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

$5 x-4 y+3=0$

For the parabola

$ y^2 = 16x $

we compare it with the standard form

$ y^2 = 4ax $

So,

$ 4a = 16 \implies a = 4 $

Hence the focus is

$ (a,0) = (4,0) $

Now we find the tangent $T$ at the point $(64,32)$.

Since the point lies on the parabola,

$ 32^2 = 16 \cdot 64 $

so it is correct.

For the parabola $y^2 = 4ax$, the equation of tangent at point $(x_1,y_1)$ is

$ yy_1 = 2a(x+x_1) $

At $(64,32)$, we get

$ y \cdot 32 = 2\cdot 4(x+64) $

$ 32y = 8x + 512 $

$ x - 4y + 64 = 0 $

Now slope of this tangent $T$ is

$ x - 4y + 64 = 0 \implies y = \frac{1}{4}x + 16 $

So slope of $T$ is

$ m_T = \frac14 $

Since $L$ is perpendicular to $T$, its slope must be

$ m_L = -4 $

Now let $(x_1,y_1)$ be another point on the parabola where tangent $L$ is drawn.

For parabola $y^2 = 4ax$, slope of tangent at point $(x_1,y_1)$ is

$ \frac{2a}{y_1} $

because tangent equation is

$ yy_1 = 2a(x+x_1) $

and rewriting,

$ y = \frac{2a}{y_1}x + \frac{2ax_1}{y_1} $

So here,

$ m_L = \frac{2a}{y_1} = \frac{8}{y_1} $

Given that $m_L = -4$, we get

$ \frac{8}{y_1} = -4 $

$ y_1 = -2 $

Since $(x_1,y_1)$ lies on the parabola,

$ y_1^2 = 16x_1 $

$ (-2)^2 = 16x_1 $

$ 4 = 16x_1 $

$ x_1 = \frac14 $

So the point is

$ \left(\frac14,-2\right) $

Now distance from this point to the focus $(4,0)$ is

$ \sqrt{\left(4-\frac14\right)^2 + (0+2)^2} $

$ = \sqrt{\left(\frac{15}{4}\right)^2 + 4} $

$ = \sqrt{\frac{225}{16} + \frac{64}{16}} $

$ = \sqrt{\frac{289}{16}} $

$ = \frac{17}{4} $

Therefore, the required distance is

$ \boxed{\frac{17}{4}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

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

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.

$ y^2=x $

chord with given mid point

$ \begin{aligned} & T=S_1 \\ & y k-\frac{x+h}{2}=k^2-h \\ & 2 k y-x-h=2 k^2-2 h \end{aligned} $

Now,

$\begin{aligned} & \left.A=\int_{y_1}^{y_2}\left(2 k y-2 k^2+h\right)-y^2\right) d y=\frac{4}{3} \\ & \left(k y^2+\left(h-2 k^2\right) y-\frac{y^3}{3}\right)_{y_1}^{y_2}=\frac{4}{3} \\ & k\left(y_2^2-y_1^2\right)+\left(h-2 k^2\right)\left(y_2-y_1\right)-\frac{1}{3}\left(y_2^3-y_1^3\right)=\frac{4}{3} \\ & \left(y_2-y_1\right)\left[k-2 k+h-2 k^2-\frac{1}{3}\left(4 k^2-2 k^2+h\right)\right]=\frac{4}{3} \\ & 2\left(h-k^2\right)^{1 / 2}\left[2 h-2 k^2\right]=4 \\ & \left(h-k^2\right)^{3 / 2}=1 \\ & x-y^2=1 \Rightarrow y^2=x-1\end{aligned}$

$\begin{aligned} & A=\int_1^4(\sqrt{x}-\sqrt{x-1}) d x \\ & =\frac{2}{3}\left[x^{3 / 2}-(x-1)^{3 / 2}\right]_1^4=\frac{2}{3}[8-3 \sqrt{3}-1] \\ & =\frac{14}{3}-2 \sqrt{3}\end{aligned}$

