Find the coordinates of a point on the parabola $y^{2} = 8 x$ whose focal distance is $4$ .

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Solution

Solution-
$y^{2} = 8 x F (2, 0)$
Let point on it be $p (2 t^{2}, 4 t)$
then
$4 = \sqrt{(2 x^{2} - 2)^{2} + (4 t)^{2}}$
$16 = 4 t^{4} + 4 - 8 t^{2} + 16 t^{2}$
$4 t^{4} + 8 t^{2} - 12 = 0$
$9 t^{4} + 2 t^{2} - 3 = 0$
$t^{2} = 1$
$t = \pm 1$
$t^{2} = - 3 - R e j e c t$
points are (2,4)& (2,-4).

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