Parabola

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

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

Equation of tangent at (2t$^2$, 4t) is

$\mathrm{ty}=\mathrm{x}+2 \mathrm{t}^2$

$\because$ It is passing through $(-4,0)$

$\therefore 0=-4+2 \mathrm{t}^2 \Rightarrow \mathrm{t}= \pm \sqrt{2}$

$\left.\begin{array}{rl} \therefore & \mathrm{A}_1=(4,4 \sqrt{2}) \\ & \mathrm{B}_1=(4,-4 \sqrt{2}) \end{array}\right\} \quad \begin{aligned} & \mathrm{OA}_1=\sqrt{48}=4 \sqrt{3} \\ & \mathrm{~A}_1 \mathrm{~B}_1=8 \sqrt{2} \end{aligned}$

Equation of altitude of $\Delta A_1 B_1 C_1$ drawn from $A_1$ is

$\begin{aligned} & y-4 \sqrt{2}=\sqrt{2}(x-4) \\ & \Rightarrow \sqrt{2} x-y=0 \quad \text{... (1)} \end{aligned}$

Equation of altitude of $\Delta A_1 B_1 C_1$ drawn from $C_1$ is

$\mathrm{x}=0\quad \text{... (2)}$

Solving (1) and (2) $\Rightarrow$ orthocentre is $(0,0)$

$\therefore$ correct options are (A), (C)

Vertex is the mid-point of focus and point of intersection directrix and axis

$ \begin{array}{l} \quad \quad\left(\frac{a+c}{2}, \frac{d+b}{2}\right)=(1, l) \\\\ \Rightarrow \quad a+c=2 \\\\ \text { and } \quad b+d=2 \\\\ \therefore a+b+c+d=4 \end{array} $

Equation of parabola having axis purlild to $ Y $-axis is

$ (x-h)^{2}=4 a(y-k) $

it passes through

$ (1,0):(1-h)^{2}=4 a(-k) $

$ (0,2): h^{2}=4 a(2-k) $

$ (-1,-1):(-1-h)^{2}=4 a(-1-k) $

On subtracting in Eqs (ii) and (iv), we get,

$1+h^{2}-2 h-1-h^{2}-2 h = -4 a K+4 a+4 a K-4 h=4 a $

$ \Rightarrow \quad h=-a $

From Eq. (iii), we get,

$ \begin{array}{rlrl} & a^{2} =4 a(2-k) \\\\ \therefore & a=4(2-k) \\\\ \Rightarrow & h=-8+4 k \end{array} $

On subtracting in Eqs (i) and (iii), we get

$ \begin{aligned} 1+h^{2}-2 h-h^{2} & =-4 a k-8 a+4 a k \\\\ 1-2 h & =-8 a \\\\ 1+2 a & =-8 a \\\\ -10 a & =1 \\\\ a & =-\frac{1}{10} \\\\ h & =\frac{1}{10} \\\\ 4 k & =h+8=\frac{1}{10}+8=\frac{81}{10} \\\\ k & =\frac{81}{40} \end{aligned} $

Put the value of $ B, h, k $ and a in above equation (i), we get

$ \begin{array}{l} \left(x-\frac{1}{10}\right)^{2}=4\left(-\frac{1}{10}\right)\left(y-\frac{81}{40}\right) \\\\ \frac{(10 x-1)^{2}}{10}=-\frac{4}{10}\left(\frac{40 y-81}{40}\right) \\\\ 100 x^{2}+1-20 x=-40 y+81 \\\\ 100 x^{2}-20 x+40 y-80=0 \\\\ 5 x^{2}-x+2 y-4=0 \\\\ a=5 \quad b=-1 c=2 \text { and } d=-4 \\\\ \frac{a d}{b c}=\frac{-20}{-2}=10 \end{array} $

Coordinate of point $ P \equiv\left(\frac{-a t^{2}}{4}, \frac{a t}{2}\right) $

$ \Rightarrow A \equiv\left(\frac{-a t^{2}}{4}, 0\right) \text { and } B \equiv\left(0, \frac{a t}{2}\right) $

