JEE Main 2019 — Parabola Question with Solution
From: JEE Main 2019 (Online) 8th April Evening Slot
Question
The tangent to the parabola y2
= 4x at the point
where it intersects the circle x2
+ y2
= 5 in the
first quadrant, passes through the point :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
Parabola y2
= 4x and circle x2
+ y2
= 5 intersect with each other.
So, x2 + 4x = 5
x2 + 5x – x – 5 = 0
x(x + 5) –1(x + 5) = 0
x = 1, –5
Intersection point in 1st quadrant is = (1, 2)
Equation of tangent to y2 = 4x at (1, 2) is
y(2) = 2 (x + 1)
y = x + 1 .....(1)
By checking each options, you can see
point lies on equation (1).
So, x2 + 4x = 5
x2 + 5x – x – 5 = 0
x(x + 5) –1(x + 5) = 0
x = 1, –5
Intersection point in 1st quadrant is = (1, 2)
Equation of tangent to y2 = 4x at (1, 2) is
y(2) = 2 (x + 1)
y = x + 1 .....(1)
By checking each options, you can see
point lies on equation (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.