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Question: If e<sup>­</sup><sub>1</sub> and e<sub>2</sub> are the eccentricities of the conic sections 16x<sup>...

If e­1 and e2 are the eccentricities of the conic sections 16x2 + 9y2 = 144 and 9x2 – 16y2 = 144, then-

A

e12+e22=3e_{1}^{2} + e_{2}^{2} = 3

B

e12+e22>3e_{1}^{2} + e_{2}^{2} > 3

C

e12+e22<3e_{1}^{2} + e_{2}^{2} < 3

D

e12e22>1e_{1}^{2} - e_{2}^{2} > 1

Answer

e12+e22<3e_{1}^{2} + e_{2}^{2} < 3

Explanation

Solution

16x2 + 9y2 = 144 9x2 – 16y2 = 144

x29\frac{x^{2}}{9} + y216\frac{y^{2}}{16} = 1 x216\frac{x^{2}}{16}y29\frac{y^{2}}{9} = 1

e1 = 1916\sqrt{1 - \frac{9}{16}} e2 = 1+916\sqrt{1 + \frac{9}{16}}

e1 = 74\frac{\sqrt{7}}{4} e2 = 54\frac{5}{4}

e12 + e22 = 716\frac{7}{16} + 2516\frac{25}{16} = 3216\frac{32}{16} = 2