Parabola

$\because y^2=4 x$

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} $

$ y^2=8 x, a=2 $

Equation of tangent

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

at ( 1,3 )

$ \begin{aligned} 3 & =m+\frac{2}{m} \Rightarrow m^2-3 m+2=0 \\ \Rightarrow \quad m & =1,2 \end{aligned} $

So, equation of tangent (put point $(1,3)$ on $y=m c+c$ )

$ \begin{aligned} y & =x+c, y=2 x+c \\ \Rightarrow \quad y & =x+2 y=2 x+1 \end{aligned} $

to find coordinates of $A$ and $B$, put the line on parabola.

$ \begin{aligned} & (x+2)^2=8 x \Rightarrow x^2-4 x+4=0 \\ & \Rightarrow \quad x=2,-2 \\ & \text { at } \quad x=2, y=\sqrt{8 \times 2}=4 \end{aligned} $

So, $\quad A=(2,4)$

and $(2 x+1)^2=8 x$

$\Rightarrow 4 x^2-4 x+1=0$

$\Rightarrow x=\frac{1}{2},-\frac{1}{2}$

at $x=\frac{1}{2}, y=2$

So, $B=\left(\frac{1}{2}, 2\right)$

Now, area of $\triangle P A B$

$ \begin{aligned} A & =\frac{1}{2}\left|\begin{array}{lll} 1 & 3 & 1 \\ 2 & 4 & 1 \\ \frac{1}{2} & 2 & 1 \end{array}\right| \\ & =\left\lvert\, \frac{1}{2}\left(\left.1(4-2)-3\left(2-\frac{1}{2}\right)+1(4-2) \right\rvert\,\right.\right. \\ & =\left|\frac{1}{2}\left(2-\frac{9}{2}+2\right)\right|=\left|\frac{1}{2}\left(4-\frac{9}{2}\right)\right| \\ & =\left|\frac{1}{2}\left(-\frac{1}{2}\right)\right|=\frac{1}{4} \end{aligned} $

Given equation of parabola is

$ y^2=16 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Comparing it with $y^2=4 a x$ we get, $a=4$

Let coordinates of end points of focal chord of a parabola (i) is, $\left(a t^2, 2 a t\right)$ and $\left(\frac{a}{t^2}, \frac{-2 a}{t}\right)$

$ \begin{aligned} & =\text { its length }=\sqrt{\left(a t^2-\frac{a}{t^2}\right)^2+\left(2 a t+\frac{2 a}{t}\right)^2} \text { (given) }\\ & =25 \\ & \Rightarrow \sqrt{\left(4\left(t^2-\frac{1}{t^2}\right)\right)^2+\left(8\left(t+\frac{1}{t}\right)\right)^2}=25 \\ & \quad \text { (putting } a=4) \\ & \Rightarrow \quad 16\left(t^2-\frac{1}{t^2}\right)^2+64\left(t+\frac{1}{t}\right)^2=625 \\ & \Rightarrow 16\left(t^4+\frac{1}{t^4}-2\right)+64\left(t^2+\frac{1}{t^2}+2\right)=625 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \begin{aligned} & \text { Let } t+\frac{1}{t}=u \text { so that }\left(t+\frac{1}{t}\right)^2=u^2 \\ & \Rightarrow \quad t^2+\frac{1}{t^2}+2=u^2 \\ & \Rightarrow \quad t^2+\frac{1}{t^2}=u^2-2 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad\left(t^2+\frac{1}{t^2}\right)^2=\left(u^2-2\right)^2 \\ & \Rightarrow \quad t^4+\frac{1}{t^4}+2=u^4+4-2 u^2 \end{aligned} $

Putting these values in Eq. (ii) and solving then we get,

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

So, Points are $A(16,16), B(1,4), C(1,-4)$, $D(16,-16)$

$\therefore$ Area of quadrilateral $A B C D=$

$ \begin{aligned} & \left.\frac{1}{2} \right\rvert\, x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1 -\left(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1\right) \mid \\ & \left.=\frac{1}{2} \right\rvert\, 16 \times 4+1 \times(-4)+1 \times(-16)+16 \times 16 -(16 \times 1+4 \times 1+(-4) \times 16+(-16)(16) \mid \\ & =300 \text { sq. units } \end{aligned} $

$y^2=4 a x$, focus $=(a, 0)$

Equation of directrix is $x=-a$.

