JEE Main 2022 · 30 Jun · Shift 1 · Mathematics Q15

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MathematicsSingle correct+4 / -1Question 15 of 90

Let the eccentricity of the ellipse $x^2+a^2y^2=25a^2$ be b times the eccentricity of the hyperbola $x^2-a^2y^2=5$ , where a is the minimum distance between the curves $y=e^x$ and $y=\log_ex$ . Then $a^2+\dfrac1{b^2}$ is equal to :

\dfrac32

\dfrac52

Answer and worked solution

Official answer: D

a. The curves are mirror images in $y=x$ ; nearest points $(0,1)$ and $(1,0)$ : $a=\sqrt2$ .

Eccentricities. Ellipse $\frac{x^2}{50}+\frac{y^2}{25}=1$ : $e^2=\frac12$ . Hyperbola $\frac{x^2}5-\frac{y^2}{5/2}=1$ : $e^2=\frac32$ . So $b^2=\frac13$ .

Value $2+3=5$ .

(D)

Topics:Ellipse Hyperbola

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JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 15