Ellipse

$\begin{aligned} & \mathrm{P} \equiv\left(3 \cos 30^{\circ}, 2 \sin 30^{\circ}\right) \equiv\left(\frac{3 \sqrt{3}}{2}, 1\right) \\ & \mathrm{Q} \equiv\left(3 \cos 60^{\circ}, 2 \sin 60^{\circ}\right) \equiv\left(\frac{3}{2}, \sqrt{3}\right) \\ & \mathrm{R}\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right), \mathrm{S}\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)\end{aligned}$

Slope of $\mathrm{PQ}=\mathrm{m}_{\mathrm{PQ}}=\frac{\sqrt{3}-1}{\frac{3}{2}-\frac{3 \sqrt{3}}{2}}=-\frac{2}{3}$

Equation of line PQ

$ y-\sqrt{3}=-\frac{2}{3}\left(x-\frac{3}{2}\right) $

$\Rightarrow 2 x+3 y=3(\sqrt{3}+1) \quad$ option (A) is correct

Now

if $\mathrm{N}_2=\left(\mathrm{x}_2, 0\right)=\left(\frac{3}{2}, 0\right)$

$\left|\mathrm{N}_2 \mathrm{Q}\right|=\sqrt{3}$ and $\left|\mathrm{N}_2 \mathrm{~S}\right|=\frac{3 \sqrt{3}}{2}$

$\Rightarrow 3\left|\mathrm{~N}_2 \mathrm{Q}\right|=2\left|\mathrm{~N}_2 \mathrm{~S}\right| \quad$ option (C) is correct

Now, if $\mathrm{N}_1=\left(\mathrm{x}_1, 0\right) \Rightarrow \mathrm{N}_1=\left(\frac{3 \sqrt{3}}{2}, 0\right)$

$\Rightarrow\left|\mathrm{N}_1 \mathrm{P}\right|=1,\left|\mathrm{~N}_1 \mathrm{R}\right|=\frac{3}{2} \quad$ option (D) is incorrect

$ \begin{aligned} & E: \frac{x^2}{6}+\frac{y^2}{3}=1 \text {, Tangent : } y=m_1 x \pm \sqrt{6 m_1^2+3} \\\\ & P: y^2=12 x, \text { Tangent: } y=m_2 x+\frac{3}{m_2} \end{aligned} $

For common tangent

$ \begin{aligned} & m=m_1=m_2, \pm \sqrt{6 m_1^2+3}=\frac{3}{m_2} \\\\ & \Rightarrow m= \pm 1 \end{aligned} $

$\Rightarrow$ Equation of common tangents $y=x+3$ and $y=-x-3$ point of contact for parabola is $\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$

$ \Rightarrow A_1 \equiv(3,6), \quad A_4(3-6) $

Let $A_2\left(x_1, y_1\right) \Rightarrow$ tangent to $E$ is $\frac{x x_1}{6}+\frac{y y_1}{3}=1$

$A_3$ is mirror image of $A_2$ in $x$-axis $\Rightarrow A_3(-2,-1)$


Intersection point of $T_1=0$ and $T_2=0$ is $(-3,0)$

Area of quadrilateral $A_1 A_2 A_3 A_4=\frac{1}{2}(12+2) \times 5=35$ square units

Given equation of ellipse

${E_1}:{{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$ ... (i)

Now, let a vertex of rectangle of largest area with sides parallel to the axes, inscribed in E₁ be (3cos$\theta $, 2sin$\theta $).

So, area of rectangle

R₁ = 2(3cos$\theta $) $ \times $ 2(2sin$\theta $) = 12sin(2$\theta $)



The area of R₁ will be maximum, $\theta = {\pi \over 4}$ and maximum area is 12 square units and length of sides of rectangle R₁ are $2a\cos \theta = \sqrt 2 a = 3\sqrt 2 $ = length of major axis of ellipse E₂ and

$2b\sin \theta = \sqrt 2 b = 2\sqrt 2 $ = length of minor axis of ellipse E₂.

So, ${E_2}:{{{x^2}} \over {{{\left( {{a \over {\sqrt 2 }}} \right)}^2}}} + {{{y^2}} \over {{{\left( {{b \over {\sqrt 2 }}} \right)}^2}}} = 1$ and

maximum area of rectangle

${R_2}:2\left( {{a \over {\sqrt 2 }}} \right)\left( {{b \over {\sqrt 2 }}} \right)$ and so on.

