JEE Main 2022 · 25 Jul · Shift 2 · Mathematics Q14

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MathematicsSingle correct+4 / -1Question 14 of 90

Let the foci of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}7=1$ and the hyperbola $\dfrac{x^2}{144}-\dfrac{y^2}\alpha=\dfrac1{25}$ coincide. Then the length of the latus rectum of the hyperbola is :

\dfrac{32}9

\dfrac{18}5

\dfrac{27}4

\dfrac{27}{10}

Answer and worked solution

Official answer: D

Foci $(\pm3,0)$ . Hyperbola $\frac{x^2}{144/25}-\frac{y^2}{\alpha/25}=1$ : $b^2=9-\frac{144}{25}=\frac{81}{25}$ .

Latus rectum $\dfrac{2b^2}a=\dfrac{162/25}{12/5}=\dfrac{27}{10}$ .

(D)

Topics:Hyperbola Ellipse

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JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 14