Ellipse

$(\mathrm{h}, \mathrm{k})$ lie on circle $x^2+y^2=1$.

so, $h^2+k^2=1--$ (1)

$(2 h+1), 3 k+2)$ lies on ellipse This means $x=2 h+1$ and $y=3 k+2$ lies on ellipse

Substitute $\mathrm{h}=\frac{x-1}{2}$ and $\mathrm{k}=\frac{y-2}{3}$ in equation (1)

$ \begin{aligned} & \left(\frac{x-1}{2}\right)^2+\left(\frac{y-2}{3}\right)^2=1 \\ & \frac{(x-1)^2}{4}+\frac{(y-2)^2}{9}=1 \end{aligned} $

This is shifted ellipse of the form $\frac{\left(x-x_1\right)^2}{-2}+\frac{\left(y-y_1\right)^2}{12}=1 \mathrm{a}=2, \mathrm{~b}=3$

eccentricity e $=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}$

$e^2=\frac{5}{9} \Longrightarrow \frac{5}{e^2}=9$ Ans

$E_1: x^2+4 y^2=1, E_2: 4 x^2+y^2=1$

Intersection point in $1^{\text {st }}$ quadrant is $P\left(\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$ tangent to $E_1$ at point $P$ is $x \cdot \frac{1}{\sqrt{5}}+4 y \cdot \frac{1}{\sqrt{5}}=1$

$ \Rightarrow \quad x+4 y=\sqrt{5}, m_1=-\frac{1}{4} $

Tangent to $E_2$ at point $P$ is $4 x \cdot \frac{1}{\sqrt{5}}+y \cdot \frac{1}{\sqrt{5}}=1$

$ \begin{aligned} & \Rightarrow \quad 4 x+y=\sqrt{5}, m_2=-4 \\ & \Rightarrow \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-\frac{1}{4}+4}{1+\frac{1}{4} \times 4}\right|=\frac{15}{8} \\ & \Rightarrow \quad 4 \tan \theta=\frac{15}{2}=07.50 \end{aligned} $

For the parabola

$ P:\ y^2=4kx $

its latus rectum is the line

$ x=k $

and the endpoints of the latus rectum are

$ (k,\,2k)\quad \text{and}\quad (k,\,-2k). $

So, if the line segment joining the points of intersection of the parabola and ellipse is the latus rectum of the parabola, then the two curves intersect at

$ (k,\pm 2k). $

Now for the ellipse

$ E:\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, $

its latus rectum is the chord through a focus perpendicular to the major axis.

Since the common chord is vertical, the major axis of the ellipse must be along the $x$-axis.

Hence the foci are

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

where

$ e=\sqrt{1-\frac{b^2}{a^2}}. $

The latus rectum of the ellipse is the vertical line through a focus:

$ x=ae. $

Its endpoints are

$ \left(ae,\ \pm \frac{b^2}{a}\right). $

But this same segment is also the latus rectum of the parabola, whose endpoints are

$ (k,\pm 2k). $

Therefore, comparing coordinates:

$ k=ae $

and

$ 2k=\frac{b^2}{a}. $

Substitute $k=ae$ into the second equation:

$ 2ae=\frac{b^2}{a}. $

So,

$ b^2=2a^2e. $

Now for the ellipse,

$ b^2=a^2(1-e^2). $

Hence,

$ a^2(1-e^2)=2a^2e. $

Dividing by $a^2$:

$ 1-e^2=2e. $

So,

$ e^2+2e-1=0. $

Since eccentricity is positive,

$ e=-1+\sqrt{2}. $

Now calculate:

$ e^2+2\sqrt{2}. $

First,

$ e^2=(\sqrt{2}-1)^2=2+1-2\sqrt{2}=3-2\sqrt{2}. $

Therefore,

$ e^2+2\sqrt{2}=3. $

So the required value is

$ \boxed{3}. $

Let the variable point be $B(h,k)$.

We are told that the circle having $AB$ as diameter touches the fixed circle $x^2+y^2=36$ internally.

If a circle has endpoints of a diameter at $A(3,0)$ and $B(h,k)$, then its equation is

$ (x-h)(x-3) + (y-k)(y-0) = 0 $

Expanding this, we get

$x^2 + y^2 - (h+3)x - ky + 3h = 0$

So, the centre of this circle is

Centre $\left(\frac{h+3}{2}, \frac{k}{2}\right)$

Now, the fixed circle $x^2+y^2=36$ has centre $C_1(0,0)$ and radius $R=6$.

