Find the locus the mid-points of the parabola $y^{2} = 4 a x$ which subtend a right at the vertex of the parabola.

Open in App

Solution

REF.Image
Let $P = (a t_{1}^{2}, 2 a t_{1})$ & $Q = (a t_{2}^{2}, 2 a t_{2})$ be the
end points of a chord.
Mid point $M = [\frac{a (t_{1}^{2} + t_{2}^{2})}{2}, a (t_{1} + t_{2})]$
Slope of $O P = \frac{2}{t_{1}}$ , Slope of $O Q = \frac{2}{t_{2}}$
Since $O P ⊥ O Q, \frac{2}{t_{1}} . \frac{2}{t_{2}} = - 1; t_{1} t_{2} = - 4$
For M, $x = \frac{a (t_{1}^{2} + t_{2}^{2})}{2} = \frac{a (t_{1} + t_{2})^{2} - 2 a t_{1} t_{2}}{2}$
$= \frac{a (t_{1} + t_{2})^{2} + 8 a}{2}$
$y = a (t_{1} + t_{2})$
Eliminating $t_{1} + t_{2}$ ,
$y^{2} = 2 a (x - 4 a)$

Suggest Corrections

Similar questions

y^{2} = 4 a x

y^{2} = 4 a x

y^{2} = 2 a (x - 4 a)

P Q

t_{1}

t_{2}

t_{1} t_{2} = - 2

y^{2} = 4 a x

y^{2} = 4 a x

y^{2} = 4 a x

α

Join BYJU'S Learning Program