JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Ellipse

Let the eccentricity of an ellipse

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

is reciprocal to that of the hyperbola

2 x^{2} - 2 y^{2} = 1

. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is ____.

Answer: 2

Solution

For the hyperbola

2 x^{2} - 2 y^{2} = 1

\frac{x ^{2}}{1/2} - \frac{y ^{2}}{1/2} = 1

e_{H} = 1 + \frac{b ^{2}}{a ^{2}} = 2

Given

e_{E} = \frac{1}{e _{H}} = \frac{1}{2}

Since the curves intersect orthogonally, they are confocal. The hyperbola has foci at

(\pm 1, 0)

.

For the ellipse:

a e_{E} = 1 \Rightarrow a \cdot \frac{1}{2} = 1 \Rightarrow a = 2

Using

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

\frac{1}{2} = 1 - \frac{b ^{2}}{2} \Rightarrow b^{2} = 1

Length of latus rectum =

\frac{2 b ^{2}}{a} = \frac{2 \times 1}{2} = 2

Square of length =

(2)^{2} = 2

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer