Ellipse

${3 \over {2{a^2}}} + {1 \over {{b^2}}} = 1$ and $1 - {{{b^2}} \over {{a^2}}} = {1 \over 3}$

$ \Rightarrow {a^2} = 3{b^2} = 3$

$ \Rightarrow {{{x^2}} \over 3} + {{{y^2}} \over 2} = 1$ ...... (i)

Its focus is (1, 0)

Now, equation of circle is

${(x - 1)^2} + {y^2} = {4 \over 3}$ ..... (ii)

Solving (i) and (ii) we get

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

$ \Rightarrow P{Q^2} = {\left( {{4 \over {\sqrt 3 }}} \right)^2} = {{16} \over 3}$

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

${e^2} = 1 - {{{a^2}} \over {{c^2}}}$

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

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

Also b = ce

$ \Rightarrow c = {b \over e}$

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

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

$ \Rightarrow {e^2} + e - 1 = 0$

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

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

x-intercept = ${3\sqrt 3 }$ sec$\theta$

y-intercept = cosec$\theta$

sum = ${3\sqrt 3 }$ sec$\theta$ + cosec$\theta$ = f($\theta$) $\theta$$\in$$\left( {0,{\pi \over 2}} \right)$

$ \Rightarrow $ f'($\theta$) = ${3\sqrt 3 }$ sec$\theta$tan$\theta$ $-$ cosec$\theta$ cot$\theta$ = 0

$ \Rightarrow $ ${{3\sqrt 3 \sin \theta } \over {{{\cos }^2}\theta }} = {{\cos \theta } \over {\sin \theta }}$

$ \Rightarrow {\tan ^3}\theta = {\left( {{1 \over {\sqrt 3 }}} \right)^3}$

$ \Rightarrow \tan \theta = {1 \over {\sqrt 3 }}$

$ \Rightarrow $ $\theta$ = ${{\pi \over 6}}$

also f'($\theta$) changes sign $-$ to + hence minimum.

${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$ ... (1)

${x^2} + {y^2} = 4b$ .... (2)

${y^2} = 3{x^2}$ .... (3)

From eq (2) and (3)

x² = b and y² = 3b

From equation (1)

${b \over {16}} + {{3b} \over {{b^2}}} = 1$

$ \Rightarrow {b^2} + 48 = 16b$

$ \Rightarrow b = 12$

Ellipse : ${x \over 2} + {y \over 1} = 1$

Line : $x + y = 1$



Using homogenisation

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

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

${x^2} + 2{y^2} = 2{x^2} + 2{y^2} + 4xy$

${x^2} + 4xy = 0$

for $a{x^2} + 2hxy + b{y^2} = 0$

$\tan \theta = \left| {{{2\sqrt {{h^2} - ab} } \over {a + b}}} \right|$

$\tan \theta = \left| {{{2\sqrt {{{(2)}^2} - 0} } \over {1 + 0}}} \right|$

$\tan \theta = - 4$

$\cot \theta = - {1 \over 4}$

$\theta = {\cot ^{ - 1}}\left( { - {1 \over 4}} \right)$

$\theta = \pi - {\cot ^{ - 1}}\left( {{1 \over 4}} \right)$

$\theta = \pi - \left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {{1 \over 4}} \right)} \right)$

$\theta = {\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)$

Given that, $AP+BP=2a$

Using distance formula,

$AP=\sqrt{(x+ae)^2+y^2}$

and $B P=\sqrt{(x-a e)^2+y^2}$

Now, $\sqrt{(x+a e)^2+y^2}+\sqrt{(x-a e)^2+y^2}=2 a$

Squaring both sides,

$\begin{gathered} (x+a e)^2+y^2+(x-a e)^2+y^2 \\ +2 \sqrt{(x+a e)^2+y^2} \sqrt{(x-a e)^2+y^2}=4 a^2 \\ \Rightarrow \quad 2 x^2+2 y^2+2 a^2 e^2-4 a^2 \\ =-2 \sqrt{(x+a e)^2+y^2} \sqrt{(x-a e)^2+y^2} \\ x^2+y^2+a^2 e^2-2 a^2=-\sqrt{(x+a e)^2+y^2} \\ \\ \cdot \sqrt{(x-a e)^2+y^2} \end{gathered}$

