Parabola

$ y^2=4 a x $

Equation of normal

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

Point of contact

$ P\left(a m^2,-2 a m\right) $

and Point $Q\left(2 a+a m^2, 0\right)$

Area of

$\begin{aligned} \triangle P F Q & =\frac{1}{2} \times\left|a+a m^2\right||-2 a m| \\\\ 120 & =a^2\left(1+m^2\right) m \\\\ a & =2, m=3\end{aligned}$

Satisfies the equation $(1)$, hence $(2,3)$ will be the correct answer.

Equation of given parabola is

y² = 4$\lambda $x ....(i)

So the end point of the latus rectum of the parabola (i), P($\lambda $, 2$\lambda $) and the given ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$, passes through point P($\lambda $, 2$\lambda $).

On differentiating the equation of parabola, w.r.t. 'x', we get

${{dy} \over {dx}} = {{2\lambda } \over y}$

$ \therefore $ Slope of tangent to the parabola at point P is m₁ = 1

Similarly, on differentiating the equation of given ellipse, ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$, w.r.t.x, we get ${{2x} \over {{a^2}}} + {{2y} \over {{b^2}}}{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = - {{{b^2}x} \over {{a^2}y}}$

$ \therefore $ Slope of tangent to the ellipse at point P is ${m_2} = - {{{b^2}} \over {2{a^2}}}$

$ \because $ It is given that the tangents are perpendicular to each other. So, ${m_1}{m_2} = - 1$

$ \Rightarrow (1)\left( { - {{{b^2}} \over {2{a^2}}}} \right) = - 1$

$ \Rightarrow {{{b^2}} \over {{a^2}}} = 2 \Rightarrow b = \sqrt 2 a$

$ \therefore $ Eccentricity of ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ will be

$e = \sqrt {1 - {{{a^2}} \over {{b^2}}}} = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}$ {$ \because $ b > a}

It is given that, the centres of circles C₁, C₂ and C₃ are co-linear,

$ \therefore $ $\left| {\matrix{ 0 & 0 & 1 \cr 3 & 4 & 1 \cr h & k & 1 \cr } } \right| = 0 \Rightarrow 4h = 3k$ ....(i)

and MN is the length of diameter of circle C₃, so

MN = $3 + \sqrt {{{(3 - 0)}^2} + {{(4 - 0)}^2}} + 4$

= $3 + 5 + 4 = 12$

So, radius of circle C₃, r = 6 ......(ii)

Since, the circle C₃ touches C₁ at M and C₂ at N, so

|C₁ C₃| = |r $-$ 3|

$ \Rightarrow $ $\sqrt {{h^2} + {k^2}} = 3$

$ \Rightarrow $ h² + k² = 9 .....(iii)

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

${h^2} + {{16{h^2}} \over 9} = 9 \Rightarrow 25{h^2} = 81$

$ \Rightarrow h = + {9 \over 5}$ and $k = + {{12} \over 5}$

So, $2h + k = {{18} \over 5} + {{12} \over 5} = 6$

Now, equation common chord XY of circles C₁ and C₂ is

C₁ $-$ C₂ = 0

$ \Rightarrow $ 6x + 8y = 18

$ \Rightarrow $ 3x + 4y = 9 ....(iv)

Now, PY² = GY² $-$ GP²

= $9 - {{81} \over {25}} = {{144} \over {25}}$

$ \Rightarrow $ PY = ${{12} \over 5}$

$ \because $ $XY = 2PY = 2 \times {{12} \over 5} = {{24} \over 5}$

Similarly, equation of ZW is 3x + 4y = 9.