The perpendicular distance from the focus $(a, 0)$ to the directrix $x+a=0$

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad \frac{|a+a|}{\sqrt{1+0}}=|2 a|=|2 a| \\ & \Rightarrow \quad|2 a|=\frac{3}{2} \\ & \Rightarrow \quad 2 a=\frac{3}{2} \text { or } 2 a=\frac{-3}{2} \\ & \Rightarrow \quad 2 a=\frac{3}{2} \Rightarrow a=\frac{3}{4} \\ & \Rightarrow \quad y-y_1=-\frac{y_1}{2 a}\left(x-x_1\right) \end{aligned}\\ &\text { Point }(4 a,-4 a)\\ &\begin{aligned} & x_1=4 a, y_1=-4 a \\ \Rightarrow \quad & y-(-4 a)=-\frac{(-4 a)}{2 a}(x-4 a) \\ \Rightarrow \quad & y+4 a=-(-2)(x-4 a) \\ \Rightarrow \quad & y+4 a=2(x-4 a) \\ \Rightarrow \quad & y+4 a=2 x-8 a \\ \Rightarrow \quad & 2 x-y-12 a=0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Substituting } a=\frac{3}{4} \\ & \Rightarrow \quad 2 x-y-12\left(\frac{3}{4}\right)=0 \\ & \Rightarrow \quad 2 x-y-9=0 \\ & 2 x-y=9 \end{aligned} $

$ \text { Given parabola } y^2=4 x $

$ \begin{aligned} &S \equiv(1,0), \quad a=1\\ &\text { Other end of focal chord } Q\\ &\begin{aligned} & & \frac{1}{S P}+\frac{1}{S Q}=\frac{1}{a} & \Rightarrow \\ \Rightarrow & & \frac{1}{5}+\frac{1}{S Q}=\frac{4}{5} & \Rightarrow S Q=\frac{5}{4} \end{aligned} \end{aligned} $

$y^2=4 x$

Equation of tangent to the parabola

$ y^2=4 a x \text { is } y=m x+\frac{a}{m} $

$\because$ Equation of tangent is

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

Given, tangents passes through $(1,4)$

$ \begin{aligned} & 4=m+\frac{1}{m} \\ \Rightarrow \quad & m^2-4 m+1=0 \\ \Rightarrow \quad & m_1+m_2=4, \quad m_1 m_2=1 \\ \because \tan \alpha & =\frac{m_1-m_2}{1+m_1 m_2} \\ & =\frac{\sqrt{\left(m_1+m_2\right)^2-4 m_1 m_2}}{1+m_1 m_2} \\ & =\frac{\sqrt{16-4}}{1+1}=\frac{\sqrt{12}}{2}=\frac{2 \sqrt{3}}{2}=\sqrt{3} \\ \therefore \quad & \alpha=\tan ^{-1} \sqrt{3}=\frac{\pi}{3} \end{aligned} $

We have parabola, $y^2=8 x$

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

$ y+x t=2 a t+a t^3 $

But $t=\frac{1}{\sqrt{2}}$

$ \begin{aligned} y+x \frac{1}{\sqrt{2}} & =4 \frac{1}{\sqrt{2}}+\frac{2}{2 \sqrt{2}} \\ \sqrt{2} y+x & =5 \end{aligned} $

It is equation of $L$.

Focus of parabola $y^2=8 x$ is $(2,0)$

Foot of perpendicular from focus to the line $L$ is

$ \begin{gathered} \frac{x-2}{1}=\frac{y-0}{\sqrt{2}}=-\left(\frac{2-5}{3}\right)=1 \\ x=3 \text { and } y=\sqrt{2} \end{gathered} $

$\because$ Foot of perpendicular $=(3, \sqrt{2})$.

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} $

From option (a).

$\begin{aligned} & x=4 \cos t, y=4 \sin t \\ & x^2=16 \cos ^2 t, y^2=16 \sin ^2 t \end{aligned}$

So, $x^2+y^2=16$, so this equation of a circle.

From option (b),

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

$ \begin{aligned} x^2 & =2-2 \cos t \quad \text{... (i)}\\ y & =\frac{1+\cos t}{2} \quad \text{... (ii)} \end{aligned}$

From Eq. (ii),

$\begin{aligned} \cos t= & 2 y-1 \\ \text { Now, } x^2 & =2-2 \cos t \\ x^2 & =2-2(2 y-1) \Rightarrow x^2=2-4 y+2 \\ x^2 & =4-4 y \end{aligned}$

So, this is represent the parabola.

Assume the vertex of the parabola is $(h, k)$ and the length of its latus rectum is $4a$. Since its axis is parallel to the $Y$-axis, the equation of the parabola can be written as:

$(x - h)^2 = 4a(y - k) \quad \text{.... (i)}$

The parabola passes through the points $(0, 4), (1, 9)$, and $(4, 5)$. Substituting these points into equation (i), we get:

$\begin{aligned} & \therefore (0 - h)^2 = 4a(4 - k) \\\\ & \Rightarrow h^2 = 4a(4 - k) \quad \text{... (ii)}\\\\ & (1 - h)^2 = 4a(9 - k) \\\\ & \Rightarrow 1 - 2h + h^2 = 4a(9 - k) \quad \text{... (iii)}\\\\ & (4 - h)^2 = 4a(5 - k) \\\\ & \Rightarrow 16 - 8h + h^2 = 4a(5 - k) \quad \text{... (iv)} \end{aligned}$

Next, we subtract Eqs. (ii) and (iii), and Eqs. (iii) and (iv), giving us:

$\begin{aligned} 1 - 2h & = 20a \quad \text{... (v)}\\\\ 15 - 6h & = -16a \quad \text{... (vi)} \end{aligned}$

By solving Eqs. (v) and (vi), we obtain $a = -\frac{3}{19}$ and $h = \frac{79}{38}$.