So, ${E_n} = {{{x^2}} \over {{{\left( {{a \over {{{(\sqrt 2 )}^{n - 1}}}}} \right)}^2}}} + {{{y^2}} \over {{{\left( {{b \over {{{(\sqrt 2 )}^{n - 1}}}}} \right)}^2}}} = 1$,

and maximum area of rectangle

${R_n} = 2\left( {{a \over {{{(\sqrt 2 )}^{n - 1}}}}} \right) + \left( {{b \over {{{(\sqrt 2 )}^{n - 1}}}}} \right)$

Now option (a),

Since, eccentricity of ellipse

${E_n} = e{'_n} = \sqrt {1 - {{{{({b_n})}^2}} \over {{{({a_n})}^2}}}} $

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

$ = \sqrt {1 - {{{b^2}} \over {{a^2}}}} = \sqrt {1 - {4 \over 9}} = {{\sqrt 5 } \over 3}$

is independent of 'n', so eccentricity of E₁₈ and E₁₉ are equal.

Option (b),

Distance between focus and centre of

${E_9} = e.{a_9} = {a \over {{{(\sqrt 2 )}^8}}}(e)$

$ = {3 \over {{2^4}}} \times {{\sqrt 5 } \over 3} = {{\sqrt 5 } \over {16}}$ unit

Option (c),

$ \because $ $\sum\limits_{n = 1}^N {(area\,of\,{R_n})} < (area\,of\,{R_1}) + (area\,of\,{R_2}) + .....\infty $

$ < 2ab + 2{{ab} \over 2} + 2{{ab} \over {{2^2}}} + .......$

$ < 2ab\left( {1 + {1 \over 2} + {1 \over {{2^2}}} + .....} \right)$

$ < 12\left( {{1 \over {1 - 1/2}}} \right)$

$ \Rightarrow \sum\limits_{n = 1}^N {(area\,of\,{R_n}} ) < 24$, for each positive integer N.

Option (d),

Length of latusrectum ${E_9} = {{2b_9^2} \over {{a_9}}} = {{2{b^2}} \over {a{{(\sqrt 2 )}^8}}}$

$ = {{2 \times 4} \over {3 \times 16}} = {1 \over 6}$ units

Hence, options (c) and (d) are correct.

We have,

Equations of circle

${x^2} + {y^2} = {1 \over 2}$

and Equations of parabola

y² = 4x



Let the equation of common tangent of parabola and circle is

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

Since, radius of circle = ${1 \over {\sqrt 2 }}$

$ \therefore $ ${1 \over {\sqrt 2 }}$ = $\left| {{{0 + 0 + {1 \over m}} \over {\sqrt {1 + {m^2}} }}} \right|$

$ \Rightarrow $ m⁴ + m² $-$ 2 = 0

$ \Rightarrow $ m =$ \pm $1

$ \therefore $ Equation of common tangents are

y = x + 1 and y = $-$x$-$1

Intersection point of common tangent at Q($-$1, 0)

$ \therefore $ Equation of ellipse

${{{x^2}} \over 1} + {{{y^2}} \over {1/2}} = 1$

where, ${a^2} = 1$, ${b^2} = 1/2$

Now, eccentricity

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

and length of latusrectum

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



$ \therefore $ Area of shaded region

$ = 2\int_{1/\sqrt 2 }^1 {{1 \over {\sqrt 2 }}} \sqrt {1 - {x^2}} dx$

$\eqalign{ & = \sqrt 2 \left[ {{x \over 2}\sqrt {1 - {x^2}} + {1 \over 2}{{\sin }^{ - 1}}x} \right]_{1/\sqrt 2 }^1 \cr & = \sqrt 2 \left[ {\left( {0 + {\pi \over 4}} \right) - \left( {{1 \over 4} + {\pi \over 8}} \right)} \right] \cr} $

$ = \sqrt 2 \left[ {\left( {0 + {\pi \over 4}} \right) - \left( {{1 \over 4} + {\pi \over 8}} \right)} \right]$

$ = \sqrt 2 \left( {{\pi \over 8} - {1 \over 4}} \right) = {{\pi - 2} \over {4\sqrt 2 }}$

Here, ${E_1}:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,\,(a > b)$

${E_2}:{{{x^2}} \over {{c^2}}} + {{{y^2}} \over {{d^2}}} = 1,\,(c < d)$

and $S:{x^2} + {(y - 1)^2} = 2$

as tangent to E₁, E₂ and S is $x + y = 3$.

