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Question

A tangent to the parabola x2=4ay meets the hyperbola x2−y2=a2 at two points P and Q, then midpoint of P and Q lies on the curve


A

y3=x(ya)

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B

y3=x2(ya)

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C

y2=x2(ya)

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D

y2=x3(ay)

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Solution

The correct option is B

y3=x2(ya)


Equation of tangent to parabola y=mxam2......(1)

equation of chord of hyperbola whose midpoint is (h, k) is

hxky=h2k2...... (2)

form (1) and (2)

mh=1k=am2h2k2k3=h2(ka)


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