Equation of $ A B \Rightarrow \frac{4 x}{-a t^{2}}+\frac{2 y}{a t}=1 $

Point $ Q\left(t_{1}\right)=\left(a t_{1}^{2}, 2 a t_{1}\right) $

Also, point $ Q $ lies on $ A B $

$ \begin{array}{l} \frac{4\left(a t_{1}^{2}\right)}{-a t^{2}}+\frac{2\left(2 a t_{1}\right)}{a t}=1 \\\\ \Rightarrow \quad-4 t_{1}^{2}+4 t_{1} t=t^{2} \\\\ \Rightarrow 4 t_{1}^{2}-4 t_{1} t+t^{2}=0 \\\\ \Rightarrow \quad\left(2 t_{1}-t\right)^{2}=0 \Rightarrow t_{1}=\frac{t}{2} \end{array} $

Focus of given parabola $ (0,3) $

Chord is also passes through $ (3,0) $

$ \Rightarrow \quad \frac{y-3}{x-0}=\frac{3-0}{0-3} $

$ y-3=-x \Rightarrow x+y=3 $

Put $ y=3-x $ in parabola $ x^{2}=12 y $

$ x^{2}=36-12 x $

$ x^{2}+12 x-36=0 $

$ (x-6)^{2}=0 \Rightarrow x=6,6 $

$ \Rightarrow $ Sum of reciprocal of Absicca

$ =\frac{1}{6}+\frac{1}{6}=\frac{1}{3} $

Given $ y^{2}=9 x $, normal drawn at $ P(9,9) $.

Slope of tangent $ 2 y \frac{d y}{d x}=9 $

$ \frac{d y}{d x}=\frac{9}{2 y} \Rightarrow\left(\frac{d y}{d x}\right)_{(9,9)}=\frac{1}{2} $

Slope of normal $ =-2 $

$ \left[\because m_{1} m_{2}=-1\right] $

Equation of normal

$ (y-9)=-2(x-9) \Rightarrow y=-2 x+27 $

Point $ Q(a, b) $ lies on both normal and parabola.

$ \begin{array}{l} \text { So, }(-2 x+27)^{2}=9 x \\\\ \Rightarrow 4 x^{2}-117 x+729=0 \\\\ x=\frac{117 \pm 45}{8} \Rightarrow x_{1}=20.25 \text { and } 9 \end{array} $

Now, $ x=20.25 $ they $ y=-2(20.25)+27 $

$ y=-135 $

Thus, $ Q(20.25,-135) $ i.e. $ (a, b) $

Now, $ 2 a+b \Rightarrow 2(20.25)+(13.5) $

$ \Rightarrow \quad 40.5-13.5 \Rightarrow 27 $

$ \therefore \quad 2 a+b=27 $

Given, $P$ and $Q$ are the extremities of a focal chord of the parabola

$ y^2=4 a x $

Also, $P=(9,9)$ and $Q=(p, q)$

General point on the parabola

$ y^2=4 a x \text { is }\left(a t^2, 2 a t\right) $

Now, $a t^2=9$ and $2 a t=9$

$ \Rightarrow a=\frac{9}{4} \text { and } t=2 $

$\because P$ and $Q$ are the extremities of a focal chord.

$ \begin{array}{ll} \therefore & p=a\left(-\frac{1}{t}\right)^2 \text { and } q=2 a\left(-\frac{1}{t}\right) \\\\ \Rightarrow & p=\frac{9}{4}\left(-\frac{1}{2}\right)^2 \text { and } q=2\left(\frac{9}{4}\right)\left(-\frac{1}{2}\right) \\\\ \Rightarrow & p=\frac{9}{16} \text { and } q=-\frac{9}{4} \\\\ \therefore & p-q=\frac{9}{16}-\left(-\frac{9}{4}\right)=\frac{9}{16}+\frac{9}{4}=\frac{45}{16} \end{array} $

The equation of normal to the parabola $y^2=4 a x$ is given by

$ y=m x-2 a m-a m^3 $

Here, $a=1$ and point is $(9,6)$

On putting $a=1, x=9$ and $y=6$ in the equation

$ \begin{aligned} & y=m x-2 a m-a m^3, \text { we get } \\\\ & 6=9 m-2 m-m^3 \Rightarrow m^3-7 m+6=0 \\\\ & \Rightarrow m=-3,1,2 \end{aligned} $

There are 3 distinct value of $m$.