Let the point of contact of tangent be $({x_1},{y_1})$ to S.

$\therefore$ $x\,.\,{x_1} + y\,.\,{y_1} - (y + {y_1}) + 1 = 2$

or $x{x_1} + y{y_1} - y = (1 + {y_1})$, same as $x + y = 3$.

$ \Rightarrow {{{x_1}} \over 1} = {{{y_1} - 1} \over 1} = {{1 + {y_1}} \over 3}$

i.e. ${x_1} = 1$ and ${y_1} = 2$

$\therefore$ $P = (1,2)$

Since, $PR = PQ = {{2\sqrt 2 } \over 3}$. Thus, by parametric form,

${{x - 1} \over { - 1/\sqrt 2 }} = {{y - 2} \over {1/\sqrt 2 }} = \pm {{2\sqrt 2 } \over 3}$

$ \Rightarrow \left( {x = {5 \over 3},y = {4 \over 3}} \right)$

and $\left( {x = {1 \over 3},y = {8 \over 3}} \right)$

$\therefore$ $Q = \left( {{5 \over 3},{4 \over 3}} \right)$ and $R = \left( {{1 \over 3},{8 \over 3}} \right)$

Now, equation of tangent at Q on ellipse E₁ is

${{x\,.\,5} \over {{a^2}\,.\,3}} + {{y\,.\,4} \over {{b^2}\,.\,3}} = 1$

On comparing with x + y = 3, we get

${a^2} = 5$ and ${b^2} = 4$

$\therefore$ $e_1^2 = 1 - {{{b^2}} \over {{a^2}}} = 1 - {4 \over 5} = {1 \over 5}$ ..... (i)

Also, equation of tangent at R on ellipse E₂ is

${{x\,.\,1} \over {{a^2}\,.\,3}} + {{y\,.\,8} \over {{b^2}\,.\,3}} = 1$

On comparing with x + y = 3, we get

${a^2} = 1,\,{b^2} = 8$

$\therefore$ $e_2^2 = 1 - {{{a^2}} \over {{b^2}}} = 1 - {1 \over 8} = {7 \over 8}$ ...... (ii)

Now, $e_1^2\,.\,e_2^2 = {7 \over {40}} \Rightarrow {e_1}{e_2} = {{\sqrt 7 } \over {2\sqrt {10} }}$

and $e_1^2 + e_2^2 = {1 \over 5} + {7 \over 8} = {{43} \over {40}}$

Also, $\left| {e_1^2 - e_2^2} \right| = \left| {{1 \over 5} - {7 \over 8}} \right| = {{27} \over {40}}$

The ellipse and hyperbola will be confocal. Therefore,

$( \pm ae,0) \equiv ( \pm 1,0)$

$ \Rightarrow \left( { \pm a \times {1 \over {\sqrt 2 }},0} \right) \equiv ( \pm 1,0)$

$ \Rightarrow a = \sqrt 2 $ and $e = {1 \over {\sqrt 2 }}$

$ \Rightarrow {b^2} = {a^2}(1 - {e^2}) \Rightarrow {b^2} = 1$

Hence, the equation of ellipse ${{{x^2}} \over 2} + {{{y^2}} \over 1} = 1$.

From this given data, we can write as

$2\cos \left( {{{B + C} \over 2}} \right)\cos \left( {{{B - C} \over 2}} \right) = 4{\sin ^2}{A \over 2}$

$\cos \left( {{{B - C} \over 2}} \right) = 2\sin (A/2)$

$ \Rightarrow {{\cos \left( {{{B - C} \over 2}} \right)} \over {\sin A/2}} = 2$

$ \Rightarrow {{\sin B + \sin C} \over {\sin A}} = 2$

$ \Rightarrow b + c = 2a$

where a is a constant. Therefore, the locus of point A is an ellipse.

