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Hyperbola

Section: Mathematics

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Question : 69 of 98

Marks: +1, -0

Let

e_{1}

e_{2}

be the eccentricities of the ellipse,

\frac{x^{2}}{25} + \frac{y^{2}}{b^{2}} = 1(b<5)

and the hyperbola

\frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = 1

respectively satisfying

e_{1} e_{2} = 1

\alpha

\beta

are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair

(\alpha, \beta)

[3 Sep 2020 Shift 2]

(8,10)
(8,12)
$\left( \frac{20}{3}, 12 \right)$
$\left( \frac{24}{5}, 10 \right)$

Solution:

For ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{b^{2}} = 1(b<5)

Let

\ e_{1}

is eccentricity of ellipse

∴

b^{2} = 25 (1 - e_{1}^{2})

....... (1)

Again for hyperbola

\frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = 1

Let

e_{2}

is eccentricity of hyperbola.

∴

b^{2} = 16 (e_{2}^{2} - 1)

.....(2)

by

(1) \& (2)

25 (1 - e_{1}^{2}) = 16 (e_{2}^{2} - 1)

Now

e_{1} \cdot e_{2} = 1 \;\;\;

(given)

\therefore 25 (1 - e_{1}^{2}) = 16 \left( \frac{1 - e_{1}^{2}}{e_{1}^{2}} \right)

or

\;\; e_{1} = \frac{4}{5} \;\;\; \therefore e_{2} = \frac{5}{4}

Now distance between foci is 2 ae

\therefore

distance for ellipse

= 2 \times 5 \times \frac{4}{5} = 8 = \alpha

distance for hyperbola

= 2 \times 4 \times \frac{5}{4} = 10 = \beta

\therefore (\alpha, \beta) \equiv (8,10)

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