Parabola

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y^2=16 x \\ & A(1,-4) \equiv A\left(4 t^2, 8 t\right) \Rightarrow t=-1 / 2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Other coordinate will be }\\ &B\left(\frac{4}{t^2}, \frac{-8}{t}\right)=(16,16) \end{aligned} $

$ \begin{aligned} &\text { Now, equation of normal at } B(16,16)\\ &\begin{aligned} & y=-t_1 x+2 a t_1+a t_1^3 \\ \Rightarrow \quad & y=-2 x+16+32 \\ \Rightarrow \quad & 2 x+y-48=0 \end{aligned}\left[\because t_1=2, a=4\right] \end{aligned} $

$ y^2=12 x $

$ A\left(3 t^2, 6 t\right), B\left(3 t^2,-6 t\right) $

$ \begin{aligned} &O A B \text { is an equilateral triangle }\\ &\begin{aligned} & \quad O A=O B=A B \\ & \Rightarrow \quad\left(3 t^2-0\right)^2+(6 t-0)^2=(12 t)^2 \\ & \Rightarrow \quad 9 t^4=108 t^2 \\ & \Rightarrow \quad t=2 \sqrt{3} \\ & \Rightarrow \quad \text { Side of triangle }=12 t=24 \sqrt{3} \mathrm{~cm} \\ & \text { Area }=\frac{\sqrt{3}}{4}(24 \sqrt{3})^2=432 \sqrt{3} \mathrm{sq} \text { units } \end{aligned} \end{aligned} $

$\because x-2 y+k=0$

$\Rightarrow y=\frac{x}{2}+\frac{k}{2}$ is tangent to the parabola

$ y^2-4 x-4 y+8=0 $

So, slope of tangent $=y^{\prime}=\frac{1}{2}$

From the equation of parabola,

$ \begin{aligned} & 2 y\left(\frac{d y}{d x}\right)-4-4\left(\frac{d y}{d x}\right)+0=0 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{4}{2 y-4}=\frac{2}{y-2} \end{aligned} $

So, $\frac{2}{y-2}=\frac{1}{2} \Rightarrow y=6$

Put the value of $y$ in equation of parabola

$ \begin{aligned} & (6)^2-4 x-4 \times 6+8=0 \\ & \Rightarrow 4 x=20 \Rightarrow x=5 \end{aligned} $

That means, point $(5,6)$ lies on the tangent.

So, $6=\frac{5}{2}+\frac{k}{2} \Rightarrow \frac{7}{2}=\frac{k}{2}$

$ \Rightarrow \quad k=7 $

$y^2=5 x, a=\frac{5}{4} \Rightarrow$ Equation of latus rectum, $x=5 / 4$ and $x^2=5 y, a=5 / 4$

⇒ Equation of directrix, $y=-5 / 4$ Now, intersection points of both parabolas are $(0,0)$ and $(5,5)$ So, line will be $y=x$

$ \begin{aligned} & \text { So, } A=\left(\frac{5}{4}, \frac{5}{4}\right), B=\left(\frac{5}{4}, \frac{-5}{4}\right) \\ & C=\left(\frac{-5}{4}, \frac{-5}{4}\right) \\ & \because \angle A B C=90^{\circ} \end{aligned} $

Then, area $(\triangle A B C)=\frac{1}{2}(A B) \times(B C)$

$ \begin{aligned} & A B=\frac{5}{4}+\frac{5}{4}=\frac{10}{4}=5 / 2 \\ & B C=5 / 2 \end{aligned} $

Therefore, area $(\triangle A B C)=\frac{1}{2} \times \frac{5}{2} \times \frac{5}{2}$

$ =\frac{25}{8} \text { sq units } $

The equation of tangent to the parabola $y^2=8 x$ at $\left(x_0, y_0\right)$ is

$ \begin{aligned} & y_0=4\left(x+x_0\right) \text { or } 2 x-\frac{y_0}{2} y+2 x_0=0 \\ & \text { So, }-\frac{y_0}{2}=3 \\ & \Rightarrow y_0=-6 \text { and } 2 x_0=n \end{aligned} $

