First slide

Parabola

Question

If the area of the triangle inscribed in the parabola
$y^{2} = 4 a x$ with one vertex at the vertex of the parabola
and other two vertices at the extremities of
a focal chord is $5 a^{2} / 2$ then the length of the focal
chord is

Moderate

A

$3 a$
B

$5 a$
C

$25 a / 4$
D

none of these

Solution

Let the vertices be $O (0, 0), A (a t^{2}, 2 a t)$
$B (\frac{a}{t^{2}}, \frac{- 2 a}{t})$ then
${\frac{1}{2} | | \begin{array}{ccc} 0 & 0 & 1 \\ 2 t^{2} & 2 a t & 1 \\ \frac{a}{t^{2}} & \frac{2 a}{t} & 1 \end{array}|}_{1} = \frac{5 a^{2}}{2} \Rightarrow 2 t^{2} - 5 t + 2 = 0$
$\Rightarrow t = 2 or 1 / 2$ so the vertices of a focal chord are
$(4 a, 4 a)$ and $(a / 4, - a)$ (Taking $t = 2$ ) and length
of this focal chord is $25 a / 4$ .

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