $B$ and $C$ : Let the two distinct points of intersection be $B$ a $C$. Using parametric coordinates $\left(a t^2, 2 a t\right)$ :

• Point $B=\left(2 t_1^2, 4 t_1\right)$

1. Use the Centroid Formula : The centroid $G(x, y)$ of a triangle with vertices $\left(x_1, y_1\right),\left(x_2, y_2\right)$, and $\left(x_3, y_3\right)$ is:

$ G=\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $

3. Use the Centroid Property : The centroid of triangle $A B C$ with vertices $A(2,0)$, $B\left(2 t_1^2, 4 t_1\right)$, and $C\left(2 t_2^2, 4 t_2\right)$ is given as $\left(\frac{7}{3}, \frac{4}{3}\right)$.

• For the $y$-coordinate :

$ \frac{4 t_1+4 t_2+0}{3}=\frac{4}{3} \Longrightarrow 4\left(t_1+t_2\right)=4 \Longrightarrow t_1+t_2=1 $

• For the $x$-coordinate :

$ \frac{2 t_1^2+2 t_2^2+2}{3}=\frac{7}{3} \Longrightarrow 2\left(t_1^2+t_2^2\right)=5 \Longrightarrow t_1^2+t_2^2=\frac{5}{2} $

4. Find the Relationship between $t_1$ and $t_2$ :
   
   Using the identity $\left(t_1+t_2\right)^2=t_1^2+t_2^2+2 t_1 t_2$ :

$ (1)^2=\frac{5}{2}+2 t_1 t_2 \Longrightarrow 1-\frac{5}{2}=2 t_1 t_2 \Longrightarrow 2 t_1 t_2=-\frac{3}{2} \Longrightarrow t_1 t_2=-\frac{3}{4} $

Now, calculate $\left(t_1-t_2\right)^2$ :

$ \left(t_1-t_2\right)^2=\left(t_1+t_2\right)^2-4 t_1 t_2=1^2-4\left(-\frac{3}{4}\right)=1+3=4 $

5. Calculate $(B C)^2$ The distance formula for $(B C)^2$ is :

$ (B C)^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2 $

$\Rightarrow $ $ (B C)^2=\left[2\left(t_2^2-t_1^2\right)\right]^2+\left[4\left(t_2-t_1\right)\right]^2 $

$\Rightarrow $ $ (B C)^2=4\left(t_2-t_1\right)^2\left(t_2+t_1\right)^2+16\left(t_2-t_1\right)^2 $

Substitute the values $\left(t_1-t_2\right)^2=4$ and $\left(t_1+t_2\right)=1$ :

$ (B C)^2=4(4)(1)^2+16(4)=16+64=80 $

Let parametric point on parabola $x^2=4 y$ is $\left(2 t, t^2\right)$.

Key concept: image of point $(h, k)$ in the line $a x+b y+c=0$ is given as $\frac{x-h}{a}=\frac{y-k}{b}=\frac{-2(a h+b k+c)}{a^2+b^2}$

So, Image of point in line $x-y-1=0$ is given as

$ \frac{x-2 t}{1}=\frac{y-t^2}{-1}=\frac{-2\left(2 t-t^2-1\right)}{1^2+(-1)^2} $

equating

$ \begin{aligned} & \frac{x-2 t}{1}=\frac{-2\left(2 t-t^2-1\right)}{2} \\ & x=-2 t+t^2+1+2 t \Rightarrow x=t^2+1 \end{aligned} $

equating

$ \begin{aligned} & \frac{y-t^2}{-1}=\frac{-2\left(2 t-t^2-1\right)}{2} \\ & y-t^2=2 t-t^2-1 \Rightarrow y=2 t-1 \end{aligned} $

from (1) and (2) eliminate parameter t to get

image of curve.

so, from (2) put $t=\frac{y+1}{2}$ in equation (1).

$ \begin{aligned} & x=\left(\frac{y+1}{2}\right)^2+1 \\ & \frac{(y+1)^2}{4}=x-1 \\ & (y+1)^2=4(x-1) \end{aligned} $

Compare with given image of curve $(y+a)^2=b(x-c)$

to find value of $\mathrm{a}, \mathrm{b}, \& \mathrm{c}$.

$ \begin{aligned} & a=1, b=4, c=1 \\ & a+b+c=1+4+1=6 . \end{aligned} $

$ y^2=4 x \Longrightarrow a=1 $

Let $A=\left(t^2, 2 t\right)$ and $B=\left(t^2,-2 t\right)$ due to symmetry.

