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

$5 x-4 y+3=0$

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

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

$ \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 $x = - 2$

$P \equiv \left( {{A \over {{m^2}}},{{2A} \over m}} \right)$ where $\left( {A = {3 \over 4},m = {{ - 1} \over 2}} \right)$

& $Q,R = \left( { \mp \,{{{a^2}{m_1}} \over {{a^2}m_1^2 + {b^2}}},{{ \mp \,.\,{b^2}} \over {\sqrt {{a^2}m_1^2 + {b^2}} }}} \right)$

Where ${a^2} = 4,{b^2} = 1$ and ${m_1} = 1$

$\therefore$ $P \equiv (3, - 3)$

$Q \equiv \left( {{{ - 4} \over {\sqrt 5 }},{{ - 1} \over {\sqrt 5 }}} \right)$ & $R\left( {{4 \over {\sqrt 5 }},{1 \over {\sqrt 5 }}} \right)$

Area $ = {1 \over 2}\left| {\matrix{ 3 & { - 3} & 1 \cr {{{ - 4} \over {\sqrt 5 }}} & {{{ - 1} \over {\sqrt 5 }}} & 1 \cr {{4 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} & 1 \cr } } \right| = {1 \over {10}}\left| {\matrix{ 3 & { - 3} & 1 \cr { - 4} & { - 1} & {\sqrt 5 } \cr 0 & 0 & {2\sqrt 5 } \cr } } \right|$

$ = {{2\sqrt 5 } \over {10}}( - 15) = 3\sqrt 5 $

Given parabola ${y^2} = 24x = 4.6x = 4ax$

$\therefore$ a = 6

If $y = mx + c$ is a tangent to the parabola then,

$c = {a \over m}$

$ \Rightarrow c = {6 \over m}$

Let midpoint of chord AB is $({x_1},{y_1})$

We know, locus of midpoint of chord for hyperbola,

$T = {S_1}$

$ \Rightarrow {{x{y_1} + {x_1}y} \over 2} - 2 = {x_1}{y_1} - 2$

$ \Rightarrow x{y_1} + {x_1}y = 2{x_1}{y_1}$

$ \Rightarrow y = \left( { - {{{y_1}} \over {{x_1}}}} \right)x + 2{y_1}$

This equation is the same tangent to the parabola.

Then comparing with $y = mx + c$ we get,

$c = 2{y_1},\,m = - {{{y_1}} \over {{x_1}}}$

$\therefore$ $2{y_1} = {6 \over { - {{{y_1}} \over {{x_1}}}}}$

$ \Rightarrow - y_1^2 = 3{x_1}$

$\therefore$ Locus of midpoint $({x_1},{y_1})$ is,

$ - {y^2} = 3x$

$ \therefore $ Length of latus rectum = 3

Equation of directrix is $x=\frac{3}{4}$

$H:{{{x^2}} \over 4} - {{{y^2}} \over 4} = 1$

Focus $(ae,0)$

$F\left( {2\sqrt 2 ,\,0} \right)$

$y = mx + c$ passes through (1, 0)

$0 = m + C$ ...... (i)

L is tangent to hyperbola

$C = \, \pm \,\sqrt {4{m^2} - 4} $

$ - m = \, \pm \,\sqrt {4{m^2} - 4} $

${m^2} = 4{m^2} - 4$

$m = {2 \over {\sqrt 3 }}$

$C = {{ - 2} \over {\sqrt 3 }}$

$T:y = {2 \over {\sqrt 3 }}x - {2 \over {\sqrt 3 }}$

$P:{y^2} = 4x$

${y^2} = 4\left( {{{\sqrt 3 y + 2} \over 2}} \right)$

${y^2} - 2\sqrt 3 y - 4 = 0$

Area

${1 \over 2}\left| {\matrix{ 0 & 0 \cr {{x_1}} & {{y_1}} \cr {2\sqrt 2 } & 0 \cr {{x_2}} & {{y_2}} \cr 0 & 0 \cr } } \right|$

$ = \left| {{1 \over 2}\left( { - 2\sqrt 2 {y_1} + 2\sqrt 2 {y_2}} \right)} \right|$

$ = \sqrt 2 \left| {{y_2} - {y_1}} \right| = \sqrt 2 \sqrt {{{({y_1} + {y_2})}^2} - 4{y_1}{y_2}} $