Solving chord and parabola

$ \begin{aligned} & y k-\frac{y+h}{2}=k^2-h \\ & 2 k y-y^2-h=2 k^2-2 h \\ & y^2-2 k y+2 k^2-h=0 \\ & y_1+y_2=2 k \\ & y_1 y_2=2 k^2-h \\ & y_2-y_1=\sqrt{4 k^2-8 k^2+4 x} \\ & =2 \sqrt{h-k^2} \end{aligned} $

Let $P \equiv(\alpha, 4 \alpha)$

Equation of tangent to $y^2=4 x$

$ \begin{array}{ll} & y=m x+\frac{1}{m} \\ & 4 \alpha=m \alpha+\frac{1}{m} \\ & m^2 \alpha-4 \alpha m+1=0 \\ & m_1+m_2=4 \\ & m_1 m_2=\frac{1}{\alpha} \\ \Rightarrow \quad & \tan \frac{\pi}{3}=\frac{\sqrt{\left(m_1+m_2\right)^2-4 m_1 m_2}}{\left|1+\frac{1}{\alpha}\right|} \\ \Rightarrow \quad & \sqrt{3}=\frac{\sqrt{16-\frac{4}{\alpha}}}{\left|1+\frac{1}{\alpha}\right|} \Rightarrow 3\left(1+\frac{1}{\alpha}\right)^2=16-\frac{4}{\alpha} \\ \Rightarrow \quad & \frac{3(\alpha+1)^2}{\alpha^2}=\frac{16 \alpha-4}{\alpha} \\ \Rightarrow \quad & 3(\alpha+1)^2=\alpha(16 \alpha-4) \\ \Rightarrow \quad & 3 \alpha^2+6 \alpha+3=16 \alpha^2-4 \alpha \\ \Rightarrow \quad & 13 \alpha^2-10 \alpha-3=0 \\ \Rightarrow \quad & \alpha_1+\alpha_2=\frac{10}{13} \end{array} $

⇒ Sum of abscissa of all such point is $\frac{10}{3}$.

Equation of normal

at point (am²,-2am)

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

Here $a=1$

Let point $p$ be ( $\alpha, \beta$ )

$ \begin{aligned} & \Rightarrow \quad \beta=m \alpha-2 m-m^3 . \\ & \Rightarrow \quad m^3+m(2-\alpha)+\beta=0 \\ & \Rightarrow \quad m_1+m_2+m_3=0 \\ & \sum m_1 m_2=2-\alpha \end{aligned} $

Cyclic

$ \begin{aligned} & m_1 m_2 m_3=-\beta \\ & A \equiv\left(m_1^2,-2 m_1\right) \\ & B \equiv\left(m_2^2,-2 m_2\right) \\ & C \equiv\left(m_3^2,-2 m_3\right) \\ & G \equiv(2,0) \\ & 2=\frac{m_1^2+m_2^2+m_3^2}{3} \\ & 6=\left(m_1+m_2+m_3\right)^2-2 \sum m_1 m_2 \\ & 6=0-2(2-\alpha) \\ & 6=-4+2 \alpha \\ & \alpha=5 \end{aligned} $

Let $P=\left(a t_1^2, 2 a t_1\right) \quad Q=\left(a t_2^2, 2 a t_2\right)$

Normal at $P$ intersect the parabola at $Q$

$ \Rightarrow \quad t_2=-t_1-\frac{2}{t_1} $

Slope of $O P=\frac{2}{t_1}$

Slope of $O Q=\frac{\mathbf{2}}{t_2}$

$ \begin{array}{ll} \Rightarrow & \frac{2}{t_1} \times \frac{2}{t_2}=-1 \Rightarrow t_1 t_2=-4 \\ \Rightarrow & t_1 t_2=-t_1^2-2 \\ \Rightarrow & t_1= \pm \sqrt{2} \\ \Rightarrow & t_1=\sqrt{2} \\ \Rightarrow & t_2=-\sqrt{2}-\frac{2}{\sqrt{2}}=-2 \sqrt{2} \\ \Rightarrow & Q \equiv\left(a t_2^2, 2 a t_2\right) \equiv(8 a,-4 \sqrt{2} a) \end{array} $

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \equiv\left(8 \times \frac{5}{4}-4 \sqrt{2} \times \frac{5}{4}\right) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \equiv(10,-5 \sqrt{2}) \end{aligned} $

