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Maximum Common Normals of y²=4ax and x²=4by | JEE Parabola

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

Question
The maximum number of common normals of y2=4axy^2 = 4ax and x2=4byx^2 = 4by may be equal to:
A22
B44
C55correct
D33
Solution
Step 1: A normal of slope mm to each parabola is
y2=4ax: y=mx−2am−am3;x2=4by: y=mx+2b+bm2.y^2 = 4ax:\ y = mx - 2am - am^3; \qquad x^2 = 4by:\ y = mx + 2b + \frac{b}{m^2}.
 Step 2: A common normal needs equal intercepts.
−2am−am3=2b+bm2⇒am5+2am3+2bm2+b=0,-2am - am^3 = 2b + \frac{b}{m^2} \Rightarrow am^5 + 2am^3 + 2bm^2 + b = 0,
 a fifth-degree equation in mm, so at most 55 common normals. Correct answer: (3)
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