So, the number of normals will be 3 .

We have, focus: $(2,3)$ directrix : $x-y+3=0$

Tangent drawn at the vertex of parabola is the equation of $A C$.

$\because A C$ is parallel to directrix.

$\therefore$ equation of $A C$ is $x-y+k=0 \ldots(i)$

and length of $B F=\frac{O F}{2}=\frac{1}{2}\left|\frac{2-3+3}{\sqrt{2}}\right|$

$B F=\frac{1}{\sqrt{2}}$

From Eq. (i), $B F=\left|\frac{2-3+k}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$

$\Rightarrow k=2$

$\therefore$ Equation of tangent is $x-y+2=0$

We have, $y^2=8 x \Rightarrow a=2$

Any tangent to parabola is

$ y=m x+\frac{a}{m} \Rightarrow m x-y+\frac{2}{m}=0 $

If it is a tangent to the circle $x^2+y^2=2$, then perpendicular from centre $(0,0)$ is equal to radius $\sqrt{2}$.

$ \begin{aligned} & \therefore \frac{\left|\frac{2}{m}\right|}{\sqrt{m^2+1}}=\sqrt{2} \Rightarrow \frac{4}{m^2}=2\left(m^2+1\right) \\\\ & \Rightarrow m^4+m^2-2=0 \\\\ & \Rightarrow\left(m^2+1\right)\left(m^2-1\right)=0 \Rightarrow m= \pm 1 \end{aligned} $

Hence, the common tangents are.

$ \begin{aligned} & y= \pm(x+2) \\\\ & x+y+2=0 \text { or } x-y+2=0 \end{aligned} $

Since, $-\frac{a}{b}>0$, so in this case,

equation of tangent : $x-y+2=0$

On comparing with $a x+b y+2=0$, we get $a=1, b=-1$

So, $3 a^2+2 b+1=3-2+1=2$

$ \begin{aligned} &\text { We have, }\\\\ &\begin{aligned} & 25\left[(x-2)^2+(y+5)^2\right]=(3 x+4 y-1)^2 \\\\ & \Rightarrow \quad(x-2)^2+(y+5)^2=\left[\frac{3 x+4 y-1}{\sqrt{3^2+4^2}}\right]^2 \\\\ & \Rightarrow \quad S L=L F . \end{aligned} \end{aligned} $

So, focus is $(2,-5)$

and directrix is $3 x+4 y-1=0 \quad \ldots \text { (i) }$

Equation of $P F$ is

$ y+5=\frac{4}{3}(x-2) \quad \left[\because m_{P F}=\frac{4}{3}\right] $

$ \begin{aligned} & 3 x+4 y-1=0 \\\\ & 4 x-3 y-23=0 \quad \ldots \text { (ii) } \end{aligned} $

On solving Eqs. (i) and (ii), we get $P\left(\frac{19}{5},-\frac{13}{5}\right) \Rightarrow$ vertex $O$ is $\left(\frac{29 *}{10},-\frac{38}{10}\right)$

Now, equation of $L L^{\prime}$ :

$ \begin{aligned} 3 x+4 y+14 & =0 \end{aligned} \quad \ldots \text { (ii) } $

$\left(\because m_{L L^{\prime}}=-\frac{3}{4}\right)$

Clearly, $\left(-\frac{2}{5},-\frac{16}{5}\right)$ on parabola as well as on $L L^{\prime}$. So; one end of latus rectum is $\left(-\frac{2}{5},-\frac{16}{5}\right)$