Squaring both sides,

$\begin{aligned} & a^4 e^4-4 a^4 e^2+4 a^4+2 a^2 x^2 e^2-4 a^2 x^2 \\ & +2 a^2 y^2 e^2-4 a^2 y^2+x^4+2 x^2 y^2+y^4 \\ & =a^4 e^4-2 a^2 x^2 e^2+2 a^2 y^2 e^2+x^4+2 x^2 y^2+y^4 \\ & \Rightarrow \quad 4\left(a^4-a^4 e^2-a^2 x^2-a^2 y^2+a^2 x^2 e^2\right)=0 \\ & \Rightarrow \quad a^2\left(a^2-a^2 e^2-x^2-y^2+x^2 e^2\right)=0 \\ & \Rightarrow a^2\left(1-e^2\right)-x^2-y^2+x^2 e^2=0 \\ \end{aligned}$

$\begin{array}{left}\Rightarrow \quad b^2-x^2-y^2+x^2\left(1-\frac{b^2}{a^2}\right)=0 \quad\left[\because b^2=a^2\left(1-e^2\right)\right] \\ \Rightarrow \quad x^2\left[1-\frac{b^2}{a^2}-1\right]-y^2+b^2=0 \\ \Rightarrow x^2\left(-\frac{b^2}{a^2}\right)-y^2=-b^2 \\ \Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\left(\text { divide by } b^2\right)\end{array}$

Let two points be $A\left(a \cos \theta_1 b \sin \theta_1\right)$ and $\left(a \cos \theta_2, b \sin \theta_2\right)$.

$\begin{aligned} & O: \operatorname{origin}(0,0) \\ & \qquad \begin{aligned} m_{O A} & =\frac{b \sin \theta_1}{a \cos \theta_1}=\frac{b}{a} \tan \theta_1 \\ m_{O B} & =\frac{b}{a} \tan \theta_2 \\ \Rightarrow \quad m_{O A} \times m_{O B} & =-1 \Rightarrow \frac{b^2}{a^2} \tan \theta_1 \tan \theta_2=-1 \end{aligned} \end{aligned}$

In an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Distance between foci $=2 a e=6$

$\begin{aligned} & \Rightarrow \quad a e=3 \\ & \text { and length of minor axis }=2 b=8 \\ & \Rightarrow \quad b=4 \\ & \because \quad(a e)^2=a^2-b^2 \\ & \Rightarrow \quad 3^2=a^2-4^2 \Rightarrow a^2=25 \\ & \text { Now, } e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5} \\ \end{aligned}$

Given ellipse, $\frac{x^2}{25}+\frac{y^2}{16}=1$

Length of semi-major axis $=5$

Length of semi-minor axis $=4$

Centre $C=(0,0)$

If $P(x, y)$ be any point on the ellipse, then

maximum value of $C P=$ Length of the semi-major axis $=5$

minimum value of $C P=$ Length of semi-minor axis $=4$

Their sum $=5+4=9$

Equation of normal at $\left( {ae,{{{b^2}} \over a}} \right)$

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

It passes through (0,–b)

$ \therefore $ $0 - {{{b^2}\left( { - b} \right)} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$

$ \Rightarrow $ $a$b = ${a^2} - {b^2}$

$ \Rightarrow $ $a$b = ${a^2}{e^2}$ [as b² = ${a^2}\left( {1 - {e^2}} \right)$]

$ \Rightarrow $ $a$²b² = ${a^4}{e^4}$

$ \Rightarrow $ ${{{{b^2}} \over {{a^2}}}}$ = e⁴

$ \Rightarrow $ ${1 - {e^2}}$ = e⁴

$ \Rightarrow $ e⁴ + e² – 1 = 0

Let foot of perpendicular is (h, k)

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

$ \therefore $ a = 2, b = $\sqrt 2 $

and e = $\sqrt {1 - {2 \over 4}} $ = ${1 \over {\sqrt 2 }}$

$ \therefore $ Focus(ae, 0) = $\left( {\sqrt 2 ,0} \right)$

Equation of tangent

y = mx + $\sqrt {{a^2}{m^2} + {b^2}} $

$ \Rightarrow $ y = mx + $\sqrt {4{m^2} + 2} $

Passes through (h, k)

(k – mh)² = 4m² + 2 .....(1)

Line perpendicular to tangent will have slope $ - {1 \over m}$.

y - 0 = $ - {1 \over m}\left( {x - \sqrt 2 } \right)$

my = -x + ${\sqrt 2 }$

$ \therefore $ (h + mk)² = 2 .....(2)

Add equation (1) and (2)

k²(1 + m²) + h² (1 + m²) = 4 (1 + m²)

h² + k² = 4

x² + y² = 4 (Auxiliary circle)

$ \therefore $ $\left( { - 1,\sqrt 3 } \right)$ lies on the locus.