$(y-2)^2=3(x-1)$

Let $y-2=Y$ and $x-1=X$

$ \Rightarrow \quad Y^2=3 X . $

and coordinates of end-point of $L, R$

$ X=a $

and $\quad Y= \pm 2 a$

$ \begin{aligned} \Rightarrow & x-1=\frac{3}{4} \text { and } y-2= \pm \frac{3}{2} \\ & x=\frac{7}{4} \text { and } y=2 \pm \frac{3}{2} \\ \Rightarrow & p=\frac{7}{4}, \quad q=\frac{7}{2} \end{aligned} $

Equation of tangent to $(y-2)^2=3(x-1)$ at $\left(\frac{7}{4}, \frac{7}{2}\right)$

$ \begin{aligned} &\begin{aligned} & y-\frac{7}{2}=m\left(x-\frac{7}{4}\right) \\ & m=\left(\frac{d y}{d x}\right)_{\frac{7}{4}, \frac{7}{2}} \\ & 2\left(y-2 \frac{d y}{d x}=3\right. \\ & \frac{d y}{d x}=\frac{3}{2\left(\frac{7}{2}-2\right)}=\frac{3 \times 2}{2 \times 3}=1 \end{aligned}\\ &\text { Equation of tangent, } y-\frac{7}{2}=x-\frac{7}{4}\\ &\begin{array}{ll} & 4 y-14=4 x-7 \\ \Rightarrow & 4 x-4 y+7=0 \end{array} \end{aligned} $

$ \begin{aligned} & y^2=7 x \\ & 4 a=7 \Rightarrow a=\frac{7}{4} \end{aligned} $

Point $(2,0)$

Equation of normal at $\left(a t^2, 2 a t\right)$

$ \begin{aligned} \Rightarrow & & y+t x & =2 a t+a t^3 \\ \Rightarrow & & 0+2 t & =2 a t+a t^3 \\ \Rightarrow & & 2 t & =\frac{7}{2} t+\frac{7}{4} t^3 \end{aligned} $

$ \begin{aligned} \Rightarrow \quad t & =0 \\ t^2 & =\frac{4}{7}\left(2-\frac{7}{2}\right)=\frac{-6}{7} \end{aligned} $

Not possible.

∴ Only one normal.

$y^2=11 x, 4 a=11$

∴ Equation of tangent

$ \begin{aligned} y & =m x+\frac{a}{m} \\ \Rightarrow \quad y & =m x+\frac{11}{4 m} \end{aligned} $

At (1, 4)

$ \begin{aligned} & 4=m+\frac{11}{4 m} \\ \Rightarrow \quad & 16 m=4 m^2+11 \\ \Rightarrow \quad & 4 m^2-16 m+11=0 \\ \Rightarrow \quad & m_1+m_2=4 \\ \Rightarrow \quad & m_1 m_2=\frac{11}{4} \\ & \left(m_1+m_2\right)^2=m_1^2+m_2^2+2 m_1 m_2 \\ \Rightarrow \quad & 16-\frac{11}{2}=m_1^2+m_2^2 \\ \Rightarrow \quad & 21=2\left(m_1^2+m_2^2\right) \end{aligned} $

Given, $y^2=3 x$

Differential w.r.t. $x$, we get

$ 2 y \frac{d y}{d x}=3 \Rightarrow \frac{d y}{d x}=\frac{3}{2 y} $

Slope of tangent at $\left(x_1, y_1\right)=m_t=\frac{3}{2 y_1}$

Slope of normal $=m_n=\frac{-2 y_1}{3}$

∴ Equation of normal

$ \begin{aligned} & y-y_1=\frac{-2 y_1}{3}\left(x-x_1\right) \\ & \text { Put }\left(x_1, y_1\right)=\left(\frac{3}{4}, \frac{3}{2}\right) \\ & \Rightarrow \quad y-\frac{3}{2}=-\frac{2\left(\frac{3}{2}\right)}{3} \cdot\left(x-\frac{3}{4}\right) \\ & \Rightarrow \quad y+x=\frac{3}{4}+\frac{3}{2}=\frac{9}{4} \end{aligned} $

$ \Rightarrow \quad 4 x+4 y=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