Substituting these values back, we find $k = \frac{9889}{912}$.

Therefore, the equation of the parabola becomes:

$\begin{aligned} & \left(x - \frac{79}{38}\right)^2 = \frac{-12}{19}(y - \frac{9889}{912}) \\\\ \Rightarrow & 19x^2 + 12y - 79x - 48 = 0 \end{aligned}$

Focus $\equiv(0,0)$

Tangent at vertex of parabola : $x-y+1=0$

The distance between the focus and tangent at the

$\text { vertex }=\left|\frac{0-0+1}{\sqrt{1^2+1^2}}\right|=\frac{1}{\sqrt{2}}$

The directrix is the line parallel to the tangent at vertex and at a distance $2 \times \frac{1}{\sqrt{2}}$ from focus.

Let the equation of directrix be

$\begin{aligned} & x-y+\lambda=0 \\ & \text { Here, } \frac{\lambda}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}} \\ & \Rightarrow \quad \lambda=2 \end{aligned}$

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

$\begin{aligned} & \text { Parabola : } y^2=4 p x \quad \text{... (i)}\\ & \text { Normal : } a x+b y=1 \quad \text{... (ii)}\\ & \Rightarrow \quad y=-\frac{a}{b} x+\frac{1}{b} \end{aligned}$

Slope, $m=-\frac{a}{b}$ and constant, $c=\frac{1}{b}$

Condition of normal says that

$\begin{aligned} & c=-2 m\left(\frac{\text { Coefficient of } x}{4}\right)-\left(\frac{\text { Coefficient of } x}{4}\right) m^3 \\ & \Rightarrow \frac{1}{b}=-2\left(\frac{-a}{b}\right)\left(\frac{4 p}{4}\right)-\left(\frac{4 p}{4}\right)\left(\frac{-a}{b}\right)^3 \\ & \Rightarrow \frac{1}{b}=\frac{2 p a}{b}+\frac{p a^3}{b^3} \\ & \Rightarrow b^2=2 p a b^2+p a^3 \\ & \Rightarrow p a^3=b^2-2 p a b^2 \end{aligned}$

Given parabola,

$y^2+4 x+2 y-8=0$ ..... (i)

Point of intersection of latusrectum and axis of parabola is foci.

Now, from Eq. (i), we get

$\begin{array}{ll} & (y+1)^2-1+4 x-8=0 \\ \Rightarrow \quad & (y+1)^2=-4 x+9 \\ \Rightarrow \quad & (y+1)^2=-4\left(x-\frac{9}{4}\right) \end{array}$

Compare with general standard form of parabola

where, foci $=(-a, 0)$

Here, $4 a=+4$

$\Rightarrow \quad a=+1$

Then, $\left(x-\frac{9}{4}\right)=-1 \Rightarrow x=-1+\frac{9}{4}=\frac{5}{4}$

and $(y+1)=0 \Rightarrow y=-1$

$\therefore \text { Focus }=\left(\frac{5}{4},-1\right)=\text { Point of intersection }$

$x=5t^2+2$ .... (i)

$y=10 t+4 \Rightarrow t=\frac{y-4}{10}$

From Eq. (i), $(x-2)=5\left(\frac{y-4}{10}\right)^2$

$\begin{array}{ll} \Rightarrow & (y-4)^2=20(x-2) \\ \Rightarrow & (7,4) \text { focus } \end{array} \quad\left\{\begin{array}{l} x-2=5 \\ y-4=0 \end{array}\right.$

Equation of parabola whose vertex is at origin and axis is $Y$-axis is

$x^2=4 a y$

This equation passes through $(6,-2)$

$\begin{aligned} & \therefore & 6^2 & =4 a(-2) \\ & \Rightarrow & \frac{-36}{8} & =a \Rightarrow a=\frac{-9}{2} \end{aligned}$

$\therefore$ Required parabola $x^2=-18 y$

Given parabola, $y^2=8 x$

Comparing with $y^2=4 a x$

We have, $a=2$

One end of focal chord $\left(\frac{1}{2}, 2\right)$

Parametric form of ends of focal chord $=\left(a t_1^2, 2 a t\right)$

$\begin{aligned} & \Rightarrow 2 a t=2 \\ & \therefore 2 \cdot 2 \cdot t=2 \Rightarrow t=\frac{1}{2} \end{aligned}$

$\begin{aligned} \text { Length of focal chord } & =a\left(t+\frac{1}{t}\right)^2 \\ & =2\left(\frac{1}{2}+2\right)^2=2 \cdot\left(\frac{5}{2}\right)^2=\frac{25}{2} \end{aligned}$