Ellipse

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

for r > 1,     ${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$

$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)} $

$ = \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}} $

$ = \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}} $

If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively,

then distance between focus and vertex,

a $-$ ae = ${3 \over 2}$ (given)

$ \Rightarrow $ $\,\,\,$ a (1 $-$ e) = ${3 \over 2}$

Length of latus rectum,

${{2{b^2}} \over a} = 4$

$ \Rightarrow $ $\,\,\,$ b² = 2a

$ \Rightarrow $ $\,\,\,$ a²(1 $-$ e²) = 2a   [As b² = a² (1 $-$ e²)]

$ \Rightarrow $ $\,\,\,$ a (1 $-$ e) ( 1 + e) = 2

Putting    a (1 $-$ e) = ${3 \over 2}$

$ \Rightarrow $ $\,\,\,$ ${3 \over 2}$ (1 + e) = 2

$ \Rightarrow $ $\,\,\,$ 3 + 3e = 4

$ \Rightarrow $ $\,\,\,$ e = ${1 \over 3}$

Centre at (0, 0)

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

at point (4, $-$ 1)

${{16} \over {{a^2}}} + {1 \over {{b^2}}}$ = 1

$ \Rightarrow $   16b² + a² = a²b²         . . . .(i)

at point ($-$ 2, 2)

${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$

$ \Rightarrow $   4b² + 4a² = a²b²          . . . .(ii)

$ \Rightarrow $   16b² + a² = 4a² + 4b²

From equations (i) and (ii)

$ \Rightarrow $   3a² = 12b²

$ \Rightarrow $   a² = 4b²

b² = a²(1 $-$ e²)

$ \Rightarrow $   e² = ${3 \over 4}$

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

e = 3/5 & 2ae = 6  $ \Rightarrow $   a = 5

$ \because $   b² = a² (1 $-$ e²)

$ \Rightarrow $   b² = 25(1 $-$ 9/25)


$ \Rightarrow $ b = 4

$ \therefore $ Area of required quadrilateral

= 4(1/2 ab) = 2ab = 40

Given e = ${1 \over 2}$ and ${a \over e}$ = 4

$ \therefore $ $a$ = 2

We have b² = $a$² (1 – e²) = $4\left( {1 - {1 \over 4}} \right)$ = 3

$ \therefore $ Equation of ellipse is

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

Now, the equation of normal at $\left( {1,{3 \over 2}} \right)$ is

${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$

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

$ \Rightarrow $ 4x – 2y = 1

Equation of tangent to ellipse

${x \over {\sqrt {27} }}$ cos$\theta $ + ${y \over {\sqrt 3 }}$sin$\theta $ = 1

Area bounded by line and co-ordinate axis

$\Delta $ = ${1 \over 2}$ . ${{\sqrt {27} } \over {\cos \theta }}.{{\sqrt 3 } \over {\sin \theta }}$ = ${9 \over {\sin 2\theta }}$

$\Delta $ = will be minimum when sin 2$\theta $ = 1

$\Delta $_min = 9

The end point of latus rectum of ellipse

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ in first quadrant is

$\left( {ae,{{{b^2}} \over a}} \right)$ and the tangent at this point

intersects $x$-axis at $\left( {{a \over e},0} \right)$ and

$y$-axis at $(0,a).$

The given ellipse is ${{x{}^2} \over 9} + {{{y^2}} \over 5} = 1$

Then ${a^2} = 9,{b^2} = 5$

$ \Rightarrow e = \sqrt {1 - {5 \over 9}} = {2 \over 3}$

$\therefore$ end point of latus rectum in first quadrant is

$L\left( {2,\,\,5/3} \right)$

Equation of tangent at $L$ is ${{2x} \over 9} + {y \over 3} - 1$

It meets $x$-axis at $A(9/2, 0)$

and $y$-axis at $B(0,3)$

$\therefore$ Area of $\Delta OAB = {1 \over 2} \times {9 \over 2} \times 3 = {{27} \over 4}$


By symmetry area of quadrilateral

$ = 4 \times \left( {Area\,\,\Delta OAB} \right) = 4 \times {{27} \over 4} = 27\,\,$ sq. units.

