JEE Main 2019 — Hyperbola Question with Solution
From: JEE Main 2019 (Online) 10th January Morning Slot
Question
The equation of a tangent to the hyperbola 4x2 – 5y2 = 20 parallel to the line x – y = 2 is :
Choose an option
Show full solutionCorrect option: C
Correct answer
Cx y + 1 = 0
Step-by-step explanation
Hyperbola
slope of tangent = 1
equation of tangent y = x
y = x 1
y = x + 1
or
y = x 1
slope of tangent = 1
equation of tangent y = x
y = x 1
y = x + 1
or
y = x 1
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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.