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JEE MAIN 2021
20-07-2021 S2
Question
Let $P$ be a variable point on the parabola $y=4 x^2+1$. Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y=x$ is :
Select the correct option:
A
$(3 x-y)^2+(x-3 y)+2=0$
B
$2(3 x-y)^2+(x-3 y)+2=0$
C
$(3 x-y)^2+2(x-3 y)+2=0$
D
$2(x-3 y)^2+(3 x-y)+2=0$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Mathematics
Medium
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