The variable circle (with diameter $AB$) has radius equal to half of $AB$, so

$r=\frac{1}{2}\left(\sqrt{(h-3)^2 + k^2}\right)$

Since the two circles touch internally, the distance between their centres equals the difference of radii:

$\therefore\ C_1C_2 = |R-r|$

Here, $C_2\left(\frac{h+3}{2},\frac{k}{2}\right)$. So,

$\sqrt{\left(\frac{h+3}{2}\right)^2 + \left(\frac{k}{2}\right)^2} = \left|6 - \frac{1}{2}\left(\sqrt{(h-3)^2 + k^2}\right)\right|$

On simplifying and writing the locus in terms of a general point $(x,y)$ (i.e., replacing $(h,k)$ by $(x,y)$), we get

$\sqrt{(x+3)^2 + y^2} + \sqrt{(x-3)^2 + y^2} = 12$

This is the standard form of an ellipse: sum of distances from two fixed points (foci) is constant $=2a$.

So,

$2a = 12 \quad \therefore\ a = 6$

Comparing, the foci are $(-3,0)$ and $(3,0)$.

The distance between the foci is $6$, so $2c=6$. Using $c=ae$, we write

$2ae = 6$

Substituting $a=6$:

$12e = 6$

So,

$e = \frac{1}{2}$

Finally,

$\therefore\ 72e^2 = 72\left(\frac{1}{4}\right) = 18$

$\begin{aligned} & E_1=\frac{x^2}{9}+\frac{y^2}{4} \Rightarrow e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \\ & E_2: \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\ & e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{a^2}{4}} \Rightarrow \frac{5}{9}=1-\frac{a^2}{4} \\ & a^2=\frac{16}{9} \\ & E_2: \frac{x^2}{16}+\frac{y^2}{4}=1 \\ & E_3: \frac{x^2}{\frac{16}{9}}+\frac{y^2}{b^2}=1 \\ & e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{b^2}{16}} \Rightarrow b^2=\frac{64}{81} \end{aligned}$

$\begin{aligned} & \mathrm{E}_3=\frac{\mathrm{x}^2}{\frac{16}{9}}+\frac{\mathrm{y}^2}{\frac{64}{81}}=1 \\ & \mathrm{~A}_1=\pi \times 3 \times 2 \Rightarrow 6 \pi \\ & \mathrm{~A}_2=\pi \times \frac{4}{3} \times 2=\frac{8 \pi}{3} \\ & \mathrm{~A}_3=\pi \times \frac{4}{3} \times \frac{8}{9}=\frac{32 \pi}{81} \\ & \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}}=6 \pi+\frac{8 \pi}{3}+\frac{32 \pi}{81}+\ldots \infty \Rightarrow \frac{6 \pi}{1-\frac{4}{9}} \Rightarrow \frac{54 \pi}{5} \\ & \therefore \frac{5}{\pi} \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}} \Rightarrow \frac{5}{\pi} \times \frac{54 \pi}{5}=54 \end{aligned}$

Given curve,

$9{x^2} + 16{y^2} = 144$

$ \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$

$ \Rightarrow {{{x^2}} \over {{4^2}}} + {{{y^2}} \over {{3^2}}} = 1$

$\therefore$ a = 4 and b = 3

So, general point on the ellipse is $ = (4\cos \theta ,3\sin \theta )$

We know,

Equation of tangent to a given ellipse at its point $(a\cos\theta ,b\sin \theta )$ is

${{x\cos \theta } \over a} + {{y\sin \theta } \over b} = 1$

$\therefore$ Here equation of tangent at point $(4\cos \theta ,3\sin \theta )$ is

${{x\cos \theta } \over 4} + {{y\sin \theta } \over 3} = 1$

When this tangent cut's x axis then y = 0.

$\therefore$ ${{x\cos \theta } \over 4} + 0 = 1$

$ \Rightarrow x = 4\sec \theta $

$\therefore$ Point of intersection at x axis is $A(4\sec \theta ,0)$.

When this tangent cut's y axis then x = 0.

$\therefore$ $0 + {{y\sin \theta } \over 3} = 1$

$ \Rightarrow y = 3\cos ec\theta $

$\therefore$ Point of intersection at y axis is $B(0,3\cos ec\theta )$

$\therefore$ Length of AB

$ = \sqrt {{{(4\sec \theta - 0)}^2} + {{(0 - 3\cos ec\theta )}^2}} $

$ = \sqrt {16{{\sec }^2}\theta + 9\cos e{c^2}\theta } $

$ = \sqrt {16(1 + {{\tan }^2}\theta ) + 9(1 + {{\cot }^2}\theta )} $

$ = \sqrt {25 + 16{{\tan }^2}\theta + 9{{\cot }^2}\theta } $

We know, $AM \ge GM$

$\therefore$ ${{16{{\tan }^2}\theta + 9{{\cot }^2}\theta } \over 2} \ge \sqrt {(16{{\tan }^2}\theta )(9{{\cot }^2}\theta )} $

$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 2(4\tan \theta )(3\cot \theta )$

$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 2 \times 4 \times 3$

$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 24$

$\therefore$ $AB = \sqrt {25 + 16{{\tan }^2}\theta + 9{{\cot }^2}\theta } $

$ \ge \sqrt {25 + 24} $

$ \ge \sqrt {49} $

$ \ge 7$

$\therefore$ Minimum length of $AB = 7$.

