If e _ 1 and e _ 2 are the eccentricities of the ellipse x ^ 2 18 + y ^ 2 4 = 1 and the hyperbola x ^ 2 9 - y ^ 2 4 = 1 respectively and ( e _ 1 , e _ 2 ) is a point on the ellipse 15 x ^ 2 + 3 y ^ 2 = k , then the

Correct answer: A. 16 e _ 1 = 1 - (4)/(18) = (7)/(9) = 7 3 e _ 1 = 1 + (4)/(9) = (13)/(9) = 13 3 Also, 15 e _ 1 ^ 2 + 3 e _ 2 ^ 2 = k ⇒ k = 15((7)/(9)) + 3((13)/(9)) ∴ k = 16

JEE Main 20209 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2020 · 9 January · Morning Shift · Mathematics · Hyperbola

If e _ 1 and e _ 2 are the eccentricities of the ellipse x ^ 2 18 +...

{e}_{1} and {e}_{2}

are the eccentricities of the ellipse

\dfrac{{x}^{2}}{18} + \dfrac{{y}^{2}}{4} = 1

and the hyperbola

\dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{4} = 1

respectively and

\left({e}_{1}, {e}_{2}\right)

is a point on the ellipse

15{x}^{2} + 3{y}^{2} = k

, then the value of

k

is equal to

Options

A $16$
B $17$
C $15$
D $14$

Correct answer

A. $16$

Step-by-step solution

${e}_{1} = \sqrt{1 - \dfrac{4}{18}} = \sqrt{\dfrac{7}{9}} = \dfrac{\sqrt{7}}{3}$
${e}_{1} = \sqrt{1 + \dfrac{4}{9}} = \sqrt{\dfrac{13}{9}} = \dfrac{\sqrt{13}}{3}$

Also,
$15{e}_{1}^{2} + 3{e}_{2}^{2} = k ⇒ k = 15\left(\dfrac{7}{9}\right) + 3\left(\dfrac{13}{9}\right)$
$∴ k = 16$

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