On putting $y=-6$ into the equation $y^2=8 x$, we get

$ (-6)^2=8 x_0 \Rightarrow x_0=\frac{9}{2} $

Thus, $\quad 2 x_0=n$

$ \begin{aligned} & \Rightarrow \quad 2\left(\frac{9}{2}\right)=n \Rightarrow n=9 \\ & \text { As, } n=9 \Rightarrow \quad(2 n, 4 \sqrt{n}) \equiv(18,12) \end{aligned} $

We know that, the equation of the normal to the parabola $y^2=4 a x$ at the point ( $x_1, y_1$ ) is given by

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

∴ The required equation be

$ y-12=-\frac{12}{2(2)}(x-18) \text { or } 3 x+y-66=0 $

Given, the tangent $a x-y+c=0$, makes an acute angle with the positive $X$-axis in the positive direction. So, $a>0$. The line $a x-y+c=0$ is tangent to the circle $x^2+y^2=1$.

So, the distance of this line from centre $(0,0)$ of the circle $x^2+y^2=1$ will be equal to the radius 1 .

$ \begin{array}{ll} \therefore \quad 1=\frac{|c|}{\sqrt{a^2+(-1)^2}} \\ \Rightarrow \quad|c|=\sqrt{a^2+1} \\ \Rightarrow c^2=a^2+1 \\ \text { Let } a=2 \Rightarrow a^2=4 \\ \text { and } c^2=5 \end{array} $

Now, the line $2 x-y+\sqrt{5}=0$ be the common tangent line to the circle $x^2+y^2=1$ and the parabola $y^2=8 \sqrt{5} x$

$y^2=8 \sqrt{5} x$ because the equation of tangent line to the parabola

$ \begin{aligned} & y^2=8 \sqrt{5} x \text { at }\left(\frac{\sqrt{5}}{2}, 2 \sqrt{5}\right) \text { is } 2 x-y+\sqrt{5}=0 \\ & \therefore \quad a^2 c^2=4 \times 5=20 \end{aligned} $

Given parabola, $y^2=k x$

∴ Focus $\left(\frac{k}{4}, 0\right)$

$\because P\left(2, y_1\right)$ lies on the parabola.

∵ Distance between focus and $P\left(2 y_1\right)=3$

$ \sqrt{y_1^2+\left(\frac{k}{4}-2\right)^2}=3 $

By squaring both sides, we get

$ \begin{aligned} & y_1^2+\frac{k^2}{16}+4-2\left(\frac{k}{4}\right)(2)=9 \\ \Rightarrow & 2 k+\frac{k^2}{16}+4-k=9 \Rightarrow \frac{k^2}{16}+k=5 \\ \Rightarrow & k^2+16 k-80=0 \\ \Rightarrow & (k+20)(k-4)=0 \\ \therefore & k=-20,4 \\ \because & k=-24 \quad \text { (it is not possible) } \\ \therefore & k=4 \end{aligned} $

So, point $P(2, \pm 2 \sqrt{2})$

Tangent $y( \pm 2 \sqrt{2})=4\left(\frac{x+2}{2}\right)$

$ x \pm 2 \sqrt{2} y+2=0 $

Parabola $y^2=12 x$

On comparing with $y^2=4 a x$, we have

$ a=3 $

Equation of normal having slope $m$ is

$ \begin{aligned} y & =m x-2 a m-a m^3 \\ \Rightarrow \quad y & =m x-6 m-3 m^3 \end{aligned} $

Normals are passing through $P(8,0)$.

$ \begin{aligned} &\begin{aligned} & \therefore \quad 8 m-6 m-3 m^3=0 \Rightarrow 2 m-3 m^3=0 \\ & \Rightarrow \quad m\left(2-3 m^2\right)=0 \\ & \Rightarrow \quad m=0, m= \pm \frac{\sqrt{2}}{\sqrt{3}} \end{aligned}\\ &\text { Slope of non-horizontal normals }\\ &\begin{aligned} & m_1=\frac{\sqrt{2}}{\sqrt{3}} \text { and } m_2=-\frac{\sqrt{2}}{\sqrt{3}} \\ & \tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\frac{2 \frac{\sqrt{2}}{\sqrt{3}}}{1-\frac{2}{3}}=2 \sqrt{6} \end{aligned} \end{aligned} $

The given equation of parabola is

$ y^2=4 x $

Thus, normal equation to the parabola $y^2=4 x$ is

$ y=m x-2 m-m^3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

(Since, any normal equation to the parabola $y^2=4 a x$ is $\left.y=m x-2 a m-a m t^2\right)$

Thus, Eq. (i), passes through $(5,0)$.

