JEE Main 2016 — Hyperbola Question with Solution
From: JEE Main 2016 (Online) 9th April Morning Slot
Question
Let a and b respectively be the semitransverse and semi-conjugate axes of a
hyperbola whose eccentricity satisfies the equation 9e2 − 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then a2 − b2 is equal to :
Choose an option
Show full solutionCorrect option: B
Correct answer
B 7
Step-by-step explanation
As S(5, 0) is the focus.
ae = 5 . . . (1)
As 5x = 9
x = is the directrix
As we know directrix =
. . . .(2)
Solving (1) and (2), we get
a = 3 and e =
As we know,
b2 = a2 (e2 1) = 9 = 16
a2 b2 = 9 16 7
ae = 5 . . . (1)
As 5x = 9
x = is the directrix
As we know directrix =
. . . .(2)
Solving (1) and (2), we get
a = 3 and e =
As we know,
b2 = a2 (e2 1) = 9 = 16
a2 b2 = 9 16 7
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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.