JEE Main 2016MathematicsHyperbolaEasyMCQ

JEE Main 2016 — Hyperbola Question with Solution

JEE Main 2016 (09 Apr Online)

Question

Let a and b respectively be the semi-transverse and semi-conjugate axes of a standard hyperbola whose eccentricity satisfies the equation 9e2-18e+5=0. If S5, 0 is a focus and 5x=9 is the corresponding directrix of this hyperbola, then a2-b2 is equal to

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Show full solutionCorrect option: A
Correct answer
A-7

Step-by-step explanation

Given that focus of hyperbola is S5, 0

⇒ae=5......i

Therefore, Directrix x=ae

⇒ ae=95  ......ii

From i and ii, we get ⇒  a2=9

⇒a=3 and e=53    

(It satisfies 9e2-18e+5=0 )

We also know that for an ellipse,

b2=a2e2-1

⇒b2=16

⇒ a2-b2=9-16= -7

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About this question

This is a previous-year question from JEE Main 2016, covering the Hyperbola chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.