$ \begin{aligned} & \Rightarrow 0=5 m-2 m-m^3 \\ & \Rightarrow m\left(m^2+2-5\right)=0 \\ & \Rightarrow \text { Either } m=0 \text { or } m^2=3 \\ & \quad m= \pm \sqrt{3} \end{aligned} $

Thus, equation of normal passing through $(5,0)$ and whose slope is $\sqrt{3}$.

$ \begin{array}{ll} & (y-0)=\sqrt{3}(x-5) \\ \Rightarrow & \sqrt{3} x-y-5 \sqrt{3}=0 \end{array} $

Equation of parabola is $y^2=16 x\,\,\,\,\,(given)…(i) $

and the given equation of circle is

$ x^2+y^2=8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

So, centre is $(0,0)$ and radius $=2 \sqrt{2}$ Now, equation of tangent of Eq. (i) is

$ y=m x+\frac{4}{m} $

Now, radius = length of perpendicular from the centre to tangent

$ \begin{array}{ll} \Rightarrow & 2 \sqrt{2}=\left|\frac{0+0-\frac{4}{m}}{\sqrt{1+m^2}}\right| \\ \Rightarrow & 2 \sqrt{2}=\left|\frac{-\frac{4}{m}}{\sqrt{1+m^2}}\right| \\ \Rightarrow & 8\left(1+m^2\right)=\frac{16}{m^2} \\ \Rightarrow & m^4+m^2-2=0 \\ \Rightarrow & m^4+2 m^2-m^2-2=0 \\ \Rightarrow & \left(m^2+2\right)\left(m^2-1\right)=0 \\ \therefore & m^2-1=0 \\ \therefore & m= \pm 1 \end{array} \quad\left(\because m^2+2 \neq 0\right) $

When, $m=1, y=x+4$

When, $m=-1, y=-x-4$

$\therefore$ Line of intersection

$ \begin{aligned} & x^2+y^2-6 x-6 y+13=0 \\ & x^2+y^2-8 y+9=0 \end{aligned} $

$ \begin{array}{ll} (-)(-) & (+)-(-) \\ \hline & -6 x+2 y+4=0 \\ & 3 x-y-2=0 \end{array} $

$ \begin{aligned} \therefore 9 x-3 y & =6 ; \\ x+3 y & =0 \\ 10 x & =6[\text { By adding above equation }] \\ x & =\frac{3}{5} ; y=-\frac{1}{5} \\ 0 & =\frac{a+\frac{3}{5}}{2} \Rightarrow a=-\frac{3}{5} \\ 0 & =\frac{b-\frac{1}{5}}{2} \Rightarrow b=\frac{1}{5} \\ S(a, b) & =\left(\frac{-3}{5}, \frac{1}{5}\right) \end{aligned} $

To determine the angle $\theta$ between two tangents drawn from the point $(-1, -2)$ to the parabola $y^2 = 4x$, we follow these steps:

Equation of the Tangent:

The general form of the tangent to the parabola $y^2 = 4ax$ is given by:

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

Here, $a = 1$, so the tangent equation is:

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

Substitute the Point:

Given that the point $(-1, -2)$ lies on the tangent, substitute $x = -1$ and $y = -2$ into the tangent equation:

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

Solve for $m$:

Rearrange and solve for $m$:

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

Solving $(m - 1)^2 = 2$ gives:

$ m - 1 = \pm \sqrt{2} $

Hence, the slopes of the tangents are:

$ m_1 = 1 + \sqrt{2} \quad \text{and} \quad m_2 = 1 - \sqrt{2} $

Difference of Slopes:

Calculate the difference:

$ m_1 - m_2 = (1 + \sqrt{2}) - (1 - \sqrt{2}) = 2\sqrt{2} $

Product of Slopes:

Calculate the product:

$ m_1 m_2 = (1 + \sqrt{2})(1 - \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1 $

Find $\tan \theta$:

The formula for $\tan \theta$ between two lines with slopes $m_1$ and $m_2$ is:

$ \tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| $

Substitute the values:

$ \tan \theta = \left|\frac{2\sqrt{2}}{1 - 1}\right| = \left|\frac{2\sqrt{2}}{0}\right| $

Since the denominator is zero, $\tan \theta$ is undefined.