The ellipse is ${x^2} + 4{y^2} = 4$,

${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$

We have ${b^2} - {a^2}(1 - {e^2})$

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

P and Q are obtained as

$P\left( {\sqrt 3 ,{{ - 1} \over 2}} \right),Q = \left( { - \sqrt 3 ,{{ - 1} \over 2}} \right)$

We have $PQ = 2\sqrt 3 $

With PQ as latus rectum two parabolas are possible their vertices being

$\left( {0,{{ - \sqrt 3 - 1} \over 2}} \right)$ and $\left( {0,{{\sqrt 3 - 1} \over 2}} \right)$

The equation of the parabola s can be obtained as

${x^2} - 2\sqrt 3 y = 3 + \sqrt 3 $ and ${x^2} + 2\sqrt 3 y = 3 - \sqrt 3 $

$\operatorname{Ar}(\Delta \mathrm{ORT})=\frac{3}{2}$

$\begin{aligned} & \left|\frac{1}{2} \times 3 \times 2 \sin \theta\right|=\frac{3}{2} \\\\ & \sin \theta=\frac{1}{2} \Rightarrow \theta=\frac{11 \pi}{6} \\\\ & \mathrm{~T}\left(\frac{3 \sqrt{3}}{2},-1\right)\end{aligned}$

Tanget at $(0,2) $,

$\frac{x(0)}{9}+\frac{y(2)}{4}=1 \Rightarrow y=2$ .........(1)

Tangent at $\left(\frac{3 \sqrt{3}}{2},-1\right) $,

$\frac{x\left(\frac{3 \sqrt{3}}{2}\right)}{9}+\frac{y(-1)}{4}=1$ ...........(2)

$\therefore$ By solving (1) & (2) $\Rightarrow \mathrm{p}=3 \sqrt{3}, \mathrm{q}=2$

$\Rightarrow$ Option $(\mathrm{A})$ is Correct.

Given: $ \frac{x^2}{4}+\frac{y^2}{3}=1 $

Let $\quad \alpha=2 \cos \phi$

Tangent at $\mathrm{E}(2 \cos \phi, \sqrt{3} \sin \phi)$

to the ellipse is $\frac{x \cos \phi}{2}+\frac{y \sin \phi}{\sqrt{3}}=1$

This intersect $x$-axis at $\mathrm{G}(2 \sec \phi, 0)$

$ \begin{aligned} & \text { Area of triangle } \mathrm{FGH}=\frac{1}{2}(2 \sec \phi-2 \cos \phi) 2 \sin \phi \\\\ & \Delta=2 \sin ^2 \phi \cdot \tan \phi \\\\ & \Delta=(1-\cos 2 \phi) \cdot \tan \phi \end{aligned} $

$ \begin{aligned} \text {I. If }\phi & =\frac{\pi}{4}, \Delta=1 \rightarrow Q \\\\ \text {II. If }\phi & =\frac{\pi}{3}, \Delta=2\left(\frac{\sqrt{3}}{2}\right)^2 \cdot \sqrt{3} \\\\ & =\frac{3 \sqrt{3}}{2} \rightarrow \mathrm{T} \end{aligned} $

$ \begin{aligned} \text {III. If }\phi & =\frac{\pi}{6}, \Delta=2\left(\frac{1}{2}\right)^2 \cdot \frac{1}{\sqrt{3}} \\\\ & =\frac{1}{2 \sqrt{3}} \rightarrow S \end{aligned} $

$ \begin{aligned} \text {IV. If }\phi=\frac{\pi}{12}, \Delta & =\left(1-\frac{\sqrt{3}}{2}\right) \cdot(2-\sqrt{3}) \\\\ & =\frac{(\sqrt{3}-1)^4}{8} \rightarrow P \end{aligned} $

We have,

x² + y² = 4

Let P(2 cos$\theta $, 2 sin$\theta $) be a point on a circle.

