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Find k so Two Normals from (2k,0) to y²=4x are Perpendicular | JEE

JEE Maths question with a full step-by-step solution.

Question
The normals are drawn from (2k,0)(2k, 0) to the curve y2=4xy^2 = 4x. One normal is the xx-axis. Then the value of kk for which the other two normals are perpendicular to each other is:
A32\dfrac{3}{2}correct
B22
C52\dfrac{5}{2}
D72\dfrac{7}{2}
Solution
Step 1: The normal y=mx−2m−m3y = mx - 2m - m^3 through (2k,0)(2k, 0) gives
m(m2−(2k−2))=0⇒m=0 or m2=2k−2.m\big(m^2 - (2k - 2)\big) = 0 \Rightarrow m = 0\ \text{or}\ m^2 = 2k - 2.
 Step 2: The other two slopes are ±2k−2\pm\sqrt{2k - 2}; perpendicular means their product is −1-1:
−(2k−2)=−1⇒k=32.-(2k - 2) = -1 \Rightarrow k = \frac{3}{2}.
 Correct answer: (1)
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