If 2p2−3q2+4pq−p=0 and a variable line px+qy=1 always touches a parabola whose axis is parallel to X-axis, then equation of the parabola is
A
(y−4)2=24(x−2)
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B
(y−3)2=12(x−1)
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C
(y−4)2=12(x−2)
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D
(y−2)2=24(x−4)
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Solution
The correct option is C(y−4)2=12(x−2) The parabola be (y−a)2=4b(x−c) Equation of tangent is (y−a)=−pq(x−c)−bqp Comparing with px+qy=1, we get cp2−bq2+apq−p=0 ∴c2=b3=a4=1⇒ the equation is (y−4)2=12(x−2).