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

$\eqalign{ & T \equiv y = mx\, \pm \,\sqrt {2{m^2} + 4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \left( {\sqrt 2 ,2\sqrt 2 - 2} \right) \cr} $

$ \Rightarrow \left( {2\sqrt 2 - 2 - m\sqrt 2 } \right) = \pm \,\sqrt {2{m^2} + 4} $

$ \Rightarrow 2{m^2} - 2m\sqrt 2 \left( {2\sqrt {2 - 2} } \right) + 4(3 - 2\sqrt 2 ) = 2{m^2} + 4$

$ \Rightarrow - 2\sqrt 2 m(2\sqrt 2 - 2) = 4 - 12 + 8\sqrt 2 $

$ \Rightarrow - 4\sqrt 2 m(\sqrt 2 - 1) = 8(\sqrt 2 - 1)$

$ \Rightarrow m = - \sqrt 2 $ and $m \to \infty $

$\therefore$ Tangents are $x = \sqrt 2 $ and $y = - \sqrt 2 x + \sqrt 8 $

$\therefore$ $P(\sqrt 2 ,0)$ and $Q(1,\sqrt 2 )$

and $S = (0, - \sqrt 2 )$

$\therefore$ ${(PS)^2} + {(QS)^2} = 4 + 9 = 13$

Equation of ellipse is : ${x^2} + 4{y^2} + 2x + 8y - \lambda = 0$

${(x + 1)^2} + 4{(y + 1)^2} = \lambda + 5$

${{{{(x + 1)}^2}} \over {\lambda + 5}} + {{{{(y + 1)}^2}} \over {\left( {{{\lambda + 5} \over 4}} \right)}} = 1$

Length of latus rectum $ = {{2\,.\,\left( {{{\lambda + 5} \over 4}} \right)} \over {\sqrt {\lambda + 5} }} = 4$.

$\therefore$ $\lambda = 59$.

Length of major axis $ = 2\,.\,\sqrt {\lambda + 5} = 16 = l$

$\therefore$ $\lambda + l = 75$.

We will consider three cases

Case 1 : Circle passes through origin, that is, p = 0 the equation of circle becomes

x² + y² + 2x + 4y = 0

x = 0 $\Rightarrow$ y² + 4y = 0 $\Rightarrow$ y(y + 4) = 0

y = 0, $-$4

y = 0 $\Rightarrow$ x² + 2x = 0 $\Rightarrow$ x(x + 2) = 0 $\Rightarrow$ x = 0, $-$ 2

Case 2 : Circle touches y axis then circle will intersect x axis at two distinct points

Put y = 0 in equation of circle, we get

x² + 2x $-$ p = 0

Now from g² $-$ c > 0 and f² $-$ c = 0

$\Rightarrow$ 1² $-$ ($-$p) > 0 and 2² $-$ ($-$p) = 0

1 + p > 0 and 4 + p = 0

p > $-$1 and p = $-$ 4

which is a contradiction. Also, for p = $-$4 we get

x² $-$ 2x + 4 = 0

$x = {{2 \pm \sqrt {4 - 4 \times 4} } \over 2} = {{2 \pm 2\sqrt 3 i} \over 2} = 1 \pm \sqrt 3 i$

They are imaginary roots.

Therefore, Case 2 is not possible.

Case 3 : Circle touches x axis. Then the circle will intersect y axis at two distinct points.

Substituting x = 0, we get

y² + 4y $-$ p = 0

Now, from g² $-$ C = 0 and f² $-$ C > 0, we have

1² $-$ ($-$p) = 0 and 2² = ($-$p) > 0

p = $-$1 and 4 + p > 0 $\Rightarrow$ p > $-$4

Therefore, the equation of circle becomes

y² + 4y + 1 = 0

$ \Rightarrow y = {{ - 4 \pm \sqrt {16 - 4} } \over 2} = {{ - 4 \pm \sqrt {12} } \over 2}$

$y = {{ - 4 \pm \sqrt {4 \times 3} } \over 2}$

$ = {{ - 4 \pm 2\sqrt 3 } \over 2} = - 2 \pm \sqrt 3 $ real values

Thus, Case 3 is possible.

Thus, for p = 0 and p = $-$1 the circle and coordinates have exactly three points in common. Hence, the correct answer is 2.

