JEE Main 2019MathematicsHyperbolaMediumMCQ

JEE Main 2019 — Hyperbola Question with Solution

JEE Main 2019 (08 Apr Shift 2)

Question

If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is:

Choose an option

Show full solutionCorrect option: D
Correct answer
D2x-y-2=0

Step-by-step explanation

Let us suppose equation of hyperbola is
x2a2-y2b2=1
e=2 

∵b2=e2-1a2

⇒b2=3a2
It is passing through 4,6

16a2-36b2=1

⇒a2=4, b2=12
So, equation of tangent is T=0

xx1a2-yy1b2=1

Tangent at point (4,6) is
⇒x-y2=1
⇒2x-y-2=0

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Hyperbola chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

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.