JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q81

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MathematicsSingle correct+4 / -1Question 81 of 90

Let a and b respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation $9e^2-18e+5=0$ . If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then $a^2-b^2$ is equal to :

−7

−5

Answer and worked solution

Official answer: B

$e=\frac53$ (the root above 1). Then $ae=5$ gives a = 3 (and $\frac ae=\frac95$ matches the directrix). $b^2=a^2(e^2-1)=16$ , so $a^2-b^2=-7$ .

Topics:Hyperbola

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 81