$ \therefore $ Tangent at P is

2 cos$\theta $x + 2 sin$\theta $y = 4

$ \Rightarrow $ x cos$\theta $ + y sin$\theta $ = 2


$ \therefore $ The coordinates at $M\left( {{2 \over {\cos \theta }},0} \right)$ and $N\left( {0,{2 \over {\sin \theta }}} \right)$

Let (h, k) is mid-point of MN

$ \therefore $ $h = {1 \over {\cos \theta }}$ and $k = {1 \over {\sin \theta }}$

$ \Rightarrow $ $\cos \theta = {1 \over h}$

and $\sin \theta = {1 \over k}$

$ \Rightarrow $ ${\cos ^2}\theta + {\sin ^2}\theta = {1 \over {{h^2}}} + {1 \over {{k^2}}}$

$ \Rightarrow 1 = {{{h^2} + {k^2}} \over {{h^2}.{k^2}}}$

${h^2} + {k^2} = {h^2}{k^2}$

$ \therefore $ Mid-point of MN lie on the curve

${x^2} + {y^2} = {x^2}{y^2}$

F₁(x, 0) and F₂(x₂, 0) are the foci of the ellipse:

${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$

Therefore, a² = 9 and b² = 8.

${b^2} = {a^2}(1 - {e^2})$

$1 - {e^2} = {8 \over 9} \Rightarrow {e^2} = 1 - {8 \over 9} = {1 \over 9} \Rightarrow e = {1 \over 3}$

The focus is

${F_1}\left( { - 3 \times {1 \over 3},0} \right)$ and ${F_2}\left( {3 \times {1 \over 3},0} \right)$

That is, F₁($-$1, 0) and F₂(1, 0).

The equation of parabola is

${y^2} = 4(O{F_2})x$

${y^2} = 4x(O{F_2} = 1)$

The point of intersection of ellipse and parabola is

${{{x^2}} \over 9} + {{4x} \over 8} = 1 \Rightarrow {{{x^2}} \over 9} + {x \over 2} = 1$

$ \Rightarrow 2{x^2} + 9x - 18 = 0$

$ \Rightarrow 2{x^2} + 12x - 3x - 18 = 0$

$ \Rightarrow 2x(x + 6) - 3(x + 6) = 0$

$ \Rightarrow x = {3 \over 2}$ (x $-$6 is rejected)

Now, ${y^2}(4){3 \over 2} = 6$

$y = \pm \sqrt 6 $

That is, the points M and N are, respectively, $M\left( {{3 \over 2},\sqrt 6 } \right)$ and $N\left( {{3 \over 2}, - \sqrt 6 } \right)$.

Let the orthocenter be (h, k).

The slope of $OM = {{k - \sqrt 6 } \over {h - (3/2)}}$

The slope of $ON = {{\sqrt 6 } \over { - 1 - (3/2)}} = {{ - 2\sqrt 6 } \over 5}$

Now, $\left( {{{k - \sqrt 6 } \over {h - (3/2)}}} \right)\left( {{{ - 2\sqrt 6 } \over 5}} \right) = - 1$

$2\sqrt 6 k - 12 = 5h - {{15} \over 2}$

$5h - 2\sqrt 6 k = {{15} \over 2} - 12 = {{ - 9} \over 2}$

The slope of $ON = {{k + \sqrt 6 } \over {h - (3/2)}}$

The slope of ${F_1}M = {{\sqrt 6 } \over {1 + (3/2)}} = {{2\sqrt 6 } \over 5}$

${{k + \sqrt 6 } \over {h - (3/2)}} \times {{2\sqrt 6 } \over 5} = - 1$

$2\sqrt 6 k + 12 = - 5h + {{15} \over 2}$

$5h + 2\sqrt 6 k = {{15} \over 2} - 12 = {{ - 9} \over 2}$

$5h + 2\sqrt 6 k = {{ - 9} \over 2}$ ....... (1)

$5h - 2\sqrt 6 k = {{ - 9} \over 2}$ ........ (2)

Solving Eqs. (1) and (2), we get

$10h = - 9 \Rightarrow h = {{ - 9} \over {10}}$ and k = 0

Hence, the orthocentre of the triangle F₁MN is $\left( {{{ - 9} \over {10}},0} \right)$.

Equation of tangent at M(3/2, $\sqrt6$) to ${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$ is

${3 \over 2}.{x \over 9} + \sqrt 6 .{y \over 8} = 1$ ....... (i)

which intersect X-axis at (6, 0).

