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Question

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
(y4)2=24(x2)
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B
(y3)2=12(x1)
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C
(y4)2=12(x2)
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D
(y2)2=24(x4)
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Solution

The correct option is C (y4)2=12(x2)
The parabola be (ya)2=4b(xc)
Equation of tangent is (ya)=pq(xc)bqp
Comparing with px+qy=1, we get cp2bq2+apqp=0
c2=b3=a4=1 the equation is (y4)2=12(x2).

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