An isosceles triangle is inscribed in the parabolay2=4ax with its base as the line joining the vertexof the parabola and positive end of the latus rectumof the parabola. If at2,2at is the vertex of thetriangle then

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2t2−8t+5=0

2t2+8t−5=0

2t2+8t+5=0

2t2−8t−5=0

answer is B.

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Detailed Solution

at22+(2at)2=at2−a2+(2at−2a)2 as vertex of theparabola is (0, 0) and positive end of the latus-rectum is(a, 2a).

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