$ = \sqrt {56} $

$ = 2\sqrt {14} $

Equation of chord $PQ$

$ \Rightarrow y \times 1 = x - 3$

$ \Rightarrow x - y = 3$

For point P & Q

Intersection of PQ with parabola $P:(6,3)\,Q:(2, - 1)$

Slope of $RQ = - 1$ & Slope of $PQ = 1$

Therefore $\angle PQR = 90^\circ \Rightarrow $ Orthocentre is at $Q:(2, - 1)$

Equation of tangent at vertex : $L \equiv x + y - a = 0$

Focus : $F \equiv (a,a)$

Perpendicular distance of L from F

$ = \left| {{{a + a - a} \over {\sqrt 2 }}} \right| = \left| {{a \over {\sqrt 2 }}} \right|$

Length of latus rectum $ = 4\left| {{a \over {\sqrt 2 }}} \right|$

Given $4\,.\,\left| {{a \over {\sqrt 2 }}} \right| = 16$

$ \Rightarrow |a| = 4\sqrt 2 $

Centre of circle ${x^2} + {y^2} - 10x - 14y + 65 = 0$ is at (5, 7).

Let the equation of tangent to ${y^2} = 8x$ is

$yt = x + 2{t^2}$

which passes through (5, 7)

$7t = 5 + 2{t^2}$

$ \Rightarrow 2{t^2} - 7t + 5 = 0$

$t = 1,{5 \over 2}$

$A = 2 \times {1^2} \times 2 \times {\left( {{5 \over 2}} \right)^2} = 25$

$B = 2 \times 2 \times 1 \times 2 \times 2 \times {5 \over 2} = 40$

$A + B = 65$

$y = mx + 2a + {1 \over {{m^2}}}$ (Equation of normal to ${x^2} = 4ay$ in slope form) through $(1, - 1)$.

$4{m^3} + 6{m^2} + 1 = 0$

$ \Rightarrow m \simeq - 1.6$

Slope of normal $ \simeq {{ - 8} \over 5} = \tan \theta $

$ \Rightarrow \cos \theta \simeq {{ - 5} \over {\sqrt {89} }},\,\sin \theta \simeq {8 \over {\sqrt {89} }}$

${x_p} = 1 + \cos \theta \simeq 1 - {5 \over {\sqrt {89} }} \in \left( {{1 \over 4},{1 \over 2}} \right)$

Equation of tangent of slope $m$ to $y$ $= x^2$

$y = mx - {1 \over 4}{m^2}$

Equation of tangent of slope $m$ to $y = - {(x - 2)^2}$

$y = m(x - 2) + {1 \over 4}{m^2}$

If both equation represent the same line

${1 \over 4}{m^2} - 2m = - {1 \over 4}{m^2}$

$m = 0,\,4$

So, equation of tangent

$y = 4x - 4$

Given curve : ${y^2} - 2x - 2y = 1$.

Can be written as

${(y - 1)^2} = 2(x + 1)$

And, the given information can be plotted as shown in figure

Tangent at $A:2y - x - 5 = 0$ {using $T = 0$}

Intersection with $y = 1$ is $x = - 3$

Hence, point P is $( - 3,1)$

Taking advantage of symmetry

Area of $\Delta PAB = 2 \times {1 \over 2} \times (1 - ( - 3)) \times (3 - 1)$

= 8 sq. units

$\tan {\pi \over 4} = \left| {{{m - 3} \over {1 + 3m}}} \right|$

$ \Rightarrow {{m - 3} \over {1 + 3m}} = \pm \,1$

$ \Rightarrow {{m - 3} \over {1 + 3m}} = + 1$ and ${{m - 3} \over {1 + 3m}} = - 1$

$ \Rightarrow m - 3 = 1 + 3m$ and $m - 3 = - 1 - 3m$

$ \Rightarrow 2m = - 4$ and $4m = 2$

$m = - 2$ and $m = {1 \over 2}$

We know, Equation of tangent to the parabola ${y^2} = 4m$ is $y = mx + {a \over m}$ and point of contact is $\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)$

$\therefore$ Equation of tangent

$y = - 2x - {a \over 2}$

and $y = {x \over 2} + 2a$

$\therefore$ Point of contact A and B are

$A\left( {{a \over {{{( - 2)}^2}}},{{2a} \over { - 2}}} \right) = A\left( {{a \over 4}, - a} \right)$

$B\left( {{a \over {{{\left( {{1 \over 2}} \right)}^2}}},{{2a} \over {\left( {{1 \over 2}} \right)}}} \right) = B\left( {4a,4a} \right)$

As points A, B and S are colinear so area of triangle formed by those 3 points are zero.

Area of $\Delta$ABS = ${1 \over 2}\left| {\matrix{ {{a \over 4}} & { - a} & 1 \cr {4a} & {4a} & 1 \cr a & 0 & 1 \cr } } \right|$

$ = {a \over 4}(4a - 0) + a(4a - a) + 1(0 - 4{a^2})$

$ = {a^2} + 3{a^2} - 4{a^2} = 0$

$\therefore$ Area of triangle is independent of value of a.