$\therefore$ Length of latus rectum $=2\left(\sqrt{\left(-\frac{2}{5}-2\right)^2+\left(-\frac{16}{5}+5\right)^2}\right)$ $=2 \times 3=6$

Let $A(a, b)$ be any point on the parabola and point $P(h, k)$ divides $F A$ in ratio $1: 2$

$ \begin{aligned} & \Rightarrow \quad h=\frac{a+6}{3} \text { and } k=\frac{b}{3} \\ & \Rightarrow \quad a=3 h-6 \text { and } b=3 k \end{aligned} $

$A(a, b)$ lies on parabola, so $b^2=12 a$

$ \begin{aligned} 9 k^2 & =12(3 h-6) \\ k^2 & =4(h-2) \end{aligned} $

Hence, locus of $P$ is $y^2=4(x-2)$.

To find the equation of the line that touches both parabolas $ y^2 = 4x $ and $ x^2 = -32y $, we start by considering the line in the form $ y = mx + c $.

For the parabola $ y^2 = 4x $, the condition for the line to be tangent is:

$ c = \frac{1}{m} $

Thus, the line can be expressed as:

$ y = mx + \frac{1}{m} $

For the parabola $ x^2 = -32y $, the condition for tangency is:

$ c = 8m^2 $

Therefore, the line in this case is:

$ y = mx + 8m^2 $

By equating the expressions for $ c $, we have:

$ \frac{1}{m} = 8m^2 $

Solving this equation for $ m $, we get:

$ m^3 = \frac{1}{8} $

$ m = \frac{1}{2} $

Substituting $ m = \frac{1}{2} $ back into the line equation, we obtain:

$ y = \frac{x}{2} + 2 $

Rewriting this equation in a standard form, we have:

$ x - 2y + 4 = 0 $

Thus, the equation of the tangent line is:

$ x - 2y + 4 = 0 $

The normal to the parabola $ y^2 = 4ax $ at the point $(x_1, y_1)$ can be expressed as:

$ y = \frac{-y_1}{2a}(x - x_1) + y_1 $

For the point $(2a, 2a\sqrt{2})$, the equation becomes:

$ \begin{aligned} y &= \frac{-2a\sqrt{2}}{2a}(x - 2a) + 2a\sqrt{2} \\ &= -\sqrt{2}(x - 2a) + 2a\sqrt{2} \\ y &= -\sqrt{2}x + 4a\sqrt{2} \end{aligned} $

Substituting this expression into $ y^2 = 4ax $, we get:

$ \begin{aligned} (-\sqrt{2}x + 4a\sqrt{2})^2 &= 4ax \\ 2(x - 4a)^2 &= 4ax \\ x^2 + 16a^2 - 8ax &= 2ax \\ x^2 + 16a^2 - 10ax &= 0 \end{aligned} $

This factors as:

$ \begin{aligned} x^2 - 8ax - 2ax + 16a^2 &= 0 \\ (x - 8a)(x - 2a) &= 0 \end{aligned} $

Thus, $ x = 8a $ or $ x = 2a $.

Now, for $ x = 8a $:

$ \begin{aligned} y &= -\sqrt{2} \times 8a + 4a\sqrt{2} \\ &= -4a\sqrt{2} \end{aligned} $

For $ x = 2a $:

$ \begin{aligned} y &= -\sqrt{2} \cdot 2a + 4a\sqrt{2} \\ &= 2a\sqrt{2} \end{aligned} $

The chord lies between points $(2a, 2a\sqrt{2})$ and $(8a, -4a\sqrt{2})$.

The slope $ m $ of this chord is:

$ m = \frac{-4a\sqrt{2} - 2a\sqrt{2}}{8a - 2a} = -\sqrt{2} $

Considering the line from the vertex $(0,0)$ to $(2a, 2a\sqrt{2})$, the slope is $\sqrt{2}$.

