Let the eccentricity of the hyperbola \frac{x^{2}}{a^{2}} \textrm{ }\textrm{ } - \textrm{ }\textrm{ } \frac{y^{2}}{b^{2}} = 1 be reciprocal to that of the ellipse x2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then

Engineering

Mathematics

Standard Hyperbola

Question

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be reciprocal to that of the ellipse x² + 4y² = 4. If the hyperbola passes through a focus of the ellipse, then

the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$ .

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

the equation of the hyperbola is $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$ .

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

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Let e₁ = eccentricity of hyperbola

e₂ = eccentricity of ellipse

$∴ e_{1} = \frac{1}{e_{2}}$

So, eccentricity of ellipse = $\frac{\sqrt{3}}{2} = e_{2}$

eccentricity of ellipse = $\frac{\sqrt{3}}{2} = e_{1}$

Now focus of ellipse is (± ae2, 0) $\equiv (\pm \sqrt{3}, 0)$

Hyperbola passes through it

So, $\frac{{(\sqrt{3})}^{2}}{a^{2}} - 0 = 1 \Rightarrow a^{2} = 3$

also b² = a² (e₁² – 1)

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

and hyperbola

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

also focus (± ae1, 0) $\equiv$ (± 2, 0)

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