So, for all value of a > 0 (already given a must be greater than 0) point A, B and S will be collinear.

As $\angle PRQ = {\pi \over 2}$

$\left( {{{{2 \over t}} \over {3 - {1 \over {{t^2}}}}}} \right)\,.\,\left( {{{ - 2t} \over {3 - {t^2}}}} \right) = - 1$

$ \Rightarrow t = \pm \,1$

$\therefore$ $P \equiv (1,2)$ & $Q(1, - 2)$

$\therefore$ for ellipse ${1 \over {{a^2}}} + {4 \over {{b^2}}} = 1$ and $ae = 1$

$ \Rightarrow {1 \over {{a^2}}} + {4 \over {{a^2}(1 - {e^2})}} = 1$

$ \Rightarrow 1 + {4 \over {(1 - {e^2})}} = {1 \over {{e^2}}}$

$ \Rightarrow (5 - {e^2}){e^2} = 1 - {e^2}$

$ \Rightarrow {e^4} - 6{e^2} + 1 = 0$

$ \Rightarrow {e^2} = {1 \over {3 - 2\sqrt 2 }} \Rightarrow {1 \over {{e^2}}} = 3 + 2\sqrt 2 $

Vertex of Parabola : (2, $-$1)

and directrix : 4x $-$ 3y = 21

Distance of vertex from the directrix

$a = \left| {{{8 + 3 - 21} \over {\sqrt {25} }}} \right| = 2$

$\therefore$ length of latus rectum = 4a = 8

Given vertex is (5, 4) and directrix 3x + y $-$ 29 = 0

Let foot of perpendicular of (5, 4) on directrix is (x₁, y₁)

${{{x_1} - 5} \over 3} = {{{y_1} - 4} \over 1} = {{ - ( - 10)} \over {10}}$

$\therefore$ $({x_1},\,{y_1}) \equiv (8,\,5)$

So, focus of parabola will be $S = (2,3)$

Let P(x, y) be any point on parabola, then

${(x - 2)^2} + {(y - 3)^2} = {{{{(3x + y - 29)}^2}} \over {10}}$

$ \Rightarrow 10({x^2} + {y^2} - 4x - 6y + 13) = 9{x^2} + {y^2} + 841 + 6xy - 58y - 174x$

$ \Rightarrow {x^2} + 9{y^2} - 6xy + 134x - 2y - 711 = 0$

and given parabola

${x^2} + a{y^2} + bxy + cx + dy + k = 0$

$\therefore$ a = 9, b = $-$6, c = 134, d = $-$2, k = $-$711

$\therefore$ $a + b + c + d + k = - 576$

Let P(at², 2at) where a = ${3 \over 2}$

T : yt = x + at² So point Q is $\left( { - a,\,at - {a \over t}} \right)$

N : y = $-$tx + 2at + at³ passes through (5, $-$8)

$-$8 = $-$5t + 3t + ${3 \over 2}$t³

$\Rightarrow$ 3t³ $-$ 4t + 16 = 0

$\Rightarrow$ (t + 2) (3t² $-$ 6t + 8) = 0

$\Rightarrow$ t = 2

So ordinate of point Q is $-$${9 \over 4}$.

$\because$ Line $y = kx + 4$ touches the parabola $y = x - {x^2}$.

So, $kx + 4 = x - {x^2} \Rightarrow {x^2} + (k - 1)x + 4 = 0$ has only one root

${(k - 1)^2} = 16 \Rightarrow k = 5$ or $-$3 but $k > 0$

So, $k = 5$.

And hence ${x^2} + 4x + 4 = 0 \Rightarrow x = - 2$

So, P($-$2, $-$6) and V is $\left( {{1 \over 2},{2 \over 4}} \right)$

Slope of $PV = {{{1 \over 4} + 6} \over {{1 \over 2} + 2}} = {5 \over 2}$

Let tangent to ${y^2} = x$ be

$y = mx + {1 \over {4m}}$

For it being tangent to circle.

$\left| {{{{1 \over 4}m} \over {\sqrt {1 + {m^2}} }}} \right| = \sqrt 2 $

$ \Rightarrow 32{m^4} + 32{m^2} - 1 = 0$

$ \Rightarrow {m^2} = {{ - 32 \pm \sqrt {{{(32)}^2} + 4(32)} } \over {64}}$

$ \Rightarrow 8{m_1}{m_2} = - 4 + 3\sqrt 2 $