JEE Main 2019 — Parabola Question with Solution
From: JEE Main 2019 (Online) 10th January Evening Slot
Question
The length of the chord of the parabola x2 4y having equation x – is -
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
x2 = 4y
x y + 4 = 0
Solving together we get
x2 = 4
x2 + 4x + 16
x2 4x 16 = 0
x1 + x2 = 2; x1x2 = = 16
Similarly,
(y 4)2 = 4y
2y2 + 32 16y = 4y
AB =
x y + 4 = 0
Solving together we get
x2 = 4
x2 + 4x + 16
x2 4x 16 = 0
x1 + x2 = 2; x1x2 = = 16
Similarly,
(y 4)2 = 4y
2y2 + 32 16y = 4y
AB =
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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.