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Slope of Normal at Q for Focal Chord of y²+4x+4y=0 | JEE

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

Question
If P(−3,2)P(-3, 2) is one end of the focal chord PQPQ of the parabola y2+4x+4y=0y^2 + 4x + 4y = 0, then the slope of the normal at QQ is:
A−12-\dfrac{1}{2}correct
B12\dfrac{1}{2}
C22
D−2-2
Solution
Step 1: Tangent at P(−3,2)P(-3, 2) to y2+4x+4y=0y^2 + 4x + 4y = 0 (via T=0T = 0):
2y+2(x−3)+2(y+2)=0⇒x+2y−1=0,2y + 2(x - 3) + 2(y + 2) = 0 \Rightarrow x + 2y - 1 = 0,
 of slope −12-\dfrac{1}{2}. Step 2: The tangent at one end of a focal chord is parallel to the normal at the other end, so the normal at QQ has slope −12-\dfrac{1}{2}. Correct answer: (1)
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