Now, So, length of perpendicular from C₃ to

$ZW = {{\left| {3\left( {{9 \over 5}} \right) + 4\left( {{{12} \over 5}} \right) - 9} \right|} \over 5} = {6 \over 5}$

So, $P{W^2} = {C_3}{W^2} - {C_3}{P^2} = 36 - {{36} \over {25}} = {{864} \over {25}}$

{$ \because $ C₃W = r = 6}

$ \Rightarrow $ $PW = {{12\sqrt 6 } \over 5}$

$ \because $ $ZW = 2PW = {{24\sqrt 6 } \over 5}$

$ \therefore $ ${{length\,of\,ZW} \over {length\,of\,XY}} = \sqrt 6 $

Now, area of

$\Delta MZN = {1 \over 2}(MN)(PZ) = {1 \over 2} \times (12)\left( {{1 \over 2}WZ} \right)$

{$ \because $ MN = 12}

= $3WZ = 3{{24\sqrt 6 } \over 5} = {{72\sqrt 6 } \over 5}$

and area of $\Delta $ZMW = ${1 \over 2}$(ZW) (MP)

= ${1 \over 2}\left( {{{24\sqrt 6 } \over 5}} \right)(MG + GP)$

$ = {{12\sqrt 6 } \over 5}\left( {3 + {9 \over 5}} \right)$

{$ \because $ MG = 3 and GP = ${{9 \over 5}}$}

$ = {{12\sqrt 6 } \over 5}\left( {{{24} \over 5}} \right)$

= ${{288\sqrt 6 } \over 5}$

$ \therefore $ ${{Area\,of\,\Delta MZN} \over {Area\,of\,\Delta ZMW}} = {{{{72\sqrt 6 } \over 5}} \over {{{288\sqrt 6 } \over 5}}} = {5 \over 4}$

$ \because $ Common tangent of circles C₁ and C₃ is C₁ $-$ C₃ = 0

$ \Rightarrow $ $({x^2} + {y^2} - 9) - \left[ {{{\left( {x - {9 \over 5}} \right)}^2} + {{\left( {y - {{12} \over 5}} \right)}^2} - 36} \right] = 0$

$ \Rightarrow $ ${{18} \over 5}x + {{24} \over 5}y + 18 = 0$

$ \Rightarrow $ 3x + 4y + 15 = 0 ......(v)

$ \because $ Tangent (v) is also touches the parabola x² = 8$\alpha $y,

$ \therefore $ $ - 2\alpha {\left( { - {3 \over 4}} \right)^2} = - {{15} \over 4}$

$ \Rightarrow $ $\alpha = {{10} \over 3}$

So combination (iv), (S) is only incorrect.

Hence, option (b) is correct.

It is given that, the centres of circles C₁, C₂ and C₃ are co-linear,

$ \therefore $ $\left| {\matrix{ 0 & 0 & 1 \cr 3 & 4 & 1 \cr h & k & 1 \cr } } \right| = 0 \Rightarrow 4h = 3k$ ....(i)

and MN is the length of diameter of circle C₃, so

MN = $3 + \sqrt {{{(3 - 0)}^2} + {{(4 - 0)}^2}} + 4$

= $3 + 5 + 4 = 12$

So, radius of circle C₃, r = 6 ......(ii)

Since, the circle C₃ touches C₁ at M and C₂ at N, so

|C₁ C₃| = |r $-$ 3|

$ \Rightarrow $ $\sqrt {{h^2} + {k^2}} = 3$

$ \Rightarrow $ h² + k² = 9 .....(iii)

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

${h^2} + {{16{h^2}} \over 9} = 9 \Rightarrow 25{h^2} = 81$

$ \Rightarrow h = + {9 \over 5}$ and $k = + {{12} \over 5}$

So, $2h + k = {{18} \over 5} + {{12} \over 5} = 6$

Now, equation common chord XY of circles C₁ and C₂ is

C₁ $-$ C₂ = 0

$ \Rightarrow $ 6x + 8y = 18

$ \Rightarrow $ 3x + 4y = 9 ....(iv)

Now, PY² = GY² $-$ GP²

= $9 - {{81} \over {25}} = {{144} \over {25}}$

$ \Rightarrow $ PY = ${{12} \over 5}$

$ \because $ $XY = 2PY = 2 \times {{12} \over 5} = {{24} \over 5}$

Similarly, equation of ZW is 3x + 4y = 9.