9x² + 16y² = 144

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

$ \therefore $ a = 4; b = 3;

Now e = $\sqrt {1 - {9 \over {16}}} = {{\sqrt 7 } \over 4}$

A and B are foci

PA + PB = 2a = 2 × 4 = 8

$e = {1 \over 2}$

$x = {a \over e} = 4$

$ \Rightarrow $ a = 2

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

$\Rightarrow {1 \over 4} = 1 - {{{b^2}} \over 4}$

${{{b^2}} \over 4} = {3 \over 4} \Rightarrow {b^2} = 3$

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

P(1, $\beta $) is on the ellipse

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

${{{\beta ^2}} \over 3} = {3 \over 4} \Rightarrow \beta = {3 \over 2}$

$ \Rightarrow P\left( {1,{3 \over 2}} \right)$

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

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

$ \Rightarrow $ $4x - 2y = 1$

Given ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ (a > b)

Length of latus rectum $ = {{2{b^2}} \over a} = 10$

$\phi (t) = {5 \over {12}} + t - {t^2}$

$ = {8 \over {12}} - {\left( {t - {1 \over 2}} \right)^2}$

$ \therefore $ $\phi {(t)_{\max }} = {8 \over {12}} = {2 \over 3}$

$ \therefore $ eccentricity (e) = ${2 \over 3}$

Also, ${e^2} = 1 - {{{b^2}} \over {{a^2}}}$

$ \Rightarrow {4 \over 9} = 1 - {{{b^2}} \over {{a^2}}}$

$ \Rightarrow {{{b^2}} \over {{a^2}}} = {5 \over 9}$

$ \Rightarrow {{{b^2}} \over a} \times {1 \over a} = {5 \over 9}$

$ \Rightarrow {5 \over a} = {5 \over 9}$

$ \Rightarrow a = 9$

$ \therefore $ ${b^2} = 5 \times 9 = 45$

$ \therefore $ ${a^2} + {b^2} = 81 + 45 = 126$

Let the equation of ellipse

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

Given 2b = ${4 \over {\sqrt 3 }}$

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

We know, Equation of tangent y = mx $ \pm $ $\sqrt {{a^2}{m^2} + {b^2}} $ ....(1)

Given tangent is x + 6y = 8

$ \Rightarrow $ y = $ - {1 \over 6}x + {8 \over 6}$ .....(2)

By comparing (1) and (2),

m = $ - {1 \over 6}$ and ${{a^2}{m^2} + {b^2}}$ = ${{16} \over 9}$

$ \Rightarrow $ ${{a^2}\left( {{1 \over {36}}} \right) + {4 \over 3}}$ = ${{16} \over 9}$

$ \Rightarrow $ ${{a^2} = 16}$

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

= $\sqrt {1 - {{{4 \over 3}} \over {16}}} $

= $\sqrt {{{11} \over {12}}} $ = ${1 \over 2}\sqrt {{{11} \over 3}} $

Let P be (x₁ , y₁)

Equation of normal at P is ${x \over {2{x_1}}} - {y \over {{y_1}}} = {1 \over 2} - 1$

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

$ \therefore $ ${{ - 1} \over {6\sqrt 2 {x_1}}} = - {1 \over 2}$

$ \Rightarrow $ x₁ = ${1 \over {3\sqrt 2 }}$

Also using 2$x_1^2$ + $y_1^2$ = 1

$ \Rightarrow $ $y_1^2$ = 1 - ${2 \over {18}}$

$ \Rightarrow $ y₁ = ${{2\sqrt 2 } \over 3}$

(0, $\beta $) is on the normal.

So 0 - ${\beta \over {{{2\sqrt 2 } \over 3}}}$ = $ - {1 \over 2}$

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

3x + 4y = 12$\sqrt 2 $

$ \Rightarrow $ y = $ - {{3x} \over 4} + 3\sqrt 2 $ is tangent to

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

$ \therefore $ c² = a²m² + b²

$ \Rightarrow $ ${\left( {3\sqrt 2 } \right)^2} = {a^2}{\left( { - {3 \over 4}} \right)^2} + 9$

$ \Rightarrow $ a² = 16

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

= $\sqrt {1 - {9 \over {16}}} $ = ${{\sqrt 7 } \over 4}$

Distance between focii = 2ae

= $2 \times 4 \times {{\sqrt 7 } \over 4}$

= $2\sqrt 7 $

Distance between foci = 2ae = 6

$ \Rightarrow $ ae = 3 .....(1)

Distance between directrices = ${{2a} \over e}$ = 12

$ \Rightarrow $ ${a \over e}$ = 6 .....(2)

from (1) and (2)

a² = 18

also a²e² = 9

$ \Rightarrow $ 18e² = 9

$ \Rightarrow $ e² = ${1 \over 2}$

We know e² = 1 - ${{{b^2}} \over {{a^2}}}$

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

$ \Rightarrow $ b² = 9

$ \therefore $ Length of latus rectum = ${{2{b^2}} \over a}$

= ${{2 \times 9} \over {\sqrt {18} }}$

= $3\sqrt 2 $

Given ellipse, $\frac{x^2}{16}+\frac{y^2}{12}=1$

Let eccentricity of ellipse be ' $e^{\prime}$

Then $b^2=a^2\left(1-e^2\right)$

Here, $b^2=12, a^2=16$

$ \begin{aligned} \therefore \quad 12 & =16\left(1-e^2\right) \\ 1-e^2 & =\frac{3}{4} \end{aligned} $

or $\quad e^2=1-\frac{3}{4}=\frac{1}{4}$ or $e=\frac{1}{2}$

If $\alpha, \beta$ are the eccentric angles of the ends of a focal chord of the ellipse, then eccentricity is given by

$ e=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)} $

Here, $\alpha=\frac{\pi}{3}, \beta=\theta$, and $e=\frac{1}{2}$

$ \frac{1}{2}=\frac{\cos \left(\frac{\pi / 3-\theta}{2}\right)}{\cos \left(\frac{\pi / 3+\theta}{2}\right)} $

Multiplying in $N^r$ and $D^r$ by $2 \sin \left(\frac{\pi / 3+\theta}{2}\right)$

$ \frac{1}{2}=\frac{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3-\theta}{2}\right)}{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3+\theta}{2}\right)} $

$ =\frac{\sin \left(\frac{\pi / 3+\theta+\pi / 3-\theta}{2}\right)+\sin \left(\frac{\pi / 3+\theta-\pi / 3+\theta}{2}\right)}{\sin (\pi / 3+\theta)} $

$ \begin{aligned} & \left\{\because 2 \sin A \cos B=\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{A-B}{2}\right)\right. \\ & \text { and } 2 \sin A \cos A=\sin 2 A\} \\ & \frac{1}{2}=\frac{\sin \frac{\pi}{3}+\sin \theta}{\sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta} \\ & \sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta=2\left(\sin \frac{\pi}{3}+\sin \theta\right) \\ & \frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=2\left(\frac{\sqrt{3}}{2}+\sin \theta\right) \end{aligned} $

$ \begin{aligned} & \text { } \frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=\sqrt{3}+2 \sin \theta \\ & \qquad \begin{array}{c}or\,\,\, \frac{\sqrt{3}}{2} \cos \theta=\frac{3}{2} \sin \theta+\sqrt{3} \\ \cos \theta=\sqrt{3} \sin \theta+2 \end{array} \\ & \text { or } \frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta=1 \\ & \text { or } \cos \frac{\pi}{3} \cos \theta-\sin \frac{\pi}{3} \sin \theta=1 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { or } \cos \left(\frac{\pi}{3}+\theta\right)=1 \\ & \text { or } \cos \left(\frac{\pi}{3}+\theta\right)=\cos 0^{\circ} \end{aligned}\\ &\text { On comparing both sides, we get }\\ &\begin{array}{rlrl} & \frac{\pi}{3}+\theta =0^{\circ} \text { or } \theta=-\frac{\pi}{3} \\ \therefore & \tan \theta=\tan \left(\frac{-\pi}{3}\right) \Rightarrow \tan \theta=-\sqrt{3} \end{array} \end{aligned} $