From $Q(3,3)$

$ \begin{aligned} y-3 & =\frac{-2 \times 3}{3}(x-3) \\ \Rightarrow \quad y+2 x & =9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{array}{ll} & x=\frac{27}{4}, y=\frac{-9}{2} \\ \therefore & R \equiv\left(\frac{27}{4}, \frac{-9}{2}\right) \end{array} $

$y^2=9 x$

$ \Rightarrow \quad a=\frac{9}{4} $

Equation of tangent to parabola

$ \begin{aligned} y & =m x+\frac{a}{m} \\ \Rightarrow \quad y & =m x+\frac{9}{4 m} \end{aligned} $

Since, tangents are drawn from point (1, 5)

$ \begin{array}{lc} \Rightarrow & \quad 5=m+\frac{9}{4 m} \\ \Rightarrow & 4 m^2-20 m+9=0 \\ \Rightarrow & m=\frac{20 \pm \sqrt{400-144}}{8}=\frac{20 \pm 16}{2} \\ \Rightarrow & m=\frac{9}{2}, \frac{1}{2} \end{array} $

$ \begin{aligned} & \therefore \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{\frac{9}{2}-\frac{1}{2}}{1+\frac{9}{4}}\right| \\ & \quad=\frac{16}{13} \approx 1.23 \\ & \Rightarrow \quad 1<\frac{16}{13}<\sqrt{3} \\ & \Rightarrow \quad \tan \frac{\pi}{4}<\tan \theta<\tan \frac{\pi}{3} \\ & \therefore \quad \frac{\pi}{4}<\theta<\frac{\pi}{3} . \end{aligned} $

Given, equation of parabola

$ \begin{aligned} & y=x^2-3 x+2 \\ & \Rightarrow \quad y=x^2-3 x+\frac{9}{4}-\frac{9}{4}+2 \\ & \Rightarrow \quad y=\left(x-\frac{3}{2}\right)^2-\frac{1}{4} \\ & \Rightarrow \quad y+\frac{1}{4}=\left(x-\frac{3}{2}\right)^2 \end{aligned} $

which is in the form of

$ \begin{aligned} & (x-h)^2=4 a(y-k) \\ & \therefore \operatorname{Vertex}(h, k)=\left(\frac{3}{2}, \frac{-1}{4}\right) \\ & \therefore \quad Z=\left(\frac{3}{2}, \frac{-1}{2}\right) \end{aligned} $

And $4 a=1 \Rightarrow a=\frac{1}{4}$

$ \begin{aligned} \therefore \text { Focus } S & =(h, k+a) \\ & =\left(\frac{3}{2}, \frac{-1}{4}+\frac{1}{4}\right)=\left(\frac{3}{2}, 0\right) \end{aligned} $

Tangent parallel to the $X$-axis

$\therefore$ Its slope is 0

Now, $y=x^2-3 x+2$

$ \begin{array}{ll} \Rightarrow & \frac{d y}{d x}=2 x-3=0 \\ \Rightarrow & 2 x=3 \Rightarrow x=\frac{3}{2} \end{array} $

$ \begin{array}{rlrl} \therefore & y =\left(\frac{3}{2}\right)^2-3\left(\frac{3}{2}\right)+2 \\ & =\frac{9}{4}-\frac{9}{2}+2 \\ \Rightarrow & y =\frac{9-18+8}{4}=\frac{-1}{4} \\ \therefore & Q =\left(\frac{3}{2},-\frac{1}{4}\right) . \end{array} $

$ \begin{aligned} &\text { The end points of the latus rectum are }\\ &\begin{aligned} (h \pm 2 a, k+a) & \\ & =\left(\frac{3}{2} \pm \frac{1}{2}, 0\right) \\ & =\left(\frac{3}{2}+\frac{1}{2}, 0\right) \text { and }\left(\frac{3}{2}-\frac{1}{2}, 0\right) \\ & =(2,0) \text { and }(1,0) \\ \therefore \quad P & =(2,0) \text { or }(1,0) . \end{aligned} \end{aligned} $

Given, equations of parabola

$ \begin{aligned} & & y^2 & =12 x \\ \therefore & & 4 a & =12 \Rightarrow a=3 \end{aligned} $

