First slide

Parabola

Question

An isosceles triangle is inscribed in the parabola
$y^{2} = 4 a x$ with its base as the line joining the vertex
of the parabola and positive end of the latus rectum
of the parabola. If $(a t^{2}, 2 a t)$ is the vertex of the
triangle then

Moderate

A

$2 t^{2} - 8 t + 5 = 0$
B

$2 t^{2} + 8 t - 5 = 0$
C

$2 t^{2} + 8 t + 5 = 0$
D

$2 t^{2} - 8 t - 5 = 0$

Solution

${(a t^{2})}^{2} + (2 a t)^{2} = {(a t^{2} - a)}^{2} + (2 a t - 2 a)^{2}$ as vertex of the
parabola is (0, 0) and positive end of the latus-rectum is
$(a, 2 a)$ .

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