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~ 15x^ 2 + 3y^ 2 = k , then the value of~

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

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Hyperbola

NTA Abhyas 2020 · 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~

15x^{2} + 3y^{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,
$15{e}_{1}^{2} + 3{e}_{2}^{2} = k ⇒ k = 15 \left(\dfrac{7}{9}\right) +3 \left(\dfrac{13}{9}\right)<br />$ ∴ k = 16 $</p>$

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