Now, So, length of perpendicular from C₃ to

$ZW = {{\left| {3\left( {{9 \over 5}} \right) + 4\left( {{{12} \over 5}} \right) - 9} \right|} \over 5} = {6 \over 5}$

So, $P{W^2} = {C_3}{W^2} - {C_3}{P^2} = 36 - {{36} \over {25}} = {{864} \over {25}}$

{$ \because $ C₃W = r = 6}

$ \Rightarrow $ $PW = {{12\sqrt 6 } \over 5}$

$ \because $ $ZW = 2PW = {{24\sqrt 6 } \over 5}$

$ \therefore $ ${{length\,of\,ZW} \over {length\,of\,XY}} = \sqrt 6 $

So, combination (ii), Q is only correct.

Hence, option (c) is correct.

The tangent is y = x + 8.

Satisfying (8, 16) in y² = 4ax, we have

16² = 4a . 8. $\therefore$ a = 8

On comparing y = mx + ${a \over m}$, we have m = 1

The point of contact is $\left( {{a \over {{m^2}}},{{2a} \over m}} \right) = (8,16)$

So its verified.

We know that on the parabola $y^2=4 a x$, the equation of tangent at $p=\left(a t^2, 2 a t\right)$ is $t y=x+ a t^2$ and Normal at $\mathrm{S}\left(a s^2, 2 a s\right)$ is $y+s x=2 a s+a s^3$ Let tangent at P and Normal at Q intersect at $\left(x_0, y_0\right)$, then

$\begin{aligned} t y_0 & =x_0+a t^2 \quad \text{.... (i)}\\ y_0+s x_0 & =2 a s+a s^3 \quad \text{... (ii)} \end{aligned}$

From equation (i) and (ii)

$\begin{aligned} & x_0=t y_0-a t^2=\frac{-y_0+2 a s+a s^3}{s} \\ \Rightarrow & \text { st } y_0-\text { at.t.s }=-y_0+2 a s+a s^3 \end{aligned}$

Given, $s t=1$

$\begin{aligned} & \therefore \quad y_0-a t=-y_0+\frac{2 a}{t}+\frac{a}{t^3} \\ & \Rightarrow \quad 2 y_0=a t+\frac{2 a}{t}+\frac{a}{t^3} \\ & \Rightarrow \quad y_0=\frac{a\left(t^4+2 t^2+1\right)}{2 t^3} \\ & \Rightarrow \quad y_0=\frac{a\left(t^2+1\right)^2}{2 t^3} \end{aligned}$

Hint:

On the parabola $y^2=4 a x$

(i) The equation of tangent at $\mathrm{P}\left(a t^2, 2 a t\right)$ is $ty=x+a t^2$.

(ii) The equation of normal at $\mathrm{S}\left(a s^2, 2 a s\right)$ is $y+s x=2 a s+a s^3$.

Given, four points $\mathrm{P}\left(a t^2, 2 a t\right), \mathrm{Q}, \mathrm{R}\left(a r^2, 2 a r\right)$ and $S\left(a s^2, 2 a s\right)$ be distinct points on the parabola $y^2=4 a x$.