Triangle $O A B$ is equilateral:

Slope of $O A=\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$

$ \frac{2 t-0}{t^2-0}=\frac{1}{\sqrt{3}} \Longrightarrow \frac{2}{t}=\frac{1}{\sqrt{3}} \Longrightarrow t=2 \sqrt{3} $

Coordinates:

$ \begin{aligned} & A=(12,4 \sqrt{3}) \\ & B=(12,-4 \sqrt{3}) \end{aligned} $

Circle with diameter $A B$ :

Center $C=\left(\frac{12+12}{2}, \frac{4 \sqrt{3}-4 \sqrt{3}}{2}\right)=(12,0)$

Radius $r=\frac{1}{2} \sqrt{(12-12)^2+(4 \sqrt{3}-(-4 \sqrt{3}))^2}=4 \sqrt{3}$

Minimum distance from origin $O(0,0)$ :

$ \begin{aligned} & d=|O C-r| \\ & O C=\sqrt{12^2+0^2}=12 \end{aligned} $

$d=12-4 \sqrt{3}=4(3-\sqrt{3})$

Let $M\left(x_1, y_1\right)$ be the locus of the midpoint of the chord of the parabola $y^2=4 x$

∴ Equation of the chord is $S_1=S_{11} \Rightarrow y y_1-2\left(x+x_1\right)=y_1^2-4 x_1$

It passes through $0(0,0)$

$ \therefore-2 x_1=y_1^2-4 x_1 $

Locus of M is $y^2=2 x$

Let $P\left(\frac{1}{2} t^2, t\right)$ be a point on this locus

$ \begin{aligned} & Q(x, y)=\left(\frac{\frac{3 t^2}{2}+0}{4}, \frac{3 t+0}{4}\right) \Rightarrow\left(\frac{3 t^2}{8}, \frac{3 t}{4}\right) \\ & \Rightarrow t^2=\frac{8 x}{3}, t=\frac{4 y}{3} \Rightarrow \frac{16 y^2}{9}=\frac{8 x}{3} \Rightarrow 2 y^2=3 x \end{aligned} $

$ \begin{aligned} & P\left(a t_1^2, 2 a t_1\right), Q\left(a t_2^2, 2 a t_2\right) \\ & \therefore t_1 t_2=-4, a=3 \\ & x_1 x_2-y_1 y_2=a^2 \times\left(t_1 t_2\right)^2-4 a^2 t_1 t_2=288 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & m_{o p} m_{P A}=-1 \frac{6 t}{3 t^2} \times \frac{6 t}{3 t^2-x}=-1 \\ & 12=-3 t^2+x \\ & x=12+3 t^2 \\ & h=\frac{x+3 t^2+0}{3} ; k=\frac{0+0+6 t}{3} \\ & 3 h=12+6 t h 2 ; k=2 t \\ & h=4+2 t^2 ; k=2 t \end{aligned}\\ &\text { Substitute } \mathrm{t} \& \text { replace } \mathrm{h} \text { by } x \& \mathrm{k} \text { by } \mathrm{y}\\ &y^2-2 x+8=0 \end{aligned} $

If $P(\alpha, \beta)$ divides focal chord internally in $5: 2$

If $A\left(t_1\right) \& B\left(t_2\right)$ are ends of focal chord $A(16,16)$

$ \begin{aligned} & t_1 t_2=-1 \\ & 2 t_2=-1 \\ & t_2=-\frac{1}{2} \\ & B=(1,-4) \\ & P(\alpha, \beta)=\left(\frac{37}{7}, \frac{12}{7}\right) \text { or }\left(\frac{82}{7}, \frac{72}{7}\right) \end{aligned} $

Minimum of $\alpha+\beta=\frac{37}{7}+\frac{12}{7}=7$

Let the vertex of the parabola $x^2 = 4y$ be the origin $O(0, 0)$.

Let $Q$ be any point on the parabola. We can parameterize the coordinates of $Q$ as $(2t, t^2)$ for some real parameter $t$. Satisfying $x^2 = (2t)^2 = 4t^2$ and $4y = 4(t^2)$.

Let $P(x, y)$ be the point that divides the line segment $OQ$ internally in the ratio $2:3$.

Using the section formula, the coordinates of $P$ are given by:

$ x = \frac{2 \cdot x_Q + 3 \cdot x_O}{2+3} = \frac{2(2t) + 3(0)}{5} = \frac{4t}{5} $

$ y = \frac{2 \cdot y_Q + 3 \cdot y_O}{2+3} = \frac{2(t^2) + 3(0)}{5} = \frac{2t^2}{5} $

To find the locus of $P$ (the conic $C$), we eliminate the parameter $t$.

From the first equation, we have $t = \frac{5x}{4}$.

Substituting this into the second equation:

$ y = \frac{2}{5} \left( \frac{5x}{4} \right)^2 $

$ y = \frac{2}{5} \cdot \frac{25x^2}{16} $

$ y = \frac{5x^2}{8} $

$ 5x^2 = 8y $

So, the equation of the conic $C$ is $5x^2 - 8y = 0$. This is a parabola.

We need to find the equation of the chord of $C$ which is bisected at the point $(x_1, y_1) = (1, 2)$.