$\therefore$ Parametric equation of parabola

$ x=3 t^3, y=2(3) t=6 t $

Any point on the parabola is $P\left(3 t^2, 6 t\right)$. and focus of parabola is $F(3,0)$ Let $Q(x, y)$ be the point that divides the line segment $F P$ in the ratio $m: n$

$ \begin{array}{ll} \therefore & x=\frac{m\left(3 t^2\right)+n(3)}{m+n}=\frac{3 m t^2+3 n}{m+n} \\ \text { And } & y=\frac{m(6 t)+n(0)}{m+n}=\frac{6 m t}{m+n} \\ \Rightarrow & y(m+n)=6 m t \\ \Rightarrow & t=\frac{y(m+n)}{6 m} \\ \therefore & x=\frac{3 m\left(\frac{y(m+n)}{6 m}\right)^2+3 n}{m+n} \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x=\frac{\frac{y^2(m+n)^2}{12 m}+3 n}{m+n} \\ & \Rightarrow \quad x=\frac{36 m n+y^2(m+n)^2}{12 m(m+n)} \\ & \Rightarrow \quad 12 m(m+n) x=36 m n+y^2(m+n)^2 \\ & \Rightarrow \quad y^2(m+n)^2=12 m(m+n) x-36 m n \\ & \Rightarrow \quad y^2=\frac{12 m(m+n)}{(m+n)} x-\frac{36 m n}{(m+n)^2} \\ & \Rightarrow \quad y^2=\frac{12 m}{m+n} x-\frac{36 m n}{(m+n)^2} \end{aligned}\\ &\text { Which is in the form of } Y^2=4 A X+B\\ &\therefore \text { Length of latusrectum }=\frac{12 m}{m+n} \end{aligned} $

Given parabola is $y^2=32 x$,

So, $4 a=32$

$ \Rightarrow \quad a=8 $

The point $P$ is $(8,16)$. This in parametric form as $\left(a t^2, 2 a t\right)$

So, $8 t^2=8$

$ \Rightarrow \quad t^2=1 \Rightarrow t= \pm 1 $

And, $2 a t=2(8) t=16 t=16$

$ \Rightarrow \quad t=1 $

So, the perimeter for $P$ is $t_1=1$.

The equation of the normal to the parabola at a point $P\left(a t_1^2, 2 a t_1\right)$ is

$ \begin{aligned} & y+t_1 x=2 a t_1+a t_1^3 \\ & \Rightarrow \quad y+(1) x=2(8)(1)+8(1)^3 \\ & \Rightarrow \quad y+x=16+8=24 \\ & \Rightarrow \quad y=24-x \end{aligned} $

Substitute this into parabola equation

$ \begin{aligned} & y^2=32 x \\ & \Rightarrow \quad(24-x)^2=32 x \\ & \Rightarrow \quad 576-48 x+x^2=32 x \\ & \Rightarrow \quad x^2-80 x+576=0 \end{aligned} $

Since, $x=8$ is one root corresponding to point $P$. Let the other root be $x_Q$.

Using the sum of roots, $8+x_Q=80$

$ \begin{array}{lrl} \Rightarrow x_Q =72 \\ \text { So, } y =24-x_Q \\ =24-72=-48 \end{array} $

So, the coordinates of $Q$ are $(72,-48)$

Now, the equation of the tangent to parabola $y^2=4 a x$ at a point $\left(x_1, y_1\right)$ is

$ \begin{aligned} & y y_1=2 a\left(x+x_1\right) \\ & \Rightarrow y(-48)=2(8)(x+72)=16(x+72 \\ & \Rightarrow-3 y=x+72 \Rightarrow x+3 y+72=0 \end{aligned} $

So, the equation of the tangent drawn at $Q$ to the parabola is $x+3 y+72=0$

Given equation is

$ \begin{array}{ll} & x^2-2 x-4 y+5=0 \\ \Rightarrow & (x-1)^2-4 y+4=0 \\ \Rightarrow & (x-1)^2=4(y-1) \end{array} $

So, $(h, k)=(1,1)$ are the vertex

And $4 a=4$

$ \Rightarrow a=1 $

Since, the parabola open upwards (because the $y$ term is positive and $x$ term is squared), the focus is $(h, k+a)=(1,1+1)=(1,2)$ and the directrix is the line $y=k-a=1-1=0$, which is the $X$-axis.

