JEE Main 2019MathematicsHyperbolaMediumMCQ

JEE Main 2019 — Hyperbola Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

The equation of a common tangent to the curves, y2=16x and xy=-4, is:

Choose an option

Show full solutionCorrect option: B
Correct answer
Bx-y+4=0

Step-by-step explanation

A tangent to the parabola y2=16x can be written as:
y=mx+4m …. (1)
It will touch the hyperbola xy= –4,  if the quadratic xmx+4m=–4 has coincident roots,
⇒m2x2+4x+4m=0, D≡16-16m3=0
⇒m=1
⇒ Equating of common tangent will be y=x+4
⇒x-y+4=0

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About this question

This is a previous-year question from JEE Main 2019, 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.