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JEE Advanced 2015Paper 2

Exam: JEE Advanced
Year: 2015
Paper: Paper 2
Subject: Mathematics
Chapter: Ellipse

JEE Advanced 2015 · Paper 2 · Mathematics · Ellipse

Let E _ 1 and E _ 2 be two ellipse whose centers are at the origin....

Let

{E}_{1}

and

{E}_{2}

be two ellipse whose centers are at the origin. The major axes of

{E}_{1}and{E}_{2}

lie along the x - axis and the y - axis, respectively. Let S be the circle

{x}^{2} + {\left(y - 1\right)}^{2} = 2 .

The straight line

x + y = 3

touches the curves

S, {E}_{1}

and

{E}_{2}

at

P, Q

and

R,

respectively. Suppose that

PQ = PR = \dfrac{2\sqrt{2}}{3} .

If

{e}_{1}

and

{e}_{2}

are the eccentricities of

{E}_{1}

and

{E}_{2},

respectively, then the correct expression(s) is(are)

Options

A ${e}_{1}^{2} + {e}_{2}^{2} = \dfrac{43}{40}$
B ${e}_{1}{e}_{2} = \dfrac{\sqrt{7}}{2\sqrt{10}}$
C $\left|{e}_{1}^{2} - {e}_{2}^{2}\right| = \dfrac{5}{8}$
D ${e}_{1}{e}_{2} = \dfrac{\sqrt{3}}{4}$

Correct answer

A. ${e}_{1}^{2} + {e}_{2}^{2} = \dfrac{43}{40}$

Step-by-step solution

Let Q be

( {x}_{1}, {y}_{1} )

So equation tangent at Q will be

x{x}_{1} + y{y}_{1} - \left(y + {y}_{1}\right) - 1 = 0

Comparing with

x + y = 3

\left({x}_{1}{y}_{1}\right) ≡ ( 1,2 )

<br />So R and Q will be<br />\left(1 \pm \dfrac{2\sqrt{2}}{3}cos \dfrac{3\pi}{4}, 2 \pm \dfrac{2\sqrt{2}}{3}sin \dfrac{3\pi}{4}\right)

⇒ Q\left(\dfrac{5}{3}, \dfrac{4}{3}\right)and\left(\dfrac{1}{3}, \dfrac{8}{3}\right)

Let Q lies on

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

So tangent at P is

\dfrac{5x}{3{a}^{2}} + \dfrac{4y}{3{b}^{2}} = 1

Comparing with

x + y = 3

{a}^{2} = 5, {b}^{2} = 4 ⇒ {e}_{1} = \dfrac{1}{\sqrt{5}}

And R lies on

\dfrac{{x}^{2}}{{a}_{1}^{2}} + \dfrac{{y}^{2}}{{b}_{1}^{2}} = 1

So tangent at

\dfrac{x}{{3a}_{1}^{2}} + \dfrac{8y}{{3b}_{1}^{2}} = 3

Comparing with

x + y = 3

⇒ {a}_{1}^{2} = 1, {b}_{1}^{2} = 8 ⇒ {e}_{2} = \sqrt{\dfrac{7}{8}}

⇒ {e}_{1}^{2} + {e}_{2}^{2} = \dfrac{43}{40}and{e}_{1}{e}_{2} = \dfrac{\sqrt{7}}{2\sqrt{10}}

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