Parabola

$ 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} $

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)

Let $\mathrm{P}_1\left(t^2, 2 t\right)$ then tangent at $\mathrm{P}_1$ will be

$ t y=x+t^2 $


Since, it passes through $(-2,1)$

$ \begin{aligned} & \Rightarrow t^2-t-2 =0 \\\\ & \Rightarrow t = 2,-1 \end{aligned} $

So, we get point $\mathrm{P}_1(4, 4)$ and $\mathrm{P}_2(1, -2)$

Now finding the equation of $\mathrm{SP}_1: 4 x-3 y-4=0$

And equation of $\mathrm{SP}_2: x-1=0$

Now finding the point by foot of point on line formula,

$ \mathrm{Q}_1: \frac{x_1+2}{4}=\frac{y_1-1}{-3}=\frac{-(-8-3-4)}{25}=\frac{3}{5} $

We get $x_1=\frac{2}{5}, y_1=\frac{-4}{5}$ and $Q_2=(1,1)$

Now using the distance formula we get,

$ \begin{aligned} & \mathrm{SQ}_1 =\sqrt{\left(1-\frac{2}{5}\right)^2+\left(\frac{4}{5}\right)^2}=1 \\\\ & \mathrm{Q}_1 \mathrm{Q}_2 =\sqrt{\frac{9}{25}+\frac{81}{25}}=\frac{3 \sqrt{10}}{5} \\\\ & \mathrm{PQ}_1 =\sqrt{\frac{144}{25}+\frac{81}{25}}=3 \\\\ & \mathrm{SQ}_2 =1 \end{aligned} $

Hence, option (B, C, D) are correct.

Given, E : y² = 8x .... (i)

and P $\equiv$ ($-$2, 4)

Now, directrix of Eq. (i) is x = $-$2


So, point P($-$2, 4) lies on the directrix of parabola y² = 8x. Hence, $\angle QPQ' = {\pi \over 2}$ (by the definition of director circle) and chord QQ' is a focal chord and segment PQ subtends a right angle at the focus.

Slope of PF = $-$1 ($\because$ PF $\bot$ QQ')

Now, slope of $QQ' = {2 \over {{t_1} + {t_2}}} = 1$

$\therefore$ $PF = \sqrt {{{(2 + 2)}^2} + {{(0 + 4)}^2}} = \sqrt {32} = 4\sqrt 2 $

Tangent to y² = 4x at (t², 2t) is

y(2t) = 2(x + t²)

$\Rightarrow$ yt = x + t² ...... (i)

Equation of normal at P(t², 2t) is

y + tx = 2t + t³

Since, normal at P passes through centre of circle S(2, 8).

$\therefore$ 8 + 2t = 2t + t³ $\Rightarrow$ t = 2, i.e., P(4, 4)

$\therefore$ $SP = \sqrt {{{(4 - 2)}^2} + {{(4 - 8)}^2}} = 2\sqrt 5 $

$\therefore$ Option (a) is correct.

Also, SQ = 2

$\therefore$ PQ = SP $-$ SQ = 2$\sqrt5$ $-$ 2

Thus, ${{SQ} \over {QP}} = {1 \over {\sqrt 5 - 1}} = {{\sqrt 5 + 1} \over 4}$

$\therefore$ Option (b) is incorrect.

Now, x-intercept of normal is

x = 2 + 2² = 6

$\therefore$ Option (c) is correct.

Slope of tangent $ = {1 \over t} = {1 \over 2}$

$\therefore$ Option (d) is correct.

In case of option (A)

x² + y² = 3 and x² = 2y

Solving we get $P(\sqrt 2 ,1)$

Equation of tangent at P on the circle

$x\sqrt 2 + y = 3$ $\therefore$ $\tan \alpha = - \sqrt 2 $

Again, ${{{Q_2}{R_2}} \over {{Q_2}M}} = \sin \left( {\alpha - {\pi \over 2}} \right)$

or, ${{2\sqrt 3 } \over {{Q_2}M}} = - \cos \alpha = - {1 \over {\sqrt 3 }}$ [$\because$ $\tan \alpha = - \sqrt 2 $]

or, ${Q_2}M = 6$

Similarly, ${Q_3}M = 6$

$\therefore$ ${Q_2}{Q_3} = 12$

In case of option (B)

${{M{R_2}} \over {{Q_2}{R_2}}} = \tan \left( {\alpha - {\pi \over 2}} \right)$

or, ${{M{R_2}} \over {2\sqrt 3 }} = - \tan \alpha = \sqrt 2 $

or, $M{R_2} = 2\sqrt 6 $

Similarly, $M{R_3} = 2\sqrt 6 $ $\therefore$ ${R_2}{R_3} = 4\sqrt 6 $

In case of option (C)

OP is the perpendicular drawn from O to R₂R₃

$\therefore$ area of $\Delta O{R_2}{R_3} = {1 \over 2} \times OP \times {R_2}{R_3} = {1 \over 2} \times \sqrt 3 \times 4\sqrt 6 =6\sqrt 2 $

In case of option (D)

PT is perpendicular drawn from P to Q₂Q₃

$\therefore$ area of $\Delta P{Q_2}{Q_3} = {1 \over 2} \times PT \times {Q_2}{Q_3} = {1 \over 2} \times \sqrt 2 \times 12 = 6\sqrt 2 $

Therefore, (A), (B), (C) are correct options.

Let $P\left( {{{t_1^2} \over 2},{t_1}} \right)$ and $Q\left( {{{t_2^2} \over 2},{t_2}} \right)$ be two distinct points on the parabola ${y^2} = 2x$.

