First slide

Hyperbola in conic sections

Question

Let a hyperbola passes through the focus of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ . The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse. Also, the product of the eccentricities of the given ellipse and hyperbola is 1. Then,

Moderate

A

the equation of the hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
B

the equation of the hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1$
C

the focus of the hyperbola is (5, 0)
D

the vertex of the hyperbola is $(5 \sqrt{3}, 0)$

Solution

For the given ellipse
$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$
we have $e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$
Hence, the eccentricity of the hyperbola is 5/3.
Let the hyperbola be
$\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$
Then $B^{2} = A^{2} (e^{2} - 1) = A^{2} (\frac{25}{9} - 1) = \frac{16}{9} A^{2}$
Therefore, the equation of the hyperbola is
$\frac{x^{2}}{A^{2}} - \frac{9 y^{2}}{16 A^{2}} = 1$
As it passes through (3, 0), we get A²=9 and B²=16' The equation is
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
The foci of the hyperbola are ( $\pm$ ae,0) = ( $\pm$ 5,0).
The vertex of the hyperbola is ( $\pm$ 3, 0)

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