First slide

Hyperbola in conic sections

Question

$The eccentricity of the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 is reciprocal to that of the ellipse$
$x^{2} + 4 y^{2} = 4 . If the hyperbola passes through a foci of the ellipse, then$

Moderate

A

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

$a focus of the hyperbola is (\sqrt{3}, 0)$
C

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

$the equation of the hyperbola is x^{2} - 3 y^{2} = 3$

Solution

$Given ellipse is \frac{x^{2}}{4} + \frac{y^{2}}{1} = 1 .$
$Eccentricity of the ellipse = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$
$Focus of the ellipse = (- \sqrt{3}, 0)$
$\begin{array}{l} Eccentricity of the hyperbola = \sqrt{1 + \frac{b^{2}}{a^{2}}} \\ So, \sqrt{1 + \frac{b^{2}}{a^{2}}} = \frac{2}{\sqrt{3}} \Rightarrow \frac{b}{a} = \frac{1}{\sqrt{3}} \end{array}$
$Since the hyperbola passes through the focus of the ellipse, \frac{3}{a^{2}} - 0 = 1$
$\begin{array}{l} \Rightarrow a^{2} = 3 and b^{2} = 1 \\ Thus, the equation of the hyperbola is \end{array}$
$\begin{array}{l} \frac{x^{2}}{3} - \frac{y^{2}}{1} = 1 or x^{2} - 3 y^{2} = 3 \\ Focus of hyperbola is (\pm 2, 0) . \end{array}$

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