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An isosceles triangle is inscribed in the parabola

y2=4ax with its base as the line joining the vertex
of the parabola and positive end of the latus rectum
of the parabola. If at2,2at is the vertex of the
triangle then

a
2t2−8t+5=0
b
2t2+8t−5=0
c
2t2+8t+5=0
d
2t2−8t−5=0

detailed solution

Correct option is B

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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