Since, focal distance of a point on a parabola is its distance from the directrix.

The point $(5,5)$ and directrix is $y=0$.

The distance from a point $\left(x_0, y_0\right)$ to a horizontal line $y=c$ is $\left|y_0-c\right|$.

$\therefore$ Focal distance $=|5-0|=5$

So, the focal distance of the point $(5,5)$ on the parabola $x^2-2 x-4 y+5=0$ is 5 .

$\because y^2=4 x$

So, $\quad a=1$

$\therefore$ The equation of the normal

$ \begin{aligned} \quad y & =m x-2 a m-a m^3 \\ \Rightarrow \quad y & =m x-2 m-m^3 \end{aligned} $

which passes through $(5,2)$

$ \begin{array}{ll} \therefore & 2=5 m-2 m-m^3 \\ \Rightarrow & m^3-3 m+2=0 \end{array} $

and the given normal is

$ y=x-3 $

$\therefore \quad m=1$, put the value

$ (1)^3-3 \times 1+2=0 $

Thus, $m=1$ is a root.

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

So, there are two normals with slopes 1 and -2

$\because$ The given normal has slope 1

So, the other normal has slope $=-2$

$ \begin{aligned} & \text { Given, } y^2=2 p x \Rightarrow 4 a=2 p \\ & \qquad a=\frac{p}{2} \\ & \therefore \text { Focus }\left(\frac{p}{2}, 0\right) \text { directrix } x+\frac{p}{2}=0 \end{aligned} $

∴ Equation of circle

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

∴ Point of intersection circle and $y^2=2 p x$.

$ \begin{aligned} & \left(x-\frac{p}{2}\right)^2+2 p x=p^2 \\ \Rightarrow & x^2+\frac{p^2}{4}-p x+2 p x=p^2 \\ \Rightarrow & x^2+p x=\frac{3 p^2}{4} \\ \Rightarrow & 4 x^2+4 p x-3 p^2=0 \\ \Rightarrow & 4 x^2+6 p x-2 p x-3 p^2=0 \\ \Rightarrow & 2 x(2 x+3 p)-p(2 x+3 p)=0 \\ \Rightarrow & (2 x-p)(2 x+3 p)=0 \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & x=\frac{p}{2}, \frac{-3 p}{2} \\ \therefore & x=\frac{p}{2} \\ & y^2=p^2 \Rightarrow y= \pm p \end{array}\\ &\text { ∴ Point of intersection, }\left(\frac{p}{2}, p\right),\left(\frac{p}{2},-p\right) \end{aligned} $

Let $P\left(t_1^2, 2 t_1\right)$ and $Q\left(t_2^2, 2 t_2\right)$ be 2 points

$ \because M \text { divides } P Q \text { in } 1: 2 $

$ \begin{aligned} & \therefore \frac{2 t_1^2+t_2^2}{3}=h, \frac{4 t_1+2 t_2}{3}=k \\ & \because \quad m_{P Q}=2 \\ & \therefore \frac{2\left(t_2-t_1\right)}{t_2^2-t_1^2}=2 \\ & \Rightarrow t_2+t_1=1 \Rightarrow t_2=1-t_1 \\ & \because k=\frac{4 t_1+2\left(1-t_1\right)}{3} \Rightarrow 3 k=2+2 t_1 \\ & \quad \frac{3 k-2}{2}=t_1 \end{aligned} $

$ \begin{aligned} & \text { and } \quad t_2=1-\frac{3 k-2}{2} \Rightarrow \frac{4-3 k}{2} \\ & \therefore \quad 3 h=2\left(\frac{3 k-2}{2}\right)^2+\left(\frac{4-3 k}{2}\right)^2 \\ & \Rightarrow 12 h=2\left(9 k^2+4-12 k\right)+\left(9 k^2+16-24 k\right) \\ & =27 k^2+24-48 k \\ & \Rightarrow 12 h=27 k^2-48 k+24 \\ & \Rightarrow 4 h=9 k^2-16 k+8 \end{aligned} $