Also, equation of tangent at N(3/2, $-$$\sqrt6$) is

${3 \over 2}.{x \over 9} - \sqrt 6 .{y \over 8} = 1$ ....... (ii)

Eqs. (i) and (ii) intersect on X-axis at R(6, 0). ........ (iii)

Also, normal at $M(3/2,\sqrt 6 )$ is

$y - \sqrt 6 = {{ - \sqrt 6 } \over 2}\left( {x - {3 \over 2}} \right)$

On solving with y = 0, we get Q(7/2, 0) ....... (iv)

The area of MQR is

$\left| {{1 \over 2}} \right|\left| {\matrix{ {3/2} & {\sqrt 6 } & 1 \cr 6 & 0 & 1 \cr {7/2} & 0 & 1 \cr } } \right| = \left| {{{\sqrt 6 } \over 2}\left( {6 - {7 \over 2}} \right)} \right| = {{5\sqrt 6 } \over 4}$

The area of the quadrilateral MF₁NF₂ is

$2(\Delta {m_1}{F_1}{F_2}) = 2\sqrt 6 $

and the required ratio is

${{5\sqrt 6 } \over {4\,.\,2\sqrt 6 }} = {5 \over 8}$

Given, a circle $x^2+y^2=2$ and a parabola $y^2=8 x$

Let $P R$ and $Q R$ are common tangents of given circle and parabola intersect at A.

Let $\mathrm{R}=\left(2 t^2, 4 t\right)$ and $S=\left(2 t^2,-4 t\right)$

Equation of tangent of parabola

$y^2=8 x \text { at } \mathrm{R}\left(2 t^2, 4 t\right) \text { is } t y=x+2 t^2$

Here, $A$ is the point of intersection of tangent

$\begin{aligned} t y & =x+2 t^2 \text { and } X \text {-axis } \\ \Rightarrow \mathrm{A} & =\left(-2 t^2, 0\right) \end{aligned}$

Here, $t y=x+2 t^2$ also touches the circle at P

$\begin{array}{lr} \therefore & \sqrt{2}=\frac{\left|0-t \times 0+2 t^2\right|}{\sqrt{1^2+(-t)^2}} \\ \Rightarrow & \sqrt{2} \cdot \sqrt{1+t^2}=2 t^2 \\ \Rightarrow & 2+2 t^2=4 t^4 \\ \Rightarrow & 2 t^4-t^2-1=0 \\ \Rightarrow & 2 t^4-2 t^2+t^2-1=0 \\ \Rightarrow & \left(2 t^2+1\right)\left(t^2-1\right)=0 \\ \Rightarrow & t^2=\frac{-1}{2}, 1 \end{array}$

$\begin{array}{ll} \Rightarrow t^2=1 \left(t^2=-\frac{1}{2} \text { is impossible }\right) \\ \Rightarrow t = \pm 1 \end{array}$

$\therefore \quad \mathrm{R}=(2,4), \mathrm{S}=(2,-4) \text { and } \mathrm{A}=(-2,0)$

Now, the equation of chord PQ w.r.t. point $(-2,0)$ is

$\begin{aligned} \Rightarrow & & -2 \cdot x+0 \cdot y-2 & =0 \\ \Rightarrow & & x & =-1 \end{aligned}$

Here, P and Q are the point of intersection of circle $x^2+y^2=2$ and the line $x=-1$

$\begin{aligned} & \therefore & (-1)^2+y^2 & =2 \\ \Rightarrow & & y & = \pm 1 \\ \Rightarrow & & P & =(-1,1) \text { and } \mathrm{Q}=(-1,-1) \end{aligned}$

Area of quadrilateral $\mathrm{PQRS}=\frac{1}{2} \times 3 \times(2+8) =15$ Sq. units.

