If the equation of directrix of the parabola y2+4y+4x+2=0 is x=pq, where p and q are co-prime, then the value of pq is
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Solution
Given : y2+4x+4y+2=0 ⇒y2+4y+4=−4x+2 ⇒(y+2)2=−4(x−12) Comparing with Y2=−4AX whose directrix is given by X=A ∴ Required equation of directrix is x−12=1 ⇒x=32 ∴pq=6