A vertical line passing through $(h, 0)$ is $x=h$ and this vertical line intersect the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at P and Q.

Put $x=h$ in the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$

$\begin{aligned} & \Rightarrow \frac{h^2}{4}+\frac{y^2}{3}=1 \\ & \Rightarrow y= \pm \sqrt{3-\frac{3 h^2}{4}} \end{aligned}$

So $\mathrm{P}=\left(h, \sqrt{3-\frac{3 h^2}{4}}\right)$ and $\mathrm{Q}=\left(h,-\sqrt{3-\frac{3 h^2}{4}}\right)$

Equation of tangent of ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at P is

$\frac{h x}{4}+\frac{y}{3} \sqrt{3-\frac{3 h^2}{4}}=1 \quad \text{... (i)}$

Equation of tangent of ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at Q is

$\frac{h x}{4}-\frac{y}{3} \sqrt{3-\frac{3 h^2}{4}}=1 \quad \text{... (ii)}$

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

On solving the equations (i) and (ii)

$\Rightarrow \quad \mathrm{R}=\left(\frac{8}{h}, 0\right)$

Given, $\Delta(h)=$ Area of $\Delta \mathrm{PQR}$

$\begin{aligned} \Rightarrow \quad \Delta(h) & =\frac{1}{2}\left(\frac{8}{h}-h\right) \cdot 2 \sqrt{3-\frac{3 h^2}{4}} \\ \Rightarrow \quad \Delta(h) & =\frac{\sqrt{3}}{2}\left(\frac{8}{h}-h\right) \sqrt{4-h^2} \\ \Rightarrow \quad \frac{d \Delta(h)}{d h}= & \frac{\sqrt{3}}{2}\left(\frac{-8}{h^2}-1\right) \sqrt{4-h^2} +\frac{\sqrt{3}}{2}\left(\frac{8}{h}-h\right) \cdot \frac{(-2 h)}{2 \sqrt{4-h^2}} \end{aligned}$

$\begin{aligned} \Rightarrow \quad \frac{d \Delta(h)}{d h}= & \frac{-\sqrt{3}}{2 \sqrt{4-h^2}} \\ & {\left[\left(\frac{8}{h^2}+1\right)\left(4-h^2\right)+\left(\frac{8}{h}-h\right) h\right] } \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{-\sqrt{3}}{2 \sqrt{4-h^2}}\left[\frac{32}{h^2}-8+4-\hbar^2+8-h^2\right] \\ & \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{-\sqrt{3}}{2 \sqrt{4-h^2}}\left[-2 h^2+\frac{32}{h^2}+4\right] \\ & \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{\sqrt{3}}{h^2 \sqrt{4-h^2}}\left[h^4-2 h^2-16\right] \\ & \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{\sqrt{3}}{h^2 \sqrt{4-h^2}}\left[\left(h^2-1\right)^2-17\right] \end{aligned}$

For, $\frac{1}{2} \leq h \leq 1, \quad \frac{d \Delta(h)}{d h}<0$

$\therefore \quad \Delta(h)$ is a decreasing function for $x \in\left[\frac{1}{2}, 1\right]$

$\Rightarrow \Delta(1) \leq \Delta(h) \leq \Delta\left(\frac{1}{2}\right)$

Given $\quad \Delta_1=\max _{\frac{1}{2} \leq h \leq 1} \Delta(h)$ and

$\begin{aligned} \Delta_2 & =\min _{\frac{1}{2} \leq h \leq 1} \Delta(h) \\ \Rightarrow \quad \Delta_1 & =\Delta\left(\frac{1}{2}\right) \text { and } \Delta_2=\Delta(1) \\ \Rightarrow \quad \Delta_1 & =\frac{\sqrt{3}}{2}\left(16-\frac{1}{2}\right) \sqrt{4-\frac{1}{4}} \text { and } \\ \Delta_2 & =\frac{\sqrt{3}}{2}(8-1) \sqrt{4-1} \\ \Rightarrow \quad \Delta_1 & =\frac{\sqrt{3}}{2} \times \frac{31}{2} \times \frac{\sqrt{5} \cdot \sqrt{3}}{2} \text { and } \Delta_2=\frac{21}{2} \\ \Rightarrow \quad \frac{8 \Delta_1}{\sqrt{5}} & =93 \text { and } 8 \Delta_2=84 \end{aligned}$

$\Rightarrow \frac{8 \Delta_1}{\sqrt{5}}-8 \Delta_2=9$

Hints:

(i) If $f(x)$ is a decreasing function in $x \in[a, b]$, then $f(b) \leq f(x) \leq f(a)$

(ii) If $\left(x_1, y_1\right)$ is a point on the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then equation of tangent at $\left(x_1, y_1\right)$ is $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$.