JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Ellipse

Let

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

a > b

. Let

E_{2}

be another ellipse such that it touches the end points of major axis of

E_{1}

and the foci of

E_{2}

are the end points of minor axis of

E_{1}

. If

E_{1}

and

E_{2}

have same eccentricities, then its value is:

Answer: (A)

Solution

The major axis of

E_{1}

ends at

(\pm a, 0)

and its minor axis at

(0, \pm b)

. The foci of

E_{2}

are

(0, \pm b)

, so the major axis of

E_{2}

lies along the

y

-axis and

E_{2} : \frac{x ^{2}}{B ^{2}} + \frac{y ^{2}}{A ^{2}} = 1

with

A > B

. Passing through

(\pm a, 0)

gives

B = a

, and the focal distance gives

A^{2} - B^{2} = b^{2}

, so

A^{2} = a^{2} + b^{2}

. Writing

t = \frac{b ^{2}}{a ^{2}}

e_{1}^{2} = 1 - t, e_{2}^{2} = 1 - \frac{B ^{2}}{A ^{2}} = 1 - \frac{a ^{2}}{a ^{2} + b ^{2}} = \frac{t}{1 + t}

Equal eccentricities give

(1 - t) (1 + t) = t

, so

t^{2} + t - 1 = 0

and

t = \frac{- 1 + 5}{2}

. Since

t^{2} = 1 - t

, the common value satisfies

e^{2} = 1 - t = t^{2}

, so

e = \frac{- 1 + 5}{2}

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option