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Find a if x+a=0 is Directrix of y²=2y+ax+2 | JEE Parabola

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

Question
If x+a=0x + a = 0 is the directrix of the parabola y2=2y+ax+2y^2 = 2y + ax + 2, then the value of aa is:
A22correct
B33
C44
D43\dfrac{4}{3}
Solution
Step 1: Reduce to standard form.
y2−2y=ax+2⇒(y−1)2=a(x+3a).y^2 - 2y = ax + 2 \Rightarrow (y-1)^2 = a\left(x + \frac{3}{a}\right).
 This is (y−1)2=4A(x−h)(y-1)^2 = 4A(x-h) with 4A=a4A = a, so its directrix is x+3a+a4=0x + \dfrac{3}{a} + \dfrac{a}{4} = 0. Step 2: Match with the given directrix x+a=0x + a = 0.
3a+a4=a⇒12+a2=4a2⇒a2=4⇒a=2.\frac{3}{a} + \frac{a}{4} = a \Rightarrow 12 + a^2 = 4a^2 \Rightarrow a^2 = 4 \Rightarrow a = 2.
 Correct answer: (1)
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