JEE Main 2019 — Parabola Question with Solution
From: JEE Main 2019 (Online) 12th April Evening Slot
Question
The equation of common tangent to the curves y2
= 16x and xy = –4, is :
Choose an option
Show full solutionCorrect option: A
Correct answer
Ax – y + 4 = 0
Step-by-step explanation
Let the equation of tangent to parabola
y2 = 16x is y = mx + ...... (1)
It is given that tangent to xy = -4 .........(2)
Solving (1) and (2) we get
Now for tangent D = 0
m = 1
Now putting value of m in Equation (1)
y = x + 4 or x – y + 4 = 0
y2 = 16x is y = mx + ...... (1)
It is given that tangent to xy = -4 .........(2)
Solving (1) and (2) we get
Now for tangent D = 0
m = 1
Now putting value of m in Equation (1)
y = x + 4 or x – y + 4 = 0
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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.