The equation of a chord of a conic $S = 0$ bisected at a point $(x_1, y_1)$ is given by $T = S_1$.

Here, $S \equiv 5x^2 - 8y$.

$S_1$ is the value of $S$ at $(1, 2)$:

$ S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11 $

$T$ is the expression obtained by replacing $x^2$ with $xx_1$ and $y$ with $\frac{y+y_1}{2}$:

$ T = 5x(1) - 8\left( \frac{y+2}{2} \right) $

$ T = 5x - 4(y+2) $

$ T = 5x - 4y - 8 $

Equating $T = S_1$:

$ 5x - 4y - 8 = -11 $

$ 5x - 4y + 3 = 0 $

The equation of the chord is $5x - 4y + 3 = 0$.

Option A

$5 x-4 y+3=0$

$\begin{aligned} & t_1 t_2=-4 \\ & 2 h=t_1^2+t_2^2 \quad 2 k=2\left(t_1+t_2\right) \\ & 2 h=k^2-2 t_1 t_2 \\ & 2 h=k^2+8 \\ & k^2=2 h-8 \\ & =2((h-4) \\ & \text { L.R. }=2\end{aligned}$

Let the roots of the quadratic equation

$ \left(k^2-15k+27\right)x^2+9(k-1)x+18=0 $

be $\alpha$ and $\beta$.

Given that one root is twice the other, let

$ \alpha=2\beta $

So the roots are in the ratio $2:1$.

For a quadratic equation $ax^2+bx+c=0$:

$ \alpha+\beta=-\frac{b}{a}, \qquad \alpha\beta=\frac{c}{a} $

Here,

$ a=k^2-15k+27,\qquad b=9(k-1),\qquad c=18 $

So,

$ \alpha+\beta=\frac{-9(k-1)}{k^2-15k+27} $

and

$ \alpha\beta=\frac{18}{k^2-15k+27} $

Now let the roots be $2t$ and $t$.

Then,

$ \alpha+\beta=3t,\qquad \alpha\beta=2t^2 $

Using the identity:

$ (\alpha+\beta)^2=9t^2 $

and

$ \alpha\beta=2t^2 $

So,

$ (\alpha+\beta)^2=\frac{9}{2}\alpha\beta $

Now substitute the values from the quadratic:

$ \left(\frac{-9(k-1)}{k^2-15k+27}\right)^2=\frac{9}{2}\cdot \frac{18}{k^2-15k+27} $

Simplify:

$ \frac{81(k-1)^2}{(k^2-15k+27)^2}=\frac{81}{k^2-15k+27} $

Cancel $81$:

$ \frac{(k-1)^2}{(k^2-15k+27)^2}=\frac{1}{k^2-15k+27} $

Cross-multiply:

$ (k-1)^2=k^2-15k+27 $

Now expand:

$ k^2-2k+1=k^2-15k+27 $

Cancel $k^2$:

$ -2k+1=-15k+27 $

$ 13k=26 $

$ k=2 $

Now the parabola is

$ y^2=6kx $

Substitute $k=2$:

$ y^2=12x $

Compare with standard form:

$ y^2=4ax $

So,

$ 4a=12 \implies a=3 $

Length of latus rectum of the parabola $y^2=4ax$ is

$ 4a $

Therefore,

$ 4a=12 $

Hence, the required length is

$ \boxed{12} $

So, the correct option is:

$ \boxed{\text{Option D}} $

$\begin{aligned} & \text{For the parabola } y^2=12x,\ \text{we have } 4a=12 \Rightarrow a=3. \\\\ & \text{So, the parametric point on } y^2=4ax \text{ is } (at^2,2at). \\\\ & \text{Let the ordinate (}y\text{-coordinate) of }P \text{ be in ratio }1:2 \text{ with }Q. \\\\ & \text{So we take }P(3t^2,6t)\ \text{and}\ Q\big(3(2t)^2,6(2t)\big)=\left(12t^2,12t\right). \\\\ & \text{Now find }PQ: \\\\ & PQ=\sqrt{(12t^2-3t^2)^2+(12t-6t)^2} \\\\ & \quad=\sqrt{(9t^2)^2+(6t)^2}=\sqrt{81t^4+36t^2}. \\\\ & \text{Given }PQ=3\sqrt{13}\Rightarrow PQ^2=117. \\\\ & \text{So, }81t^4+36t^2=117 \\\\ & \Rightarrow 9t^4+4t^2-13=0 \\\\ & \Rightarrow (t^2-1)(9t^2+13)=0. \\\\ & \text{Since }t^2\ge 0,\ 9t^2+13\ne 0,\ \text{so }t^2=1 \Rightarrow t=1. \\\\ & \text{Hence }P(3,6),\ Q(12,12). \\\\ & \text{Focus of }y^2=12x\text{ is }S(a,0)=(3,0). \\\\ & \text{Slope of }QS=\frac{12-0}{12-3}=\frac{12}{9}=\frac{4}{3}. \\\\ & \text{Line }PS \text{ is vertical since }P(3,6)\text{ and }S(3,0)\text{ have same }x. \\\\ & \text{So the angle }\alpha \text{ at the focus is the angle between }QS \text{ and the vertical.} \\\\ & \text{If slope of }QS \text{ is }\frac{4}{3},\ \text{then the angle it makes with the vertical has} \\\\ & \tan\alpha=\frac{3}{4}. \\\\ & \Rightarrow \sin\alpha=\frac{3}{5}. \end{aligned}$