Given $e{q^n}$ of ellipse can be written as

${{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2$

Now, equation of any variable tangent is

$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)$

where $m$ is slope of the tangent

So, equation of perpendicular line drawn-

from center to tangent is

$y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Eliminating $m,$ we get

$\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}$

$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}$

$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$

From the given equation of ellipse, we have

$a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}} $

$ \Rightarrow e = {{\sqrt 7 } \over 4}$

Now, radius of this circle $ = {a^2} = 16$

$ \Rightarrow Focii = \left( { \pm \sqrt 7 ,0} \right)$

Now equation of circle is

${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16$

${x^2} + y{}^2 - 6y - 7 = 0$

Given equation of ellipse is $2{x^2} + {y^2} = 4$

$ \Rightarrow {{2{x^2}} \over 4} + {{{y^2}} \over 4} = 1 \Rightarrow {{{x_2}} \over 2} + {{{y^2}} \over 4} = 1$

Equation of tangent to the ellipse ${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$ is

$y = mx \pm \sqrt {2{m^2} + 4} \,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

( as equation of tangent to the ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$

is $y=mx+c$ where $c = \pm \sqrt {{a^2}{m^2} + {b^2}} $ )

Now, Equation of tangent to the parabola

${y^2} = 16\sqrt 3 x$ is $y = mx + {{4\sqrt 3 } \over m}\,\,\,\,\,\,\,\,...\left( 2 \right)$

( as equation of tangent to the parabola

${y^2} = 4ax$ is $y = mx + {a \over m}$ )

On comparing $(1)$ and $(2),$ we get

${{4\sqrt 3 } \over m} = \pm \sqrt {2{m^2} + 4} $

Squaring on both the sides, we get

$16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2}$

$ \Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \Rightarrow 2{m^4} + 4{m^2} - 48 = 0$

$ \Rightarrow {m^4} + 2{m^2} - 24 = 0 \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0$

$ \Rightarrow {m^2} = 4$ ( as ${m^2} \ne - 6$ ) $ \Rightarrow m = \pm 2$

$ \Rightarrow $ Equation of common tangents are $y = \pm 2x \pm 2\sqrt 3 $

Thus, statement - $1$ is true.

Statement - $2$ is obviously true.

Equation of circle is ${\left( {x - 1} \right)^2} + {y^2} = 1$

$ \Rightarrow $ radius $=1$ and diameter $=2$

$\therefore$ Length of semi-minor axis is $2.$

Equation of circle is ${x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}$

$ \Rightarrow $ radius $=2$ and diameter $=4$

$\therefore$ Length of semi major axis is $4$

We know, equation of ellipse is given by

${{{x^2}} \over {\left( {Major\,\,\,axi{s^{\,\,2}}} \right)}} + {{{y^2}} \over {\left( {Minor\,\,\,axi{s^{\,\,2}}} \right)}} = 1$

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

$ \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 4} = 1$

$ \Rightarrow {x^2} + 4{y^2} = 16$

Let the ellipse be ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$

It press through $(-3, 1)$ so ${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$

Also, ${b^2} = {a^2}\left( {1 - 2/5} \right)$

$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Solving $(i)$ and $(ii)$ we get ${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$

So, the equation of the ellipse is $3{x^2} + 5{y^2} = 32$

The given ellipse is ${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$

So $A=(2,0)$ and $B = \left( {0,1} \right)$

If $PQRS$ is the rectangular in which it is inscribed, then

$P = \left( {2,1} \right).$

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

be the ellipse circumscribing the rectangular $PQRS$.

Then it passes through $P\,\,(2,1)$

$\therefore$ ${4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)$

Also, given that, it passes through $(4,0)$

$\therefore$ ${{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16$

$ \Rightarrow {b^2} = 4/3$ $\left[ {\,\,} \right.$ substituting ${{a^2} = 16\,\,}$ in $\left. {e{q^n}\left( a \right)\,\,} \right]$

$\therefore$ The required ellipse is ${{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1$

or ${x^2} + 12y{}^2 = 16$

Perpendicular distance of directrix from focus

$ = {a \over e} - ae = 4$

$ \Rightarrow a\left( {2 - {1 \over 2}} \right) = 4$

$ \Rightarrow a = {8 \over 3}$

$\therefore$ Semi major axis $=8/3$

$2ae = 6 \Rightarrow ae = 3;\,\,2b = 8 \Rightarrow b = 4$

${b^2} = {a^2}\left( {1 - {e^2}} \right);16 = {a^2} - {a^2}{e^2}$

$ \Rightarrow a{}^2 = 16 + 9 = 25$

$ \Rightarrow a = 5$

$\therefore$ $e = {3 \over a} = {3 \over 5}$

as $\angle FBF' = {90^ \circ }$

$ \Rightarrow F{B^2} + F'{B^2} = FF{'^2}$

$\therefore$ ${\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + \left( {\sqrt {{a^2}{e^2} + {b^2}} } \right) = {\left( {2ae} \right)^2}$

$ \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2}$

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



Also ${e^2} = 1 - {b^2}/{a^2} = 1 - {e^2}$

$ \Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}$

$e = {1 \over 2}.\,\,$ Directrix, $x = {a \over e} = 4$

$\therefore$ $a = 4 \times {1 \over 2} = 2$

$\therefore$ $b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $

Equation of elhipe is

${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$