Also given PQ is the focal chord of parabola

$\begin{aligned} & y^2=4 a x \\ & \therefore \quad \mathrm{Q}=\left(\frac{a}{t^2},-\frac{2 a}{t}\right) \end{aligned}$

Given, $\mathrm{K}=(2 a, 0)$ and PK is parallel to QR

$\therefore$ Slope of line PK $=$ Slope of line QR

$\begin{aligned} & \frac{2 a t-0}{a t^2-2 a}=\frac{2 a r+\frac{2 a}{t}}{a r^2-\frac{a}{t^2}} \\ \Rightarrow & \frac{t}{t^2-2}=\frac{r+\frac{1}{t}}{r^2-\frac{1}{t^2}} \\ \Rightarrow & \frac{t}{t^2-2}=\frac{r+\frac{1}{t}}{\left(r+\frac{1}{t}\right)\left(r-\frac{1}{t}\right)} \\ \Rightarrow & \frac{t}{t^2-2}=\frac{1}{r-\frac{1}{t}} \\ \Rightarrow & t r-1=t^2-2 \\ \Rightarrow & \quad r=\frac{t^2-1}{t} \end{aligned}$

Hint:

(i) If PQ is a focal chord of parabola $y^2=4 a x$ and $\mathrm{P}=\left(a t^2, 2 a t\right)$, then $\mathrm{Q}=\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

(ii) If two lines are parallel, then their slopes are equal.

(iii) The slope of line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$.

Given, a parabola $y^2=16 x$

Let $\mathrm{F}\left(x_0, y_0\right)=\left(4 t^2, 8 t\right)$

$\Rightarrow \quad x_0=4 t^2, y_0=8 t$

Equation of tangent at F is $t y=x+4 t^2$

G is the point of intersection of tangent $t y=x +4 t^2$ and $Y$-axis

$\therefore \mathrm{G}=(0,4 t)$

Given, $\mathrm{G}=\left(0, y_1\right)$

$\therefore \quad y_1=4 t$

Let $\Delta$ be the area of $\Delta$ EGF

$\begin{aligned} & \Rightarrow \quad \Delta=\frac{1}{2} 4 t^2 \cdot\left(3-y_1\right) \\ & \Rightarrow \quad \Delta=2 t^2(3-4 t) \\ & \Rightarrow \quad \frac{d \Delta}{d t}=12 t-24 t^2=0 \text { at } t=0, \frac{1}{2} \\ & \Rightarrow \frac{d^2 \Delta}{d t^2}=12-48 t<0 \text { at } t=\frac{1}{2} \end{aligned}$

Hence, $\Delta$ is maximum at $t=\frac{1}{2}$

$\therefore \quad \Delta_{\max }=\frac{1}{2}, y_0=8 t=4, y_1=4 t=2$

Given, $y=m x+3$ intersect the parabola $y^2=16 x$ at $\mathrm{F}\left(4 t^2, 8 t\right)$

$\therefore 8 t=4 m t^2+3$

Put $t=\frac{1}{2}$

$\begin{aligned} & \Rightarrow \quad 4=m+3 \\ & \Rightarrow m=1 \end{aligned}$

Hints :

(i) $\frac{d y}{d x}=0$ at $x=\alpha$ and $\frac{d^2 y}{d x^2}<0$ at $x=\alpha$, then the function has point of local maxima at $x=a$.

(ii) The parametric point of the parabola $y^2= 4 a x$ is $\left(a t^2, 2 a t\right)$

(iii) The equation of tangent of parabola $y^2=4 a x$ at $\left(a t^2, 2 a t\right)$ is $t y=x+a t^2$.

Given, PQ be a focal chord of the parabola $y^2=4 a x$

Let $\mathrm{P}=\left(a t^2, 2 a t\right)$ and $\mathrm{Q}\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$

Equation of tangent at P is $t y=x+a t^2\quad \text{... (i)}$

Equation of tangent at Q is $-\frac{y}{t}=x+\frac{a}{t^2}\quad \text{... (ii)}$

Let $R$ is the point of intersection of tangents at $P$ and $Q$.

On solving Equation (i) and (ii)

$\Rightarrow \mathrm{R}=\left(-a, a\left(t-\frac{1}{t}\right)\right)$

Given, the tangents of parabola $y^2=4 a x$ at P and $Q$ meet at a point lying on the line $y=2 x +a$.