The angle $\theta$ between two lines with slopes $ m_1 $ and $ m_2 $ is calculated by:

$ \begin{aligned} \tan \theta &= \frac{m_1 - m_2}{1 + m_1 \cdot m_2} \\ &= \frac{\sqrt{2} + \sqrt{2}}{1 - \frac{2}{2}} \\ &= \frac{2\sqrt{2}}{0} \end{aligned} $

Since $\tan\theta = \tan \frac{\pi}{2}$, this implies:

$ \theta = \frac{\pi}{2} = 90^{\circ} $

Given the equation of the parabola:

$ y^2 = 12x $

By comparing this with the standard form $ y^2 = 4ax $, we find:

$ a = 3 $

Consider the coordinates of the points $ P $ and $ Q $ on the parabola:

$ P = (at_1^2, 2at_1) = (3t_1^2, 6t_1) $

$ Q = (at_2^2, 2at_2) = (3t_2^2, 6t_2) $

Given that the ordinates of $ P $ and $ Q $ are in the ratio 1:2, we have:

$ \frac{6t_2}{6t_1} = 2 $

This simplifies to:

$ t_2 = 2t_1 $

Let $ R(x, y) $ be the point of intersection of the normals at $ P $ and $ Q $.

For a point $t$ on the parabola $ y^2 = 4ax $, the point of intersection of the normals at $ t_1 $ and $ t_2 $ is given by:

$ \left(a(t_1^2 + t_2^2 + t_1t_2 + 2), -at_1t_2(t_1 + t_2)\right) $

Substituting the values, we have:

$ x = 3(t_1^2 + t_2^2 + t_1t_2 + 2) $

Substitute $ t_2 = 2t_1 $:

$ x = 3(t_1^2 + (2t_1)^2 + t_1 \cdot 2t_1 + 2) $

$ x = 3(t_1^2 + 4t_1^2 + 2t_1^2 + 2) = 3(7t_1^2 + 2) $

$ x = 21t_1^2 + 6 $

From this, we can express:

$ t_1^2 = \frac{x-6}{21} $

Now, calculate $ y $:

$ y = -3t_1 \cdot 2t_1(3t_1) $

$ y = -18t_1^3 $

Substituting $ t_1^2 = \frac{x-6}{21} $:

$ t_1 = \left(\frac{x-6}{21}\right)^{1/2} $

Therefore:

$ y = -18\left(\frac{x-6}{21}\right)^{3/2} $

Hence, the locus of the point of intersection of the normals at $ P $ and $ Q $ is:

$ y + 18\left(\frac{x-6}{21}\right)^{3/2} = 0 $

To find the common tangent to the circle $x^2+y^2=9$ and the parabola $y^2=8x$, we start with the equation of the tangent to the parabola. The general form for the tangent to $y^2 = 8x$ is given by:

$ y = mx + \frac{2}{m} $

This tangent must also touch the circle $x^2 + y^2 = 9$. The condition for this line to be tangent to the circle is:

$ \left|\frac{2}{m \sqrt{1+m^2}}\right| = 3 $

Simplifying the equation, we get:

$ \frac{4}{m^2(1+m^2)} = 9 $

Solving this, we arrive at the polynomial equation:

$ 9m^4 + 9m^2 - 4 = 0 $

By factoring, we find:

$ 9m^4 + 12m^2 - 3m^2 - 4 = 0 $

This can be factored further into:

$ (3m^2 + 4)(3m^2 - 1) = 0 $

We obtain:

$ m^2 = \frac{1}{3} $

Note that the solution $3m^2 + 4 = 0$ is rejected since it doesn't provide a valid, real number for $m^2$.