Hint:

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

(ii) Equation of tangent of the parabola $y^2=4 a x$ at $\left(a t^2, 2 a t\right)$ is $t y=x+a t^2$

(iii) If a line $a x+b y+c=0$ touches the circle $\left(x-x_1\right)^2+\left(y-y_1\right)^2=R^2$, then

$\mathrm{R}=\frac{\left|a x_1+b y_1+\mathrm{C}\right|}{\sqrt{a^2+b^2}}$

Let the equation of ellipse $\mathrm{E}_2$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Now, it is given that $\mathrm{E}_2$ passes through $(0,4)$

$\begin{aligned} \Rightarrow \quad \frac{0^2}{a^2}+\frac{4^2}{b^2} & =1 \\ b^2 & =16 \end{aligned}$

Also, $\mathrm{E}_2$ passes through $( \pm 3, \pm 2)$

$\begin{aligned} & \Rightarrow \quad \frac{3^2}{a^2}+\frac{2^2}{b^2}=1 \\ & \Rightarrow \quad \frac{9}{a^2}+\frac{4}{16}=1 \\ & \Rightarrow \quad \frac{9}{a^2}=\frac{3}{4} \\ & \Rightarrow \quad a^2=12 \\ & \text { Now, } \quad b^2>a^2 \\ & \Rightarrow \quad b>a \\ & \therefore \quad a^2=b^2\left(1-e^2\right) \\ & \Rightarrow \quad 12=16\left(1-e^2\right) \\ & \Rightarrow \quad \frac{12}{16}=1-e^2 \\ & \Rightarrow \quad e^2=1-\frac{3}{4} \\ & \Rightarrow \quad e^2=\frac{1}{4} \\ & \Rightarrow \quad e=\frac{1}{2} \\ \end{aligned}$

Equation of chord of contact is ${x \over 3} + y = 1$, i.e. $x = 3(1 - y)$

Solving it with the ellipse ${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$

${(1 - y)^2} + {{{y^2}} \over 4} = 1 \Rightarrow 4({y^2} - 2y + 1) + {y^2} = 4$

$\therefore$ $y = 0 \Rightarrow 5{y^2} - 8y = 0$

$\therefore$ $y = 0,\,8/5$

Correspondingly $x = 3,\, - 9/5$

Points are $(3,0)$ and $\left( { - {9 \over 5},{8 \over 5}} \right)$

From the given condition,

$\sqrt {{{(x - 3)}^2} - {{(y - 4)}^2}} = {{|x + 3y - 3|} \over {\sqrt {1 + 9} }}$

$ \Rightarrow 10\{ ({x^2} - 6x + 9) + ({y^2} - 8y + 16)\} = {(x + 3y - 3)^2}$

$ \Rightarrow 10{x^2} + 10{y^2} - 60x - 80y + 250 = {x^2} + 9{y^2} + 9 + 6xy - 6x - 18y$

$ \Rightarrow 9{x^2} + {y^2} - 6xy - 54x - 62y + 241 = 0$

Equation of AB is

$y - 0 = {{{8 \over 5}} \over { - {9 \over 5} - 3}}(x - 3) = {8 \over { - 24}}(x - 3)$

$ \Rightarrow y = - {1 \over 3}(x - 3)$

$ \Rightarrow x + 3y = 3$ ...... (i)

Equation of the straight line perpendicular to AB through P is $3x - y = 5$.

Equation of PA is $x - 3 = 0$.

The equation of straight line perpendicular to PA through $B\left( {{{ - 9} \over 5},{8 \over 5}} \right)$ is $y = {8 \over 5}$.

Hence, the orthocentre is $\left( {{{11} \over 5},{8 \over 5}} \right)$.

The normal is

$4x\sec \phi - 2y\cos ec\phi = 12$

Now, the points Q and M are given by

$Q \equiv (3\cos \phi ,0)$

$M \equiv (\alpha ,\beta )$

Therefore, $\alpha = {{3\cos \phi + 4\cos \phi } \over 2} = {7 \over 2}\cos \phi \Rightarrow \cos \phi = {2 \over 7}\alpha $

and $\beta = \sin \phi ;{\cos ^2}\phi + {\sin ^2}\phi = 1$.

Therefore, ${4 \over {49}}{\alpha ^2} + {\beta ^2} = 1 \Rightarrow {4 \over {49}}{x^2} + {y^2} = 1$

Hence, the rectum is $x = \pm 2\sqrt 3 $.

Hence,

${{48} \over {49}} + {y^2} = 1 \Rightarrow y = \pm {1 \over 7}$

$\left( { \pm 2\sqrt 3 , \pm {1 \over 7}} \right)$

Hence, the locus of M intersects the latus rectum of the given ellipse at the points

Equation of line AM is $x + 3y - 3 = 0$.

