JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Hyperbola

Let the eccentricity of the hyperbola

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

be reciprocal to that of the ellipse

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

. If the hyperbola passes through a focus of the ellipse, then

Answer: (B), (D)

Solution

The ellipse

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

can be written as

\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1

, with

a^{2} = 4, b^{2} = 1

.

For the ellipse:

b^{2} = a^{2} (1 - e^{2})

, so

1 = 4 (1 - e^{2})

, giving

e = \frac{3}{2}

.

The foci are at

(\pm 3, 0)

.

For the hyperbola, the eccentricity is

\frac{2}{3}

, so

b^{2} = a^{2} (\frac{4}{3} - 1) = \frac{a ^{2}}{3}

, giving

3 b^{2} = a^{2}

.

Since the hyperbola passes through

(3, 0)

\frac{3}{a ^{2}} = 1

, so

a^{2} = 3

and

b^{2} = 1

.

The equation is

\frac{x ^{2}}{3} - y^{2} = 1

, or

x^{2} - 3 y^{2} = 3

.

The foci are at

(\pm a e, 0) = (\pm 3 \cdot \frac{2}{3}, 0) = (\pm 2, 0)

.

Options (B) and (D) are correct.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · One or more correct options