Given ellipse

$ 2 x^2+y^2=2 \text { or } x^2+\frac{y^2}{2}=1 $

Here, $\quad a^2=1, b^2=2$

Equation of tangent

$ x+2 y+k=0 \text { or } 2 y=-x-k $

or       $ y=-\frac{x}{2}-\frac{k}{2} $

Here,    $ c=\frac{-k}{2}, m=-\frac{1}{2} $

From the condition of tangent

$ \begin{aligned} c & = \pm \sqrt{a^2 m^2+b^2} \\ \Rightarrow \frac{-k}{2} & = \pm \sqrt{(1)\left(\frac{1}{4}\right)+2} \\ \frac{-k}{2} & = \pm \sqrt{\frac{9}{4}} \Rightarrow \frac{-k}{2}= \pm \frac{3}{2} \end{aligned} $

or $k= \pm 3$

But $k>0$, so, $k=3$

So, equation of tangent, $y=-\frac{x}{2}-\frac{3}{2}$ and point is $\left(\frac{1}{\sqrt{2}}, 1\right)$

The equation of the normal to the ellipse at point ( $x_1, y_1$ ) is given by

$ \begin{aligned} & \frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2 \\ \Rightarrow \quad & \frac{1 \times x}{1 / \sqrt{2}}-\frac{2 \times y}{1}=1-2 \\ & \sqrt{2} x-2 y=-1 \text { or } \sqrt{2} x-2 y+1=0 \end{aligned} $

$a \alpha^2+b \beta^2+c \alpha \beta+d=0$ is the transformed equation of

$ 4 x^2+\sqrt{3} x y+5 y^2-4=0 $

where,

$ \begin{align*} & \alpha=\frac{\sqrt{3}}{2} x+\frac{y}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \beta=\frac{-x}{2}+\frac{\sqrt{3} y}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

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

$ \begin{aligned} x & =\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2} \\ \Rightarrow \quad y & =\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta \end{aligned} $

Putting the value of $x$ and $y$ in

$ 4 x^2+\sqrt{3} x y+5 y^2-4=0, $

$ \begin{aligned} &\text { we get }\\ &\begin{aligned} & 4\left(\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2}\right)^2 +\sqrt{3}\left(\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2}\right)\left(\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta\right) +5\left(\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta\right)^2-4=0 \end{aligned} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & 5 \alpha^2+4 \beta^2+\sqrt{3} \alpha \beta-4=0 \\ \therefore & a=5, b=4, c=\sqrt{3}, d=-4 \\ \therefore & c(a+b+d)=\sqrt{3}(5+4-4)=5 \sqrt{3} \end{array} $

Given ellipse,

$ x^2+2 y^2=2 \Rightarrow \frac{x^2}{2}+\frac{y^2}{1}=1 $

Point $P(\sqrt{2} \cos \theta, \sin \theta)$ lie in ellipse Tangent at $P(\sqrt{2} \cos \theta, \sin \theta)$ on ellipse is,

$ x \cos +\sqrt{2} \sin \theta y=\sqrt{2} $

Intercept on line is, $A\left(\frac{\sqrt{2}}{\cos \theta}, 0\right)$ and $ B\left(0, \frac{1}{\sin \theta}\right) $

Let $(h, k)$ is mid-point of the $A B$.

$ \begin{array}{rlrl} \therefore & h =\frac{\sqrt{2}}{2 \cos \theta} \text { and } k=\frac{1}{2 \sin \theta} \\ \Rightarrow & \cos \theta =\frac{1}{\sqrt{2} h} \text { and } \sin \theta=\frac{1}{2 k} \end{array} $

Squaring and adding, we get

$ \frac{1}{2 h^2}+\frac{1}{4 k^2}=1 $

∴ Locus is, $\frac{1}{2 x^2}+\frac{1}{4 y^2}=1$

Given ellipse, $\frac{x^2}{16}+\frac{y^2}{12}=1$

End point of latus rectum

$ =\left( \pm a e, \frac{2 b^2}{a}\right)=(2,3) $

Equation of tangent at $(2,3)$ of ellipse

$\frac{x^2}{16}+\frac{y^2}{12}=1$ is,

$ \frac{2 x}{16}+\frac{3 y}{12}=1 \Rightarrow \frac{x}{8}+\frac{y}{4}=1 $

Area of quadrilateral

$ \begin{aligned} A B C D & =4 \times \frac{1}{2} \times O A \times O B \\ & =4 \times \frac{1}{2} \times 8 \times 4=64 \end{aligned} $

Since the foci of given ellipse lie on the $Y$-axis, it is vertical ellipse.

Let the required equation be $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$,

where $a^2>b^2$.

Foci are $(0, \pm c)=(0, \pm 1) \Rightarrow c=1$

$a=$ length of the semi-major axis

$ \begin{aligned} & =\frac{1}{2} \times \sqrt{5}=\frac{\sqrt{5}}{2} \\ \therefore \quad a^2 & =\frac{5}{4} \end{aligned} $

Now, $c^2=a^2-b^2 \Rightarrow 1=\left(\frac{\sqrt{5}}{2}\right)^2-b^2$

$ \Rightarrow \quad b^2=\frac{5}{4}-1=\frac{1}{4} $

∴ The required equation is

$ \begin{aligned} & \frac{x^2}{1 / 4}+\frac{y^2}{5 / 4}=1 \Rightarrow 4 x^2+\frac{4 y^2}{5}=1 \\ \Rightarrow & 20 x^2+4 y^2=5 \end{aligned} $

Equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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

Length of major axis is the diameter of circle

$ \begin{array}{rlrl} x^2+y^2 =18 \\ \therefore 2 a=2(3 \sqrt{2})^2 \Rightarrow a & =3 \sqrt{2} \\ e^2 & =1-\frac{b^2}{a^2} \\ \Rightarrow & \frac{8}{9} =1-\frac{b^2}{18} \Rightarrow b^2=2 \end{array} $

∴ Equation of ellipse is $\frac{x^2}{18}+\frac{y^2}{2}=1$

Equation of circle is $x^2+y^2=18$

Let $(h, k)$ be the pole, then equation of polar with respect to ellipse is

$ \frac{x h}{18}+\frac{y k}{2}=1 $

Since this is the tangent of circle

$ \begin{aligned} \therefore 3 \sqrt{2}=\left|\frac{1}{\sqrt{\frac{h^2}{(18)^2}+\frac{k^2}{4}}}\right| \\ \frac{h^2}{18^2}+\frac{k^2}{4}= & \frac{1}{18} \end{aligned} $

$ \begin{array}{r} \frac{h^2}{18}+\frac{9 k^2}{2}=1 \\ \therefore \text { Locus is } \frac{x^2}{18}+\frac{9 y^2}{2}=1 \end{array} $

Let the equation of ellipse be,

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

It passes through $(3 \sqrt{2}, \sqrt{10})$

$ \therefore \quad \frac{18}{a^2}+\frac{10}{b^2}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, foci are ( $\pm 4,0$ )

$ \begin{array}{ll} \therefore & a e=4 \Rightarrow a^2 e^2=16 \\ \Rightarrow & a^2\left(1-\frac{b^2}{a^2}\right)=16 \Rightarrow a^2-b^2=16 \\ \Rightarrow & b^2=a^2-16 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

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

$ \begin{aligned} & \frac{18}{a^2}+\frac{10}{a^2-16}=1 \\ \Rightarrow & 18 a^2-288+10 a^2=a^4-16 a^2 \\ \Rightarrow & a^4-44 a^2+288=0 \\ \Rightarrow & \left(a^2-36\right)\left(a^2-8\right)=0 \end{aligned} $

$ \begin{aligned} &\begin{array}{lll} \Rightarrow & a^2=36,8 & \\ \Rightarrow & a^2=36 & {\left[\because a^2 \neq 8\right. \text { as }} \\ & & \left.b^2=a^2-16\right] \end{array}\\ &\text { Now, } \quad a^2 e^2=16\\ &e^2=\frac{16}{a^2}=\frac{16}{36} \Rightarrow e=\frac{4}{6}=\frac{2}{3} \end{aligned} $

We know that, the product of the length of the perpendicular drawn from the foci to the tangent of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $b^2$.