$\therefore \mathrm{R}$ must be lies on the line $y=2 x+a$

$\begin{aligned} & \Rightarrow \quad a\left(t-\frac{1}{t}\right)=-2 a+a \\ & \Rightarrow \quad t-\frac{1}{t}=-1 \quad \text{... (iii)} \end{aligned}$

Let the length of chord PQ is L

$\begin{aligned} & \Rightarrow \mathrm{L}=\sqrt{\left(a t^2-\frac{a}{t^2}\right)^2+\left(2 a t+\frac{2 a}{t}\right)^2} \\ & \Rightarrow \mathrm{L}=a \sqrt{\left(t-\frac{1}{t}\right)^2\left(t+\frac{1}{t}\right)^2+4\left(t+\frac{1}{t}\right)^2} \\ & \Rightarrow \mathrm{L}=a \sqrt{\left(t-\frac{1}{t}\right)^2\left(\left(t-\frac{1}{t}\right)^2+4\right)+4\left(\left(t-\frac{1}{t}\right)^2+4\right)} \\ & \Rightarrow \mathrm{L}=a \sqrt{1 \cdot(1+4)+4 \cdot(1+4)} \\ & \Rightarrow \mathrm{L}=5 a \end{aligned}$

Hints:

If $P Q$ is a focal chord of parabola $y^2=4 a x$, then $\mathrm{P}=\left(a t^2, 2 a t\right)$ and $\mathrm{Q}=\left(\frac{a}{t^2}, \frac{-2 a}{t}\right)$

Equation of tangent at P is $t y=x+a t^2\quad \text{.... (i)}$

Equation of tangent at Q is $\frac{-y}{t}=x+\frac{a}{t^2}\quad \text{... (ii)}$

On solving equations (i) and (ii)

$\Rightarrow \mathrm{R}=\left(-a, a\left(t-\frac{1}{t}\right)\right)$

Given, the tangents of parabola $y^2=4 a x$ at P and Q meet at a point lying on the line $y=2 x+a$

$\begin{aligned} & \Rightarrow \quad a\left(t-\frac{1}{t}\right)=-2 a+a \\ & \Rightarrow \quad\left(t-\frac{1}{t}\right)=-1 \quad \text{... (iii)} \end{aligned}$

$\Rightarrow$ Slope of line $\mathrm{OP}=\frac{2}{t}$ and Slope of line OQ $=-2 t$

Given, $\angle \mathrm{POQ}=\theta$

$\begin{aligned} & \Rightarrow \quad \tan \theta=\frac{\frac{2}{t}-(-2 t)}{1+\frac{2}{t}(-2 t)} \\ & \Rightarrow \quad \tan \theta=\frac{-2}{3}\left(t+\frac{1}{t}\right) \\ & \Rightarrow \quad \tan \theta=\frac{-2}{3} \sqrt{\left(t-\frac{1}{t}\right)^2+4} \\ & \Rightarrow \quad \tan \theta=\frac{-2 \sqrt{5}}{3} \end{aligned}$

Hints:

(i) If slope of two lines are $m_1$ and $m_2$, then the angle of intersection between them is equal to $\tan ^{-1}\left(\frac{m_1-m_2}{1+m_1 m_2}\right)$

(ii) If PQ is a focal chord of parabola $y^2=4 a x$, then $\mathrm{P}=\left(a t^2, 2 a t\right)$ and $\mathrm{Q}=\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

Now, ${y^2} = 4x$ and Q will lie on it.

$ \Rightarrow {(4k)^2} = 4 \times 4h$

$ \Rightarrow {k^2} = h$

$ \Rightarrow {y^2} = x$ (replacing h by x and k by y)

The intersection point of $y = 0$ with first line is $B( - p,0)$.

The intersection point of $y=0$ with second line is $A( - q,0)$.

The intersection point of the two lines is $C(pq,(p + 1)(q + 1))$.

The altitude from C to AB is $x = pq$.

The altitude from B to AC is

$y = - {q \over {1 + q}}(x + p)$

Solving these two equations, we get $x = pq$ and $y = - pq$.