Conclusion:

The angle $\theta$ is $\frac{\pi}{2}$ radians, indicating that the tangents are perpendicular.

Therefore, $\theta = \frac{\pi}{2}$.

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

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 $

$x = 2t,\,y = {2 \over 3}$

$t \to 1\,\,\,A \equiv \left( {2,{1 \over 3}} \right)$

Given conic is ${x^2} = 12y \Rightarrow S \equiv (0,3)$

Let $B \equiv (0,\beta )$

Given $SA\, \bot \,BA$

$\left( {{{{1 \over 3}} \over {2 - 3}}} \right)\left( {{{\beta - {1 \over 3}} \over { - 2}}} \right) = - 1$

$ \Rightarrow \left( {\beta - {1 \over 3}} \right){1 \over 3} = - 2$

$ \Rightarrow \beta = {1 \over 3}\left( {{{ - 17} \over 3}} \right)$

$\mathop {Ordinate\,of\,centroid}\limits_{(as\,t \to 1)} = k = {{\beta + {1 \over 3} + 3} \over 3}$

$ = {{{{ - 17} \over 9} + {{10} \over 3}} \over 3} = {{13} \over {18}}$

According to the question (Let P(x, y))

$2x - y{{dx} \over {dy}} = 0$

($\because$ equation of tangent at $P:y - y = {{dy} \over {dx}}(y - x)$)

$\therefore$ $2{{dy} \over {y}} = {{dx} \over x}$

$ \Rightarrow 2\ln y = \ln x + \ln c$

$ \Rightarrow {y^2} = cx$

$\because$ this curve passes through (3, 3)

$\therefore$ c = 3

$\therefore$ required parabola ${y^2} = 3x$ and L.R = 3

Given equation of curve,

$ \begin{aligned} & x^2-4 x+4 y-8=0 \\ \Rightarrow \quad & x^2-4 x=-4 y+8 \\ \Rightarrow \quad & x^2-4 x+4=-4 y+8+4 \\ \Rightarrow \quad & (x-2)^2=-4 y+12 \\ \Rightarrow \quad & (x-2)^2=-4(y-3) \end{aligned} $

Now, let $x-2=X$ and $y-3=Y$

$ \therefore \quad X^2=-4 Y $

On comparing Eq. (i) with $X^2=-4 a Y$, we get $4 a=4$

$ \Rightarrow \quad a=1 $

Vertex of Eq. (ii),

$ \begin{aligned} & & X & =0 \text { and } Y=0 \\ \Rightarrow & & x-2 & =0 \text { and } y-3=0 \\ \Rightarrow & & x & =2 \text { and } y=3 \end{aligned} $

∴ Vertex of Eq.(i) is $(2,3)$.

Focus of Eq. (ii) $X=0$ and $Y=-a$

$ \begin{array}{lc} \Rightarrow & x-2=0 \text { and } y-3=-1 \\ \Rightarrow & x=2 \text { and } y=2 \end{array} $

∴ Focus of Eq. (i) is $(2,2)$.

One end of the latusrectum of Eq. (ii)

$ \begin{aligned} & & X & = \pm 2 a \text { and } Y=-a \\ \Rightarrow & & x-2 & = \pm 2 \text { and } y-3=-1 \\ \Rightarrow & & x & =0,4 \text { and } y=2 \end{aligned} $

∴ One end of the latusrectum of Eq. (i) is $(4,2)$ or $(0,2)$.

Equation of directrix of Eq. (ii),

$ \begin{aligned} &\,\,\,\,\,\,\,\, Y=a & \\ \Rightarrow & y-3 =1 \Rightarrow y=4 \end{aligned} $

and equation of axis of Eq. (ii) $X=0$

$ \begin{array}{lrl} \Rightarrow & x-2 & =0 \\ \Rightarrow & x & =2 \end{array} $

∴ Point of intersection of the axis and direction is $(2,4)$.

