JEE Main 2019 — Parabola Question with Solution
From: JEE Main 2019 (Online) 12th April Evening Slot
Question
The tangents to the curve y = (x – 2)2 – 1 at its points of intersection with the line x – y = 3, intersect at the point :
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
Let the coordinates of C be (h, k)
So the chord of contact of C w.r.t y = (x-2)2 - 1
y = x2 - 4x + 3 is T = 0
= xh + 3 - 2(x+h)
2(h-2)x - y = -(6-4h-k)
On comparing it with x - y = 3
2 (h -2) = 1 h =
6 - 4h - k = -3 k = -1
C = (, -1)
So the chord of contact of C w.r.t y = (x-2)2 - 1
y = x2 - 4x + 3 is T = 0
= xh + 3 - 2(x+h)
2(h-2)x - y = -(6-4h-k)
On comparing it with x - y = 3
2 (h -2) = 1 h =
6 - 4h - k = -3 k = -1
C = (, -1)
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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.