For the parabola

$ y^2=8x $

we compare it with the standard form

$ y^2=4ax $

So,

$ 4a=8 \implies a=2 $

Hence:

• focus $S=(2,0)$

• directrix is

$ x=-2 $

Since the directrix cuts the $x$-axis at point $A$, we get

$ A=(-2,0) $

Now let $B(\alpha,\beta)$ be a point on the parabola such that the slope of $AB$ is $\dfrac35$.

The slope of $AB$ is

$ \frac{\beta-0}{\alpha-(-2)}=\frac{\beta}{\alpha+2} $

Given this is $\dfrac35$, so

$ \frac{\beta}{\alpha+2}=\frac35 $

Thus,

$ 5\beta=3(\alpha+2) \tag{1} $

Also, since $B$ lies on the parabola $y^2=8x$,

$ \beta^2=8\alpha \tag{2} $

Now substitute from (1). From (1),

$ \beta=\frac{3(\alpha+2)}{5} $

Put into (2):

$ \left(\frac{3(\alpha+2)}{5}\right)^2=8\alpha $

$ \frac{9(\alpha+2)^2}{25}=8\alpha $

$ 9(\alpha+2)^2=200\alpha $

$ 9(\alpha^2+4\alpha+4)=200\alpha $

$ 9\alpha^2+36\alpha+36=200\alpha $

$ 9\alpha^2-164\alpha+36=0 $

Now solve:

$ 9\alpha^2-164\alpha+36=0 $

Discriminant:

$ D=164^2-4\cdot 9\cdot 36=26896-1296=25600 $

$ \sqrt D=160 $

So,

$ \alpha=\frac{164\pm 160}{18} $

Hence,

$ \alpha=\frac{324}{18}=18 \quad \text{or} \quad \alpha=\frac{4}{18}=\frac29 $

Given $\alpha>1$, we take

$ \alpha=18 $

Then from $\beta^2=8\alpha$,

$ \beta^2=144 \implies \beta=\pm 12 $

Since slope is $\dfrac35>0$, $\beta$ must be positive. Therefore

$ B=(18,12) $

Now $BC$ is a focal chord of the parabola.

For parabola $y^2=4ax$, a point in parametric form is

$ (at^2,\,2at) $

Here $a=2$, so point is

$ (2t^2,\,4t) $

For point $B=(18,12)$,

$ 2t^2=18 \implies t^2=9 \implies t=3 $

and

$ 4t=12 \implies t=3 $

So $B$ corresponds to parameter $t=3$.

If $BC$ is a focal chord, then the other endpoint corresponds to parameter $-\dfrac1t$.

Thus for $t=3$, point $C$ corresponds to parameter $-\dfrac13$.

So,

$ C=\left(2\left(\frac{1}{9}\right),\,4\left(-\frac13\right)\right) =\left(\frac29,\,-\frac43\right) $

Now find area of $\triangle ABC$.

Coordinates are:

$ A=(-2,0),\quad B=(18,12),\quad C=\left(\frac29,-\frac43\right) $

Using determinant formula,

$ \text{Area} = \frac12\left| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right| $

So,

$ \text{Area} = \frac12\left[ -2\left(12+\frac43\right) +18\left(-\frac43\right) +\frac29(0-12) \right] \Bigg| $

Now simplify:

$ 12+\frac43=\frac{40}{3} $

So,

$ -2\cdot \frac{40}{3}=-\frac{80}{3} $

Also,

$ 18\cdot \left(-\frac43\right)=-24 $

and

$ \frac29(-12)=-\frac{24}{9}=-\frac83 $

Therefore,

$ \text{Area} = \frac12\left|-\frac{80}{3}-24-\frac83\right| $

$ = \frac12\left|-\frac{88}{3}-24\right| = \frac12\left|-\frac{88}{3}-\frac{72}{3}\right| $

$ = \frac12\left|-\frac{160}{3}\right| = \frac{80}{3} $

Hence,

$ 6\times \text{Area} = 6\cdot \frac{80}{3}=160 $

Therefore, the required answer is

$ \boxed{160} $

So, the correct option is B.