The circle with PQ as diameter passes through the vertex O(0, 0) of the parabola.

Clearly, PO $\bot$ OQ

So, slope of PO $\times$ slope of OQ = $-$1

or, ${{{t_1} - 0} \over {{{t_1^2} \over 2} - 0}} \times {{{t_2} - 0} \over {{{t_2^2} \over 2} - 0}} = - 1$

or, ${2 \over {{t_1}}} \times {2 \over {{t_2}}} = - 1$

or, ${t_1}{t_2} = - 4$

By question, area of $\Delta OPQ = 3\sqrt 2 $

or, ${1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr {{{t_1^2} \over 2}} & {{t_1}} & 1 \cr {{{t_2^2} \over 2}} & {{t_2}} & 1 \cr } } \right| = 3\sqrt 2 $

or, ${1 \over 2}\left| {{{t_1^2{t_2}} \over 2} - {{{t_1}t_2^2} \over 2}} \right| = 3\sqrt 2 $

or, $\left| {{t_1}{t_2}} \right|\left| {{t_1} - {t_2}} \right| = 12\sqrt 2 $

or, $\left| {{t_1} + {4 \over {{t_1}}}} \right| = 3\sqrt 2 $ [$\because$ ${t_1}{t_2} = - 4$]

or, ${t_1} + {4 \over {{t_1}}} = 3\sqrt 2 $ [$\because$ P lies in first quadrant]

or, $t_1^2 - 3\sqrt 2 {t_1} + 4 = 0$

or, ${t_1} = {{3\sqrt 2 \pm \sqrt {18 - 4 \times 1 \times 4} } \over 2}$

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

$ = 2\sqrt 2 ,\sqrt 2 $

$\therefore$ coordinates of $P = (4,2\sqrt 2 )$ or $(1,\sqrt 2 )$.

Therefore, (A) and (D) are correct options.

The equation of normal is

y = mx $-$ 2m $-$ m³

As (9, 6) lies on it, 6 = 9m $-$ 2m $-$ m³3 $-$ 7m + 6 = 0

The roots are m = 1, 2, $-$3. So the normal are

y = x $-$ 3, y = 2x $-$ 12, y = $-$3x + 33.

Let A $\equiv$ (t$_1^2$, 2t₁) and B $\equiv$ (t$_2^2$, 2t₂)

The centre of the circle = $\left( {{{t_1^2 + t_2^2} \over 2},{t_1} + {t_2}} \right)$

As the circle touches the x-axis thus ${t_1} + {t_2} = \pm \,r$

Slope of $AB = {2 \over {{t_1} + {t_2}}} = \pm \,{2 \over r}$

We have $G \equiv (h,k)$.

$ \Rightarrow h = {{2a + a{t^2}} \over 3},k = {{2at} \over 3}$

$ \Rightarrow \left( {{{3h - 2a} \over a}} \right) = {{9{k^2}} \over {4{a^2}}}$

Therefore, the required parabola is

${{9{y^2}} \over {4{a^2}}} = {{(3x - 2a)} \over a} = {3 \over a}\left( {x - {{2a} \over 3}} \right)$

$ \Rightarrow {y^2} = {{4a} \over 3}\left( {x - {{2a} \over 3}} \right)$

Hence, the vertex $ \equiv \left( {{{2a} \over 3},0} \right)$; focus $ \equiv (a,0)$.

Let’s say the common tangent line has the equation $y = m x + c$.

We want this line to touch both parabolas: $y = x^2$ and $y = -(x-2)^2$.

1. Tangent to $y = x^2$:

For the line to be a tangent, it must touch the curve at just one point. Set $y = x^2$ and $y = m x + c$ equal:

$x^2 = m x + c$

Bring all terms to one side:

$x^2 - m x - c = 0$

This quadratic has one solution (double root) if its discriminant is zero. The discriminant is $b^2 - 4ac$ in $ax^2 + bx + c = 0$.

Here, $a = 1$, $b = -m$, $c = -c$.

Set discriminant to zero:

$(-m)^2 - 4 \times 1 \times (-c) = 0$

$m^2 + 4c = 0$

So, $c = -\frac{m^2}{4}$

That means the tangent equation is:

$y = m x - \frac{m^2}{4}$

2. Same line tangent to $y = -(x-2)^2$:

The line must also touch the second parabola at one point. Set their equations equal:

$-(x-2)^2 = m x - \frac{m^2}{4}$

Expand $-(x-2)^2$:

$-(x^2 - 4x + 4) = -x^2 + 4x - 4$

So:

$-x^2 + 4x - 4 = m x - \frac{m^2}{4}$

Bring all terms to one side:

$x^2 + (m - 4)x + \left(4 - \frac{m^2}{4}\right) = 0$

Again, for a tangent, discriminant is zero:

$(m - 4)^2 - 4 \left(4 - \frac{m^2}{4}\right) = 0$

Calculate this step-by-step:

$(m - 4)^2 = m^2 - 8m + 16$

$4 \times 4 = 16$

$4 \times \left(-\frac{m^2}{4}\right) = -m^2$

So:

$(m - 4)^2 - 16 + m^2 = 0$

Substituting $(m - 4)^2$:

$m^2 - 8m + 16 - 16 + m^2 = 0$

$2m^2 - 8m = 0$

$m(m - 4) = 0$

This gives two possible values: $m = 0$ and $m = 4$.

3. Find the equations of the tangents:

If $m = 0$:

$y = 0$

If $m = 4$:

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

So the common tangents are $y = 0$ and $y = 4x - 4$. $y = 4x - 4$ can also be written as $y = 4(x - 1)$.