$ \begin{aligned} & \Rightarrow 4 h=9\left(k^2-\frac{16}{9} k+\left(\frac{8}{9}\right)^2\right)-9\left(\frac{8}{9}\right)^2+8 \\ & \Rightarrow 4 h+\frac{64}{9}-8=9\left(k-\frac{8}{9}\right)^2 \\ & \Rightarrow 4 h-\frac{8}{9}=9\left(k-\frac{8}{9}\right)^2 \\ & \Rightarrow 4\left(h-\frac{2}{9}\right)=9\left(k-\frac{8}{9}\right)^2 \\ & \therefore \text { Locus } \\ & \Rightarrow\left(x-\frac{2}{9}\right)=\frac{9}{4}\left(y-\frac{8}{9}\right)^2 \\ & \therefore \text { Vertex }\left(\frac{2}{9}, \frac{8}{9}\right) \end{aligned} $

$y^2=15 x$

Equation of normal at $\left(\frac{15}{2}, \frac{15}{\sqrt{2}}\right)$

$ \begin{aligned} &\begin{aligned} & & 2 y \frac{d y}{d x} & =15 \\ & \Rightarrow & \frac{d y}{d x} & =\frac{15}{2 y} \\ & \Rightarrow & m_N & =\frac{-1}{\frac{d y}{d x}}=\frac{-2 y}{15} \quad\left(\because y=\frac{15}{\sqrt{2}}\right) \\ & \Rightarrow & m_N & =\frac{-2 \times 15 / \sqrt{2}}{15}=-\sqrt{2} \end{aligned}\\ &\text { Equation of normal }\\ &\begin{aligned} & y-\frac{15}{\sqrt{2}}=-\sqrt{2}\left(x-\frac{15}{2}\right) \\ \Rightarrow & \sqrt{2} y-15=-2\left(x-\frac{15}{2}\right) \\ \Rightarrow & \sqrt{2} y-15=-2 x+15 \\ \Rightarrow & 2 x+\sqrt{2} y=30 \\ \Rightarrow & \frac{2 x+\sqrt{2} y}{30}=1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Homogenisation of } y^2=15 x \text { w.r.t } L\\ &\begin{array}{ll} & y^2=15 x\left(\frac{2 x+\sqrt{2} y}{30}\right) \\ \Rightarrow \quad & 2 y^2=2 x^2+2 \sqrt{2} x y \\ \Rightarrow \quad & 2 x^2+2 \sqrt{2} x y-2 y^2=0 \\ \because \quad & a+b=0 \\ \therefore \quad & \theta=\pi / 2 \\ \text { Now, } & \sin \frac{\theta}{3}+\cos \frac{2 \theta}{3}-\sec \frac{4 \theta}{3} \\ & =\sin \frac{\pi}{6}+\cos \frac{\pi}{3}-\sec \left(\frac{2 \pi}{3}\right) \\ & =1 / 2+1 / 2+2=3 \end{array} \end{aligned} $

Given, parabola is $y^2=x$ and $t_1=2 t_2=-4$ and $t_3=6$.

So, intersection of tangents at $t_1$ and $t_2$, then tangents are $2 y=x+1$ and $-4 y=x+4$

Solving these two equations, we get $y=\left(\frac{-1}{2}\right)$ and $x=-2$

So, point $L=\left(-2,-\frac{1}{2}\right)$

Now, intersection of tangents at $t_2=-4$ and $t_3=6$, then tangents are

$ -4 y=x+4,6 y=x+9 $

Solving these two,

So, print $M=\left(-6, \frac{1}{2}\right)$

Now, intersection of tangents at $t_1=2$ and $t_3=6$, so tangents are $2 y=x+1$ and $6 y=x+9$

Solving these two equation, we get $x=3, y=2$

So, point $N=(3,2)$

Area of triangle with vertices $L\left(-2,-\frac{1}{2}\right), M\left(-6, \frac{1}{2}\right)$ and $N(3,2)$ is

$ \begin{aligned} \text { Area } & =\left|\frac{1}{2}\right| \begin{array}{ccc} -2 & -\frac{1}{2} & 1 \\ -6 & \frac{1}{2} & 1 \\ 3 & 2 & 1 \end{array}| | \\ & =\left|\frac{-15}{2}\right|=7.5 \end{aligned} $