Hence, the locus of orthocentre is $x + y = 0$.

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

$(2 y-3)=-(x-1)^{2} \Rightarrow$ parabola is symmetric with respect to $x=1$

Also statement 1 is true statement 2 is correct explanation of statement 1.

The tangent to the curve $y=e^x$ drawn at the point ($c,e^c$) is given as:

$(y-e^c)=e^c(x-c)$ ..... (i)

The equation of the line joining the points

$(c-1),(e^{c-1})$ and $(c+1,e^{c+1})$

$y-e^{c-1}=\frac{e^c(e-e^{-1})}{2}(x-c+1)$ ..... (ii)

On subtracting (i) from (ii)

${e^c} - {e^{c - 1}} = {{{e^c}(e - {e^{ - 1}})} \over 2}(x - c + 1) - {e^c}(x - c)$

${e^c} - {e^{c - 1}} = (x - c){e^c}\left( {{{e - {e^{ - 1}}} \over 2} - 1} \right) + {e^c}\left( {{{e - {e^{ - 1}}} \over 2}} \right)$

$x - c = {{{{(e - 1)}^2}} \over {2 - {{(e - 1)}^2}}} < 0$

$ \Rightarrow x < c$

We first obtain the coordinate of P & Q.

Towards this end we solve

${x^2} + {y^2} = 9$ ..... (i)

${y^2} = 8x$ ...... (ii)

from (i) and (ii), we get

${x^2} + 8x - 9 = 0$

$(x + 9)(x - 1) = 0 \Rightarrow x = 1$ [$\therefore$ $x + 9 > 0$]

${y^2} = 8 \Rightarrow y = \, \pm 2\sqrt 2 $

Thus, coordinates of P are $(1,2\sqrt 2 )$ and Q are $(1, - 2\sqrt 2 )$ equation of tangents to the circle at P and Q respectively.

$x + 2\sqrt 2 y = 9$

$x - 2\sqrt 2 y = 9$

On solving these we get coordinates of R as (9, 0)

Equations of tangents to parabola at P and Q are

$2\sqrt 2 y = 4(x + 1)$

$ - 2\sqrt 2 y = 4(x + 1)$

These lines intersect each other at S($-1,0$)

Let $\Delta_1$ = area of $\Delta$PQS = $\frac{1}{2}$ (PQ)(ST)

$=\frac{1}{2}(2)(4\sqrt2)$

$=4\sqrt2$

And $\Delta_2$ = area of $\Delta$PQR = $\frac{1}{2}$(PQ)(TR)

$=\frac{1}{2}(4\sqrt2)8$

$=16\sqrt2$

$\Delta_1:\Delta_2=4\sqrt2:16\sqrt2$

= 1 : 4

Equation of perpendicular bisector of SR is x = 4

Midpoint of PR is (5, $\sqrt2$) ..... (i)

Slope of PR is ${{0 - 2\sqrt 2 } \over {9 - 1}} = {{ - \sqrt 2 } \over 4}$

$\therefore$ Equation of perpendicular bisector of PR is

$y - \sqrt 2 = {4 \over {\sqrt 2 }}(x - 5)$ ...... (ii)

Solving equation (i) and (ii) we get

$x = 4,y = - \sqrt 2 $

Circumcentre is (4, $-\sqrt2$)

Radius of circumcircle is = 3$\sqrt3$ units

We have

PR = RQ = 6$\sqrt2$

PQ = 4$\sqrt2$

$\Delta_2$ = area of $\Delta$PQR

= 16$\sqrt2$

$r = {{{\Delta _2}} \over s}$

$ = {{16\sqrt 2 } \over {{1 \over 2}(6\sqrt 2 + 6\sqrt 2 + 4\sqrt 2 )}} = 2$ units

Radius of incircle of $\Delta$PQR = 2 units

Axis of parabola along line $y=x$ distance of vertex from origin is $\sqrt{2}$. It implies parabola lies in first Quadrant.