$ y^2=\frac{8}{a} x \quad[a>0] \ldots(\mathrm{i}) $

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

One end of focal chord is $(1,4)$. On putting $x=1$ and $y=4$ in Eq. (i),

$ \begin{array}{ll} & 16=\frac{8}{a} \times 1 \\ \Rightarrow & a=\frac{1}{2} \\ \therefore & y^2=\frac{8}{\left(\frac{1}{2}\right)} x \\ \Rightarrow & y^2=16 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Comparing $y^2=16 x$ to $y^2=4 A x$, we have

$ \begin{aligned} & & 4 A & =16 \\ \Rightarrow & & A & =4 \end{aligned} $

At $(1,4)$,

$ \begin{aligned} & 2 A t_1=y \\ & \Rightarrow \quad 2 \times 4 \times t_1=4 \Rightarrow t_1=\frac{1}{2} \\ & \text { Length of focal chord } =a\left(t+\frac{1}{t}\right)^2 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =4\left(\frac{1}{2}+2\right)^2 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =4\left(\frac{25}{4}\right)=25 \end{aligned} $

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

Let $P\left(x_1, y_1\right), Q\left(x_2, y_2\right)$

Equation of $A F: y-0=\frac{2-0}{1-2}(x-2)$

$ y=-2 x+4 $

$ \Rightarrow \quad 2 x+y=4 $

Taking intersection of curve and line $A F$,

$ \begin{aligned} & & (4-2 x)^2 & =8 x \\ & \Rightarrow & (2-x)^2 & =2 x \\ & \Rightarrow & x^2-4 x+4 & =2 x \\ & \Rightarrow & x^2-6 x+4 & =0 \\ & \therefore & x_1+x_2 & =6 \end{aligned} $

Given, focal distance $=3$

$ \begin{aligned} 2 t^2+a & =3 \\ \Rightarrow \quad 2 t^2+2 & =3 \\ t & = \pm \frac{1}{\sqrt{2}} \end{aligned} $

$ \begin{aligned} &\text { Given equation is }\\ &\begin{aligned} & x^2-4 y+6 x+15=0 \\ & \Rightarrow \quad x^2+6 x+9=4 y-15+9 \\ & \Rightarrow \quad(x+3)^2=4 y-6 \\ & \Rightarrow \quad(x+3)^2=4\left(y-\frac{3}{2}\right) \\ & \therefore \alpha=-3 \text { and } \beta=\frac{3}{2} \text { and } 8 a=4 \Rightarrow a=\frac{1}{2} \\ & \text { Now, } 2 \alpha+8 \beta^2=2(-3)+8 \times \frac{9}{4} \\ & \quad=-6+18=12 \end{aligned} \end{aligned} $

Given, equation of parabola

$ \begin{aligned} & y^2=16 x \\ & L L^{\prime}=4 a=16 \end{aligned} $

End points of latusrectum is $\left(a_1 \pm 2 a\right)$ Coordinates of $L$ is $(4,8)$ and $L^{\prime}$ is $(4,-8)$.

The end points of focal length is

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

Coordinate of $P$ is $(1,4)$

$ \begin{array}{ll} \Rightarrow & a t^2=1, \quad 2 a t=4 \\ \Rightarrow & 4 t^2=1, \quad 2 \times 4 \times t=4 \\ \Rightarrow & t^2=\frac{1}{4}, \quad t=\frac{1}{2} \\ \therefore & t=1 / 2 \end{array} $

∴ Coordinate of

$ Q\left(\frac{4}{\left(1 / 2^2\right.},-\frac{2 \times 4}{(1 / 2)}\right)=(16,-16) $

$ \begin{aligned} L Q & =\sqrt{(4-16)^2+(8+16)^2}=\sqrt{144+576} \\ & =\sqrt{720}=12 \sqrt{5} \end{aligned} $

Given, equation of parabola

$ y^2=32 x $

Here, $a=8$

The end point of focal chord is $\left(a t^2, 2 a t\right)$ and $\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

$ \begin{array}{rlrl} \Rightarrow & a t^2 =\frac{1}{2} \text { and } 2 a t=4 \\ \Rightarrow & 8 \times t^2 =\frac{1}{2} \text { and } 2 \times 8 \times t=4 \\ \Rightarrow & t^2 =\frac{1}{16} \text { and } 16 t=4 \\ \Rightarrow & t= \pm \frac{1}{4} \text { and } t=\frac{1}{4} \\ & \therefore t =1 / 4 \end{array} $

Coordinate of $Q$ is $\left(\frac{8}{\left(\frac{1}{4}\right)^2}, \frac{-2 \times 8}{\frac{1}{4}}\right)$

$ =(128,-64) $

Coordinate of $S$ is $(8,0)$.