Also, length $=9 \quad \therefore b^2=9$

Equation of tangent

$ \begin{aligned} y & =\frac{-3}{4} x+3 \sqrt{2} \\ \therefore \quad 3 \sqrt{2} & = \pm \sqrt{a^2\left(\frac{-3}{4}\right)^2+b^2} \\ 18 & =\frac{9}{16} a^2+9 \Rightarrow 2-1=\frac{a^2}{16} \\ \Rightarrow \quad a^2 & =16 \\ \therefore \quad e & =\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4} \end{aligned} $

Let ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1(a < b)$ is the equation of ellipse, focii $(0, \pm 2)$

Given 2a = 4 $ \Rightarrow $ a = 2

e² = 1 - ${{{a^2}} \over {{b^2}}}$ $ \Rightarrow $ b²e² = b² - a²

4 = b² - 4

b² = 8

$ \because $ equation of ellipse is ${{{x^2}} \over 4} + {{{y^2}} \over 8} = 1$

then it passes through $\left( {\sqrt 2 ,2} \right)$

Equation of ellipse is ${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$

Normal at P(2 cos $\theta $, $\sqrt 3 \sin \theta $) is 2x sin$\theta $ - $\sqrt 3 y\,cos\theta $ = sin $\theta $ cos $\theta $ as the normal is parallel to 2x + y = 4

$ \Rightarrow $ ${2 \over {\sqrt 3 }}\tan \theta = - 2$

$ \Rightarrow \tan \theta = - \sqrt 3 \,\,......\,(i)$

tangent at P(2 cos $\theta $, $\sqrt 3 \sin \theta $) is

$\sqrt 3 $ x cos $\theta $ + 2y sin $\theta $ = 2 $\sqrt 3 $

Passes through (4, 4)

$ \Rightarrow $ 4$\sqrt 3 $ cos $\theta $ + 8 sin $\theta $ = 2$\sqrt 3 $ ....... (ii)

From (i) and (ii)

$ \Rightarrow P\left( { - 1,{3 \over 2}} \right)\,\& \,Q(4,4)$

$ \Rightarrow PQ = \sqrt {25 + {{25} \over 4}} = {{5\sqrt 5 } \over 2}$

$3{x^2} + 5{y^2} = 32$

6x + 10yy' = 0

$ \Rightarrow y' = {{ - 3x} \over {5y}}$

$ \Rightarrow y{'_{(2,2)}} = - {3 \over 5}$

Tangent $(y - 2) = - {3 \over 5}(x - 2) \Rightarrow Q\left( {{{16} \over 3},0} \right)$

Normal $(y - 2) = {5 \over 3}(x - 2) \Rightarrow R\left( {{{4} \over 5},0} \right)$

$ \therefore $ Area = ${1 \over 2}(QR) \times 2 = QR = {{68} \over {15}}$

Equation of tangent at $\left( {3, - {9 \over 2}} \right)$ to ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ is

${{3x} \over {{a^2}}} - {{{y^9}} \over {2{b^2}}} = 1$ which is equivalent to x – 2y = 12

${3 \over {{a^2}}} = {{ - 9} \over {2{b^2}( - 2)}} = {1 \over {12}}$   (On comparing)

${a^2} = 3 \times 12$ and ${b^2} = {{9 \times 12} \over 4}$

$ \Rightarrow $ a = 6 and b = $3\sqrt 3 $

So latus rectum = ${{2{b^2}} \over a} = {{2 \times 27} \over 6} = 9$

Point P($\alpha $, $\beta $) is on the parabola y² = x

$ \therefore $ ${\beta ^2} = \alpha $ ...........(1)

Equation of tangent to the parabola y² = x

at ($\alpha $, $\beta $) is T = 0

$\beta y = {{x + \alpha } \over 2}$

$ \Rightarrow $ $2\beta y = x + \alpha $

$ \Rightarrow $ $y = {1 \over {2\beta }}x + {\alpha \over {2\beta }}$

For this straight line, m = ${1 \over {2\beta }}$ and c = ${\alpha \over {2\beta }}$

This tangent is also a tangent to ellipse x² + 2y² = 1.

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

$ \therefore $ ${a^2} = 1$ and ${b^2} = {1 \over 2}$.

We know the condition of tangency on ellipse is

${c^2} = {a^2}{m^2} + {b^2}$

$ \Rightarrow $ ${\left( {{\alpha \over {2\beta }}} \right)^2} = 1{\left( {{1 \over {2\beta }}} \right)^2} + {1 \over 2}$

$ \Rightarrow $ ${{{\alpha ^2}} \over {4{\beta ^2}}} = {1 \over {4{\beta ^2}}} + {1 \over 2}$

$ \Rightarrow $ ${\alpha ^2} = 1 + 2{\beta ^2}$

$ \Rightarrow $ ${\alpha ^2} = 1 + 2\alpha $

$ \Rightarrow $ ${\alpha ^2} - 2\alpha - 1 = 0$

$ \Rightarrow $ $\alpha = 1 + \sqrt 2 $ and $\alpha = 1 - \sqrt 2 $

Focus (0, be) = (0, 5$\sqrt 3$)

$ \therefore $ be = 5$\sqrt 3$

$ \Rightarrow $ b²e² = 75

As here b > a

so e² = $1 - {{{a^2}} \over {{b^2}}}$

$ \therefore $ b²$\left( {1 - {{{a^2}} \over {{b^2}}}} \right)$ = 75

$ \Rightarrow $ b² - a² = 75 .......(1)

Given that,

difference of the lengths of major axis and minor axis is 10.

$ \therefore $ 2b - 2a = 10

$ \Rightarrow $ b - a = 5 .......(2)

From (1),

(b + a)(b - a) = 75

$ \Rightarrow $ (b + a)5 = 75

$ \Rightarrow $ (b + a) = 15 .......(3)

From (2) and (3) we get,

b = 10, a = 5

$ \therefore $ Length of its latus rectum = ${{{2{a^2}} \over b}}$

= ${{2 \times 25} \over {10}}$ = 5

Given, Equation of ellipse 4x² + y² = 8

We know equation of tangent at any point (x₁, y₁) is

4xx₁ + yy₁ = 8

$ \therefore $ Equation of tangent at point (1, 2) is

4x + 2y = 8

$ \Rightarrow $ 2x + y = 4

$ \therefore $ Slope of this tangent = -2

Equation of tangent at point (a, b) is

4ax + by = 8

Slope of this tangent = $ - {{4a} \over b}$

As tangent at (1, 2) and (a, b) are perpendicular

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

$ \Rightarrow $ b = -8$a$

As point (a, b) lies on the ellipse

$ \therefore $ $4{a^2} + {b^2} = 8$

$ \Rightarrow $ $4{a^2} + 64{a^2} = 8$

$ \Rightarrow $ ${a^2} = {8 \over {68}}$ = ${2 \over {17}}$

b² = a²e² . . . . . . (i)

${1 \over 2}$ S'B.SB = 8

S'B.SB = 16

a²e² + b² = 16 . . . . .(ii)

b² = a² (1 $-$ e²) . . . . .(iii)

using (i), (ii), (iii) a = 4

      b = $2\sqrt 2 $

      e = ${1 \over {\sqrt 2 }}$

$ \therefore $  $\ell $ (L.R) $=$ ${{2{b^2}} \over a} = 4$

${{2{b^2}} \over a} = 8$ and 2ae $=$ 2b

$ \Rightarrow $  ${b \over a}$ = e and 1 $-$ e² = e² $ \Rightarrow $ e $=$ ${1 \over {\sqrt 2 }}$

$ \Rightarrow $  b = 4$\sqrt 2 $ and a $=$ 8

So equation of ellipse is ${{{x^2}} \over {64}} + {{{y^2}} \over {32}} = 1$

Equation of general tangent on ellipse

${x \over {a\,\sec \theta }} + {y \over {b\cos ec\theta }} = 1$

$a = \sqrt 2 ,\,\,b = 1$

$ \Rightarrow {x \over {\sqrt 2 \sec \theta }} + {y \over {\cos ec\theta }} = 1$

Let the midpoint be (h, k)

$h = {{\sqrt 2 \sec \theta } \over 2} \Rightarrow \cos \theta = {1 \over {\sqrt 2 h}}$

and $k = {{\cos ec\theta } \over 2} \Rightarrow \sin \theta = {1 \over {2k}}$

$ \because $  ${\sin ^2}\theta + {\cos ^2}\theta = 1$

$ \Rightarrow $  ${1 \over {2{h^2}}} + {1 \over {4{k^2}}} = 1$

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

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

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}}} $

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.

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 = ${1 \over 3}$

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 }}$

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

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

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