Perpendicular distance of line from origin is $3/\sqrt {10} $.

Length of AM is

$2\sqrt {9 - {9 \over {10}}} = 2 \times {9 \over {\sqrt {10} }}$

The required area of the rectangle is

${1 \over 2} \times 2 \times {9 \over {\sqrt {10} }} \times {3 \over {\sqrt {10} }} = {{27} \over {10}}$ sq. unit

(P) We have ${1 \over {{k^2}}} = 4\left( {1 + {{{h^2}} \over {{k^2}}}} \right) \Rightarrow 1 = 4({k^2} + {h^2})$

Hence, ${h^2} + {k^2} = {\left( {{1 \over 2}} \right)^2}$ , which is a circle.

(Q) If $|z - {z_1}| - |z - {z_2}| = k$, where $k < |{z_1} - {z_2}|$, the locus is a hyperbola.

(R) Let $t = \tan \alpha $. Hence,

$x = \sqrt 3 \cos 2\alpha $ and $y = \sin 2\alpha $

or $\cos 2\alpha = {x \over {\sqrt 3 }}$ and $\sin 2\alpha = y$

${{{x^2}} \over 3} + {y^2} = {\sin ^2}2\alpha + {\cos ^2}2\alpha = 1$, which is an ellipse.

(S) If eccentricity is $[1,\infty]$, then the conic can be a parabola (if $e=1$) and a hyperbola if $e\in(1,\infty)$.

(T) Let $z = x + iy;x,y \in R$. Hence,

${(x + 1)^2} - {y^2} = {x^2} + {y^2} + 1$

$ \Rightarrow {y^2} = x;$ which is a parabola

Given that,

${C_1}:{y^2} = 4x$

${C^2}:{x^2} + {y^2} - 6x + 1 = 0$

Putting ${y^2} = 4x$ in ${x^2} + {y^2} - 6x + 1 = 0$, we get

${x^2} + 4x - 6x + 1 = 0 \Rightarrow {(x - 1)^2} = 0$

$x = 1$

putting $x = 1$ in ${y^2} = 4x$ we get $y = \, \pm ~2$

So, the curves touches each other at two points (1, 2) and (1, $-$2),

Equation of tangent to ellipse.

$\frac{x \cos \theta}{5}+\frac{y \sin \theta}{2}=1$

For all real values of $\theta$

This line also touches the given circle which implies that

$\left|\frac{-10}{\sqrt{4 \cos ^{2} \theta+25 \sin ^{2} \theta}}\right|=4$

$\Rightarrow 4\left(4 \cos ^{2} \theta+25 \sin ^{2} \theta\right)=25$

$\Rightarrow 16 \cos ^{2} \theta+100 \sin ^{2} \theta=25$

$\Rightarrow \quad 84 \sin ^{2} \theta+16=25$

$\Rightarrow \sin ^{2} \theta=\frac{9}{84}$ and $\cos ^{2} \theta=\frac{75}{84}$

Since, the tangent is in first quadrant, its slope must negative. Hence, the equation of tangent is

$\Rightarrow y=\left(\frac{-2}{5} \frac{\cos \theta}{\sin \theta}\right) x+\frac{2}{\sin \theta}$

$\Rightarrow y=\frac{-2}{5} \times \sqrt{\frac{75}{9}} x+\frac{2}{3} \sqrt{84}$

$\Rightarrow y=\frac{-2}{\sqrt{3}} x+\frac{4}{\sqrt{3}} \sqrt{7}$

$\Rightarrow 2 x+\sqrt{3} y=4 \sqrt{7}$

Therefore, the common tangent is

$2 x+\sqrt{3} y=4 \sqrt{7}$

The common tangent meet the axes in the points.

A $(2 \sqrt{7}, 0)$ and $B\left(0,4 \frac{\sqrt{7}}{\sqrt{3}}\right)$

Hence, length $\mathrm{AB}=\sqrt{(0-2 \sqrt{7})^{2}+\left(\frac{4 \sqrt{7}}{\sqrt{3}}\right)^{2}-0}$

$=\sqrt{28+\frac{112}{3}}=\sqrt{\frac{196}{3}}=\frac{14}{\sqrt{3}} \text { units }$