JEE Main 2022 · 30 Jul · Shift 2 · Mathematics Q75

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MathematicsSingle correct+4 / -1Question 75 of 90

If foci of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{12}=1$ and the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}9=1$ coincide, then the eccentricity of the hyperbola, with foci $(0,\pm\sqrt{10})$ passing through the point (a, a + 1), is

\sqrt2

\dfrac3{\sqrt5}

\sqrt{\dfrac52}

Answer and worked solution

Official answer: B

First part. Ellipse foci $(\pm4,0)$ : $a^2+12=16$ , $a=2$ , so the point is (2, 3).

Hyperbola $\frac{y^2}{B^2}-\frac{x^2}{A^2}=1$ with $A^2+B^2=10$ through (2, 3): $B^4-23B^2+90=0$ , $B^2=5$ . $e=\frac{\sqrt{10}}{\sqrt5}=\sqrt2$ .

(B)

Topics:Hyperbola Ellipse

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JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 75