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

Question

If e1 and e2 are the eccentricities of the conic sections 16 x2+9 y2=144 and 9 x2-16 y2=144, then (A) e 12- e 22=1 (B) e 12+ e 22=3 (C) e 12+ e 22>3 (D) e 12+ e 22<3

Tardigrade Admin · Accepted Answer

e 12=1-(144/16) × (9/144)=(7/16) e 22=1+(44/16) × (9/144)=(25/16) ⇒ e 12+ e 22=2

Q. If $e_{1}$ and $e_{2}$ are the eccentricities of the conic sections $16 x^{2} + 9 y^{2} = 144$ and $9 x^{2} - 16 y^{2} = 144$ , then

$e_{1}^{2} - e_{2}^{2} = 1$

$e_{1}^{2} + e_{2}^{2} = 3$

$e_{1}^{2} + e_{2}^{2} > 3$

$e_{1}^{2} + e_{2}^{2} < 3$

$e_{1}^{2} = 1 - \frac{144}{16} \times \frac{9}{144} = \frac{7}{16}$
$e_{2}^{2} = 1 + \frac{44}{16} \times \frac{9}{144} = \frac{25}{16}$
$\Rightarrow e_{1}^{2} + e_{2}^{2} = 2$

Q. If e1​ and e2​ are the eccentricities of the conic sections 16x2+9y2=144 and 9x2−16y2=144, then

e12​−e22​=1

e12​+e22​=3

e12​+e22​>3

e12​+e22​<3