parabola passes through $(1,-1)$ so

$-1 = 1 + p + q$

$p + q = -2$

$q = -2 - p \,\,\,\ldots (1)$

Distance from x-axis

$\frac{-D}{4a} = \frac{-(p^2 - 4q)}{4(1)} = \frac{4q - p^2}{4}$

$= \frac{4(-2-p)-p^2}{4} \qquad \text{from (1)}$

$= \frac{-8-4p-p^2}{4} = \frac{-(p^2+4p+8)}{4} = \frac{-((p+2)^2+4)}{4}$

Minimum at $p=-2 \rightarrow q=0$

$p^2 + q^2 = 4 + 0 = 4$

$\begin{aligned} &\begin{aligned} & P M=P F \\ & \Rightarrow \quad \frac{|2 x+y+2|}{\sqrt{5}}=\sqrt{(x+2)^2+(y-1)^2} \end{aligned}\\ &\text { Now abscissa of } P \text { is }-2 \Rightarrow x=-2\\ &\begin{aligned} & \left|\frac{y-2}{\sqrt{5}}\right|=\sqrt{0+(y-1)^2} \Rightarrow \frac{|y-2|}{\sqrt{5}}=|y-1| \\ & \Rightarrow(y-2)^2=5(y-1)^2 \\ & \Rightarrow 4 y^2-6 y+1=0 \Rightarrow \text { Sum of ordinates } \\ & =-\left(\frac{-6}{4}\right)=\frac{3}{2} \end{aligned} \end{aligned}$

$\begin{aligned} & y^2=4\left(\frac{1}{4}\right)(x-2) \\ & y=m(x-2)+\frac{1}{4 m} \text { passes through }(-2,0) \\ & \Rightarrow 0=-4 m+\frac{1}{4 m} \Rightarrow 16 m^2=1 \\ & \Rightarrow m= \pm \frac{1}{4} \\ & m=\frac{1}{4} \text { in first quadrant } \Rightarrow \text { contact point }(6,2) \end{aligned}$

$ \begin{aligned} & \Rightarrow \text { Area }=\frac{1}{2} \times(1) \times 4+\int_2^6\left[\left(\frac{x+2}{4}\right)-\sqrt{x-2}\right] d x \\ & \quad=2+\frac{2}{3}=\frac{8}{3} \end{aligned}$

Equation of directrix

$\Rightarrow y=-x$

By definition of parabola,

$\begin{aligned} & P M=P F \\ & \left|\frac{1+K}{\sqrt{2}}\right|=\sqrt{(1-2)^2+(K-2)^2} \\ & \frac{(1+K)^2}{2}=1+K^2+4-4 K \\ & 1+K^2+2 K=10+2 K^2-8 K \\ & K^2-10 K+9=0 \\ & (K-9)(K-1)=0 \\ & \therefore \quad K=1 \text { or } K=9 \end{aligned}$

The circle will have centre on $x=y$ line since parabolas are symmetric about $y=x$ line

The slope of tangent at closest point $y^2=x-2$

$\Rightarrow 2 y\left(\frac{d y}{d x}\right)=1 \Rightarrow y=\frac{1}{2} \Rightarrow$ point will be $\left(\frac{9}{4}, \frac{1}{2}\right)$

Similarly, on $x^2=y-2 \Rightarrow 2 x=\frac{d y}{d x}=1 \Rightarrow\left(\frac{1}{2}, \frac{9}{4}\right)$

$\begin{aligned} & 2 r=\sqrt{\left(\frac{9}{4}-\frac{1}{2}\right)^2+\left(\frac{1}{2}-\frac{9}{4}\right)^2}=\sqrt{2} \cdot\left|\frac{9}{4}-\frac{1}{2}\right| \\ & =\frac{14}{8} \sqrt{2}=\frac{7 \sqrt{2}}{4} \\ & \Rightarrow r=\frac{7 \sqrt{2}}{8} \end{aligned}$

Given the parabola $ y^2 = 16x $, the point $ P $ on the focal chord $ PQ $ is $ (1, -4) $. We need to determine how the focus divides the chord.

First, express the coordinates for the point $ P $ using the parametric form of the parabola. The general parametric equations for the parabola are:

$ P(a t^2, 2 a t) \equiv P(4t^2, 8t) \equiv (1, -4) $

The equation $ 8t = -4 $ leads to solving for $ t $:

$ t = -\frac{1}{2} $

Now, find the coordinates of point $ Q $ using the relation $ t_1 \cdot t_2 = -1 $. Therefore:

$ Q\left(\frac{a}{t^2}, \frac{-2a}{t}\right) $

Given the focus $ S $ of the parabola is at $ (4, 0) $, we can find the lengths of $ PS $ and $ QS $:

The length $ PS $ is calculated as:

$ PS = a + a t^2 $

And $ QS $ is given by:

$ QS = a + \frac{a}{t^2} = \frac{a t^2 + a}{t^2} $

Since the focus divides the chord in a specific ratio $ m:n $, and $\gcd(m, n) = 1$, we find:

$ \frac{PS}{QS} = t^2 = \frac{1}{4} = \frac{m}{n} $

Thus, the squares of $ m $ and $ n $ sum up to:

$ m^2 + n^2 = 17 $

$\begin{aligned} & P T: t y=x+a t^2 \\ & P S=P T \\ & M_t=\frac{1}{t}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ & t=\sqrt{3} \\ & \alpha=a t=\sqrt{3} \quad(a=1) \\ & \therefore 5 \alpha^2=15 \end{aligned}$

Let intersection points of these two parabolas are $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right) \& \mathrm{~B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$

$\because$ equation of parabola I and II are given below

$\begin{aligned} & \therefore(\mathrm{x}-4)^2+(\mathrm{y}-3)^2=\mathrm{x}^2 \quad\text{..... (1)}\\ & \&(\mathrm{x}-4)^2+(\mathrm{y}-3)^2=\mathrm{y}^2 \quad\text{..... (2)} \end{aligned}$

Here $A\left(x_1, y_1\right) \& B\left(x_2, y_2\right)$ will satisfy the equation

Also from equations (1) & (2), we get $x=y$ .......(3)

Put $x=y$ in equation (1)

We get $x^2-14 x+25=0$

$\begin{aligned} & x_1+x_2=14 \\ & x_1 x_2=25 \\ & \therefore A B^2=\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2 \\ & =2\left(x_1-x_2\right)^2 \\ & =2\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right] \\ & =192 \end{aligned}$

$\begin{aligned} & \mathrm{A}\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right) \& \mathrm{C}\left(\frac{\mathrm{a}}{\mathrm{t}_1^2},-\frac{2 \mathrm{a}}{\mathrm{t}_1}\right) \\ & \text { Length } \mathrm{AC}=\mathrm{a}\left(\mathrm{t}_1+\frac{1}{\mathrm{t}_1}\right)^2=\frac{25}{4}, \mathrm{t}_1+\frac{1}{\mathrm{t}_1}= \pm \frac{5}{2} \\ & \Rightarrow \mathrm{t}_1=2 \text { or } \frac{1}{2}, \mathrm{~A}\left(\frac{1}{2}, 1\right), \mathrm{D}\left(\frac{1}{4},-1\right), \mathrm{B}(4,4), \mathrm{C}(4,-4) \\ & \text { So, area of trapezium }=\frac{1}{2}(8+2)\left(4-\frac{1}{4}\right)=\frac{75}{4} \end{aligned}$

Equation of axis

$\begin{aligned} & y-3=2\left(x-\frac{3}{2}\right) \\ & y-2 x=0 \end{aligned}$

foot of directrix

$\begin{aligned} & \quad y-2 x=0 \\ & \& \quad \Rightarrow(0,0)\\ & 2 y+x=0 \end{aligned}$

$\begin{aligned} & \text { Focus }=(3,6) \\ & \operatorname{PS}^2=P^2 \\ & (x-3)^2+(y-6)^2=\left(\frac{x+2 y}{\sqrt{5}}\right)^2 \\ & 4 x^2+y^2-4 x y-30 x-60 y+225=0 \\ & \Rightarrow \alpha=4, \beta=1, \gamma=4 \Rightarrow \alpha+\beta+\gamma=9 \end{aligned}$

$\begin{aligned} &\text { Normal at } \mathrm{P}\\ &\begin{gathered} y+t x=2 t+t^3 \\ \uparrow \\ (a, 0) \end{gathered} \end{aligned}$

$\begin{aligned} & \mathrm{at}=2 \mathrm{t}+\mathrm{t}^3 \\ & \mathrm{a}=2+\mathrm{t}^2 \\ & \mathrm{R}\left(2+\mathrm{t}^2, 0\right) \\ & \mathrm{PR}=4 \Rightarrow 4+4 \mathrm{t}^2=16 \end{aligned}$

$\begin{aligned} & \quad 4 \mathrm{t}^2=12 \Rightarrow \mathrm{t}^2=3 \\ & \mathrm{a}=5, \mathrm{R}(5,0) \end{aligned}$

Focus $(1,0)$

$(1,0) \&(5,0)$ will be the end points of diameter

$\Rightarrow E q^{\mathrm{n}}$ of circle is

$\begin{aligned} & (x-1)(x-5)+y^2=0 \\ & x^2+y^2-6 x+5=0 \end{aligned}$

$\begin{aligned} & 3 x-2 y+12=0 \\ & 4 y=3 x^2 \\ & \therefore 2(3 x+12)=3 x^2 \\ & \Rightarrow x^2-2 x-8=0 \\ & \Rightarrow x=-2,4 \\ & \mathrm{~m}_{\mathrm{OA}}=-3 / 2, \mathrm{~m}_{\mathrm{OB}}=3 \\ & \tan \theta=\left(\frac{\frac{-3}{2}-3}{1-\frac{9}{2}}\right)=\frac{9}{7} \\ & \theta=\tan ^{-1}\left(\frac{9}{7}\right) \text { (angle will be acute) } \end{aligned}$

