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 $