JEE Main 2019MathematicsParabolaMediumMCQ

JEE Main 2019 — Parabola Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

The length of the chord of the parabola x2=4y having equation x-2y+42=0 is

Choose an option

Show full solutionCorrect option: A
Correct answer
A63 units

Step-by-step explanation

Solving the equations of the given parabola and the line, we get,

2y-422=4y⇒y2-10y+16=0

⇒y1-y2=y1+y22-4y1y2=100-64=6



⇒ Length of the chord =AB=x1-x22+y1-y22=2y1-y22+y1-y22=3y1-y2

=63 units

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Parabola 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 Parabola 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.