$\begin{aligned} &\begin{aligned} & (4,4 \sqrt{3}) \text { lies on } \mathrm{y}^2=4 \mathrm{ax} \\ & \Rightarrow 48=4 \mathrm{a} .4 \\ & \quad 4 \mathrm{a}=12 \\ & \Rightarrow \mathrm{y}^2=12 \mathrm{x} \text { is equation of parabola } \end{aligned}\\ &\text { Now, parameter of } P \text { is } t_1=\frac{2}{\sqrt{3}} \Rightarrow \text { Parameters of } Q \end{aligned}$

$\begin{aligned} &\text { is } \mathrm{t}_2=-\frac{\sqrt{3}}{2} \Rightarrow \mathrm{Q}\left(\frac{9}{4},-3 \sqrt{3}\right)\\ &\text { Area of trapezium PQNM }\\ &\begin{aligned} & =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PM}+\mathrm{QN}) \\ & =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PS}+\mathrm{QS}) \\ & =\frac{1}{2} \mathrm{MN} \cdot \mathrm{PQ} \\ & =\frac{1}{2} 7 \sqrt{3} \cdot \frac{49}{4}=(343) \frac{\sqrt{3}}{8}=3 \end{aligned} \end{aligned}$

The given parabola is $ y = x^2 + px - 3 $.

Intersection with the y-axis:

At $ x = 0 $, we find $ y = -3 $.

Thus, the parabola intersects the y-axis at the point $ (0, -3) $.

Circle Equation:

We are given the circle has its center at $(-1, -1)$ and it passes through the points where the parabola intersects the axes. The radius can be found using the distance from the center to any given point the circle passes through. Using $ (0, -3) $:

$ \text{Radius} = \sqrt{(0 + 1)^2 + (-3 + 1)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{5} $

Therefore, the equation of the circle is:

$ (x + 1)^2 + (y + 1)^2 = 5 $

This simplifies to:

$ x^2 + 2x + 1 + y^2 + 2y + 1 = 5 $

$ x^2 + y^2 + 2x + 2y + 2 = 5 $

$ x^2 + y^2 + 2x + 2y - 3 = 0 $

Intersection with the x-axis:

When $ y = 0 $, solving the quadratic $ x^2 + 2x - 3 = 0 $ gives:

$ (x + 3)(x - 1) = 0 $

So, $ x = -3 $ or $ x = 1 $.

Thus, the intersection points on the x-axis are $(-3, 0)$ and $(1, 0)$.

Vertices of Triangle $\triangle PQR$:

The vertices of the triangle formed are $ P = (0, -3) $, $ Q = (-3, 0) $, and $ R = (1, 0) $.

Area of $\triangle PQR$:

Use the determinant formula to find the area of the triangle:

$ \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} 0 & -3 & 1 \\ -3 & 0 & 1 \\ 1 & 0 & 1 \end{array} \right| $

Calculate the determinant:

$ = \frac{1}{2} \left( (0)(0)(1) + (-3)(1)(1) + (1)(-3)(1) - (1)(0)(1) - (-3)(1)(0) - (0)(-3)(1) \right) $

Simplify:

$ = \frac{1}{2} \left( 0 - 3 + 3 + 0 \right) = \frac{1}{2} (12) = 6 $

Thus, the area of $\triangle PQR$ is 6.

Let centre be (0, k)

Now radius is $r=6-k$

Also, $6-k=\left|\frac{k}{2}\right|$

$\begin{aligned} & \Rightarrow 6-k=\frac{k}{2} \\ & \Rightarrow 12-2 k=k \\ & \Rightarrow k=4 \end{aligned}$

Radius, $r=6-4=2$

So circle will be

$\begin{aligned} & (x)^2+(y-k)^2=4 \\ & x^2+(y-4)^2=4 \end{aligned}$

$(2,4)$ satisfies this equation.

$y^2=12 x$

Chord $P Q$ having mid-point $(x_1, y_1)=(4,1)$ equation of chord $P Q$

$\begin{aligned} & T=S_1 \\ & y y_1-12 \frac{\left(x+x_1\right)}{2}=y_1^2-12 x_1 \\ & y-6(x+4)=1-12 \times 4 \\ & y-6 x-24=-47 \\ & y-6 x+23=0 \end{aligned}$

From option (4) $x=\frac{1}{2}$ & $y=-20$

$-20-6 \times \frac{1}{2}+23=0$

Equation of normal to parabola

$\mathrm{y=m x-2 m-m^3}$

this normal passing through center of circle $(2,8)$

$\begin{aligned} & 8=2 \mathrm{~m}-2 \mathrm{~m}-\mathrm{m}^3 \\ & \mathrm{~m}=-2 \end{aligned}$

So point $\mathrm{P}$ on parabola $\Rightarrow\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(4,4)$

And $\mathrm{C}=(2,8)$

$\begin{aligned} & \mathrm{PC}=\sqrt{4+16}=\sqrt{20} \\ & \mathrm{~d}^2=20 \end{aligned}$

For each value of t, we can find the coordinates of the points P and Q on the parabola using the parametric form of the parabola $y^2=4ax$, where a is 9. The parametric form is given by $x=at^2$ and $y=2at$.