Equation of directrix is line passing through the origin and perpendicular on $y=x$ i.e, $y=-x \Rightarrow x+y=0$

Now by definition of parabola. Parabola is a locus of a point which moves in such a way its distance from fixed point and from fixed line are equal where fixed point is called focus and fixed line is called directrix.

i.e, if $\mathrm{P}(x, y)$ be any point on parabola then Distance of P from directrix $=$ Distance of P from focus

$ \begin{aligned} & \Rightarrow & \left(\frac{x+y}{\sqrt{1^2+1^2}}\right)^2 & =(x-2)^2+(y-2)^2 \\ & \Rightarrow & (x+y)^2 & =2\left[x^2+y^2-4 x-4 y+8\right] \\ & \Rightarrow & x^2+y^2+2 x y & =2 x^2+2 y^2-8 x-8 y+16 \\ & \Rightarrow & x^2+y^2-2 x y & =8(x+y-2) \\ & \Rightarrow & (x-y)^2 & =8(x+y-2) \end{aligned} $

$ y^2=4 x $

normal on parabola

$ \begin{aligned} 2 y \frac{d y}{d x} & =4 \\ \frac{d y}{d x} & =\frac{2}{y} \\ \frac{-d x}{d y} & =\frac{-y}{2} \end{aligned} $

Equation of normal

$ \begin{aligned} y & =\left(\frac{-y}{2}\right)(x-3) \\ \Rightarrow 2 & =3-x \\ \Rightarrow x & =1 \text { or } y=0 \end{aligned} $

Points $(\mathrm{P}, \mathrm{Q}, \mathrm{R})$ are $(0,0)(1,2)$ and $(1,-2)$

Base $\mathrm{QR}=4$ and Height $=1$

(i) area of $\triangle \mathrm{PQR}=0.5 \times 4 \times 1$

(ii) circum radius $=\frac{a b c}{4 a}=\frac{20}{4 \times 2}=\frac{5}{2}$

(iii) centroid of $\triangle \mathrm{PQR}$

$ =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\left(\frac{2}{3}, 0\right) $

(iv) circum centre: since, the triangle is an isosceles triangle and circum centre lies $\frac{5}{2}$ distance apart (radius)

From origin.

Therefore, circumcentre $\left(\frac{5}{2}, 0\right)$

$ 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 $

Given : The equation of parabola is y² = 16x

Equation chord is

2x + y = p ..... (1)

Equation of chord with middle point (h, k) is given as

$ky - 16\left( {{{x + h} \over 2}} \right) = {k^2} - 16h$

$ \Rightarrow ky - 8(x + h) = {k^2} - 16h$

$ \Rightarrow ky - 8x - 8h = {k^2} - 16h$

$ \Rightarrow ky - 8x - 8h - {k^2} + 16h = 0$

$ \Rightarrow - 8x + ky + 8h - {k^2} = 0$

$ \Rightarrow 8x - ky = 8h - {k^2}$ ..... (2)

Comparing above equation with equation of chord, we get

$2x + y = p$

Dividing Eq. (2) by 4, we get

${{8x} \over 4} - {{ky} \over 4} = {{8h} \over 4} - {{{k^2}} \over 4}$

$2x - {{ky} \over 4} = 2h - {{{k^2}} \over 4}$

On comparing, we get

$ \Rightarrow {{ - k} \over 4} = 1$ and $p = 2h - {{{k^2}} \over 4}$

$ \Rightarrow k = - 4$ and $p = 2h - {{{{( - 4)}^2}} \over 4} = 2h - 4$

$ \Rightarrow p = 2h - 4 \Rightarrow 2h - p = 4$

Only p = 2 and h = 3 satisfies this equation. Therefore, p = 2, h = 3 and k = $-$4.

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 $y = 4(x - 1)$.