Thus, the tangent can be expressed as:

$ y = \pm \frac{1}{\sqrt{3}} x \pm 2 \sqrt{3} $

This translates to:

$ \sqrt{3} y = \pm(x + 6) $

The resulting equations are:

$ x \pm \sqrt{3} y + 6 = 0 $

Thus, the common tangents are:

$ x + \sqrt{3} y + 6 = 0 \quad \text{or} \quad x - \sqrt{3} y + 6 = 0 $

To find where the normal at the point $(2, -4)$ on the parabola $y^2 = 8x$ intersects the same parabola again, we proceed as follows:

Equation of the Normal:

The equation of the normal to the parabola $y^2 = 8x$ at a point $(x_1, y_1)$ is given by $y - y_1 = -\frac{2x_1}{y_1}(x - x_1)$.

Here, the point is $(2, -4)$, so $x_1 = 2$ and $y_1 = -4$.

Substituting into the normal equation:

$ y + 4 = -\frac{4}{-4}(x - 2) $

$ y + 4 = x - 2 $

$ x - y = 6 $

Thus, $x = y + 6$.

Substitute Back into the Parabola's Equation:

Substituting $x = y + 6$ into the parabola equation $y^2 = 8x$:

$ y^2 = 8(y + 6) $

$ y^2 = 8y + 48 $

$ y^2 - 8y - 48 = 0 $

Solve the Quadratic Equation:

Factor the quadratic:

$ (y - 12)(y + 4) = 0 $

The solutions are $y = 12$ and $y = -4$.

Find the Corresponding $x$ Values:

For $y = 12$, $x = 12 + 6 = 18$.

We already know for $y = -4$, $x = 2$, which is the original point.

Conclusion:

The other intersection point is $(\alpha, \beta) = (18, 12)$.

Thus, $\alpha + \beta = 18 + 12 = 30$.

To solve the problem of transforming the equation $ y^2 = 4ax $ when the coordinate axes are rotated by $ 45^\circ $ anticlockwise about the origin, we start by expressing the new coordinates $(x, y)$ in terms of the original coordinates $(X, Y)$.

Given that the coordinates are rotated $ 45^\circ $ anticlockwise, we can use the following transformation equations:

$ x = X \cos 45^\circ - Y \sin 45^\circ = \frac{X}{\sqrt{2}} - \frac{Y}{\sqrt{2}} = \frac{X-Y}{\sqrt{2}} $

$ y = X \sin 45^\circ + Y \cos 45^\circ = \frac{X}{\sqrt{2}} + \frac{Y}{\sqrt{2}} = \frac{X+Y}{\sqrt{2}} $

Substitute these expressions into the given equation $ y^2 = 4ax $:

$ \left(\frac{X+Y}{\sqrt{2}}\right)^2 = 4a \left(\frac{X-Y}{\sqrt{2}}\right) $

Simplifying the equation, we get:

$ \frac{(X+Y)^2}{2} = 4a \frac{(X-Y)}{\sqrt{2}} $

Multiply through by 2 to eliminate the denominator:

$ (X+Y)^2 = 4\sqrt{2}a(X-Y) $

Thus, the transformed equation is:

$ (x+y)^2 = 4\sqrt{2}a(x-y) $

This is the equation of the curve after rotating the axes by $ 45^\circ $ anticlockwise.

Focus of the parabola $y^2=42 x$ is ($\alpha$, 0)

Here, $a=\frac{1}{4}$ and $K=0$

So, equation of parabola is $y^2=x$

As, the line $x=2 y+3$ cuts the parabola

$ \begin{aligned} y^2 & =x \\ y^2 & =2 y+3 \\ \Rightarrow y^2-2 y-3 & =0 \\ y^2-3 y+y-3 & =0 \\ y & =3,-1 \end{aligned} $

If $y=3$, then $x=9$ and

If $y=-1$, then $x=1$

The two intersecting points

$P$ and $Q$ are $(9.3)$ and $(1,-1)$

$ \begin{aligned} P Q & =\sqrt{(9-1)^2+(3+1)^2} \\ & =\sqrt{64+16}=\sqrt{80} \\ & =4 \sqrt{5} \\ & =16 a \sqrt{5} \quad \left[\because a=\frac{1}{4}\right]\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 $x=\frac{3}{4}$