Let's consider $t=3$ first.

1. For $t=3$, the coordinates of P are given by $P(at^2, 2at) = P(81, 54)$.

2. Since $t_1t_2=-1$ for a focal chord, the parameter value for Q is $t_2=-1/t_1=-1/3$. So the coordinates of Q are given by $Q(at_2^2, 2at_2) = Q(1, -6)$.

Now, let's find the coordinates of the point M that divides the line segment PQ in the ratio 3:1.

1. The x-coordinate of M is given by $\frac{3Q_x+P_x}{4} = \frac{3*1+81}{4} = 21$.

2. The y-coordinate of M is given by $\frac{3Q_y+P_y}{4} = \frac{3*(-6)+54}{4} = 9$.

So, the coordinates of M are M$(21, 9)$.

We can repeat these steps for $t=1/3$ to find the other set of points P, Q, and M. However, since the ordinate of P is positive and PQ makes an acute angle with the positive x-axis, $t=3$ is the appropriate choice in this context.

The slope of the line PQ is $m_{PQ} = \frac{-6 - 54}{1 - 81} = \frac{-60}{-80} = \frac{3}{4}$. The slope of the line perpendicular to PQ is $m_{\perp PQ} = -1/m_{PQ} = -\frac{4}{3}$. Thus, the equation of the line through M and perpendicular to PQ is $(y - 9) = -\frac{4}{3}(x - 21)$, or $4x + 3y = 111$.

Now, we check which of the given points do NOT lie on this line.

Option A : $(6, 29)$ Substitute these values into the equation :

$4\times6 + 3\times29 = 111?$

$24 + 87 = 111?$

$111 = 111$

So, option A does lie on the line.

Option B : $(-3, 43)$ Substitute these values into the equation :

$4\times(-3) + 3\times43 = 111?$

$-12 + 129 = 111?$

$117 = 111$

So, option B does NOT lie on the line.

Option C : $(3, 33)$ Substitute these values into the equation :

$4\times3 + 3\times33 = 111?$

$12 + 99 = 111?$

$111 = 111$

So, option C does lie on the line.

Option D : $(-6, 45)$ Substitute these values into the equation :

$4\times(-6) + 3\times45 = 111?$

$-24 + 135 = 111?$

$111 = 111$

So, option D does lie on the line.

Therefore, $(-3, 43)$ which does not lie on the line passing through M and perpendicular to the line PQ.

Let a tangent on ${y^2} = 16x$ be $y = mx + {4 \over m}$

For common to ${x^2} + {y^2} = 8$

${4 \over m} = 2\sqrt 2 (1 + {m^2})$

$ \Rightarrow {2 \over {{m^2}}} = 1 + {m^2} \Rightarrow m = \, \pm 1$

Taking one of the tangent $y = x + 4$

Point of tangency with ${y^2} = 4x$

${x^2} + 8 + 16 = 4x \Rightarrow x = 4$ & $y = 8$

$\therefore$ $Q(4,8)$

and for ${x^2} + {y^2} = 8$

$2{x^2} + 8x + 8 = 0$

${x^2} + 4x + 4 = 8 \Rightarrow x = - 2,y = 2 \Rightarrow R = ( - 2,2)$

${(QR)^2} = {6^2} + {6^2} = 72$

Let point of intersection be ($\alpha,1$)

$\alpha x^2+2b\alpha+c=0$ ..... (i)

and $d\alpha^2+2e\alpha+f=0$ .... (ii)

$\Rightarrow a\alpha^2+2\sqrt{ac}\alpha+c=0$ ($\because$ $b^2=ac$)

${\left( {\sqrt a \alpha + \sqrt c } \right)^2} = 0$

$\alpha = - \sqrt {{c \over a}} $

Put the value of $\alpha$ in (ii),

$d{c \over a} - 2e\sqrt {{c \over a}} + f = 0$

${d \over a} - {{2e} \over {\sqrt {ac} }} + {f \over c} = 0$

${d \over a} + {f \over c} = 2{e \over b}$

${d \over a},{e \over b},{f \over c}$ are in A.P.

Equation of normal

$y = - tx + 2.{1 \over {16}}t + {1 \over {16}}{t^3}$

$33 = {t \over 8} + {{{t^3}} \over {16}}$

$ \Rightarrow {t^3} + 2t = 528$

$t = 8$

$(a{t^2},2at) = (4,1)$

Distance from $x = - 2$