Let the eccentricity of the hyperbola x²/a² - y²/b² = 1 be reciprocal to that of the ellipse x² + 4y² = 4. If the hyperbola passes through a focus of the ellipse, then [JEE Adv 2011 P1]

the equation of the hyperbola is {x²}{3} - {y²}{2} = 1.

a focus of the hyperbola is (2, 0).

the eccentricity of the hyperbola is {{5}}{3}.

the equation of the hyperbola is x² - 3y² = 3.

Test Index

Hyperbola

Section: Mathematics

Question : 11 of 15

Marks: +1, -0

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

x^2 + 4y^2 = 4

Solution:

Ellipse is

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

= 1.

1^2

2^2(1-e^2)

⇒ e =

\frac{\sqrt{3}}{2}

Therefore, the eccentricity of the hyperbola is

\frac{2}{\sqrt{}}

b^2

a^2\left(\frac{4}{3} - 1\right)

⇒

3b^2

a^2

Foci of the ellipse are

(\sqrt{3},0)

and

(-\sqrt{3},0)

Hyperbola passes through

(-\sqrt{3},0)

\frac{3}{a^2}

= 1 ⇒

a^2

= 3 and

b^2

= 1.

Therefore, the equation of hyperbola is

x^2 - 3y^2

= 3.

Focus of hyperbola is (ae, 0) =

\left(\sqrt{3} \times \frac{2}{\sqrt{3}},0\right)

= (2 , 0).

Go to Question:

Prev Question

Next Question