JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Ellipse

Let

E_{1}

and

E_{2}

be two ellipses 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} + (y - 1)^{2} = 2

. The straight line

x + y = 3

touches the curves

S, E_{1}

and

E_{2}

P, Q

and

R

, respectively. Suppose that

P Q = P R = \frac{2 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)

Answer: (A), (B)

Solution

The line

x + y = 3

is tangent to circle

S : x^{2} + (y - 1)^{2} = 2

at point

P

. The distance from center

(0, 1)

to the line is

\frac{∣0 + 1 - 3∣}{2} = 2

, confirming tangency. Using the tangency conditions and the given distances

P Q = P R = \frac{2 2}{3}

, we can set up equations for the ellipses

E_{1} : \frac{x ^{2}}{a _{1}^{2}} + \frac{y ^{2}}{b _{1}^{2}} = 1

and

E_{2} : \frac{x ^{2}}{b _{2}^{2}} + \frac{y ^{2}}{a _{2}^{2}} = 1

. Solving these yields

e_{1}^{2} = 1 - \frac{b _{1}^{2}}{a _{1}^{2}}

and

e_{2}^{2} = 1 - \frac{b _{2}^{2}}{a _{2}^{2}}

. Detailed calculation confirms

e_{1}^{2} + e_{2}^{2} = \frac{43}{40}

(A) and

e_{1} e_{2} = \frac{7}{2 10}

(B).

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options