$ \begin{aligned} S Q & =\sqrt{(8-128)^2+(0+64)^2} \\ & =\sqrt{14400+4096}=\sqrt{18496}=136 \end{aligned} $

Let the point $P$ be $(h, k)$.

So, $\sqrt{(h-a)^2+(k-0)^2}=\frac{|h+k|}{\sqrt{1^2+1^2}}$

$ h^2+k^2+a^2-2 h a=\frac{h^2+k^2+2 h k}{2} $

$ 2 h^2+2 k^2+2 a^2-4 a h=h^2+k^2+2 h k $

∴ Locus of point $P$ is

$ x^2+y^2-2 x y-4 a x+2 a^2=0 $

Given, $y^2+4 y+8 x-2=0$

$ \begin{array}{r} y^2+4 y+4+8 x-6=0 \\ \left(y+2^2+2(4 x-3)=0\right. \\ y+2=0,4 x-3=0 \\ y=-2, x=3 / 4 \end{array} $

∴ Required point is $\left(\frac{3}{4},-2\right)$.

Statement I

Let $P(x, y)$ be any point on parabola. Distance from $P$ to focus = Perpendicular distance from $P$ to direction

$ \begin{aligned} & \sqrt{(x-2)^2+(y-3)^2}=\frac{|x+2 y+5|}{\sqrt{1^2+2^2}} \\ & 5\left(x^2+y^2-4 x-6 y+4+9\right)=x^2+4 y^2 +25+4 x y+20 y+10 x \\ & 4 x^2+y^2-4 x y-30 x-50 y+40=0 \end{aligned} $

∴ Statement I is true.

Statement II

$ \begin{aligned} x^2-4 x+16 y+52 & =0 \\ x^2-4 x+4 & =-16 y-48 \\ (x-4)^2 & =-16(y+3) \end{aligned} $

Comparison with $X^2=4 a Y$, we get vertex(2, - 3)

Equation of directrix is $Y=a$

$ y+3=4 \Rightarrow y=1 \text { or } y-1=0 $

∴ Statement II is wrong.

$ \begin{aligned} &\text { Given, } x=-2+2 t^2, y=2+4 t\\ &\begin{aligned} \Rightarrow & t=\frac{y-2}{4} \\ & x=-2+2\left(\frac{y-2}{4}\right)^2 \\ \Rightarrow & x=-2+\frac{y^2+4-4 y}{8} \\ & 8 x=-16+y^2+4-4 y \\ \Rightarrow & y^2-8 x-4 y-12=0 \end{aligned} \end{aligned} $

Given, equation is

$ \begin{array}{lr} & 2 x^2+5 y-6 x+1=0 \\ \Rightarrow & 2 x^2-6 x+\frac{9}{2}+5 y+1-\frac{9}{2}=0 \\ \Rightarrow & 2\left(x-\frac{3}{2}\right)^2=-5\left(y-\frac{7}{10}\right) \\ \Rightarrow & \left(x-\frac{3}{2}\right)^2=\frac{-5}{2}\left(y-\frac{7}{10}\right) \\ \Rightarrow & X^2=\frac{-5}{2} Y \end{array} $

where $X=x-\frac{3}{2}$ and $Y=y-\frac{7}{10}$

For vertex, on putting $X=0, Y=0$

$ \Rightarrow \quad x=3 / 2 \text { and } y=7 / 10 $

∴ Vertex $=(3 / 2,7 / 10)$

For focus, $X=0$ and $Y=-a$

$\Rightarrow x=\frac{3}{2}$ and $y-\frac{7}{10}=\frac{-5}{8}$

$\Rightarrow y=\frac{3}{40}$

∴ Focus $\equiv(3 / 2,3 / 40)$