Find the locus of the middle points of all chords of the parabola $y^{2} = 4 a x$ which are drawn through the vertex.

2 y^{2} = a x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y^{2} = 2 a x

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y^{2} = 4 a x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y^{2} = a x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $y^{2} = 2 a x$
Vertex is $V (0, 0)$ and any point on the parabola is $P (a t^{2}, 2 a t)$
Now mid point of chord $V P$ is $R (\frac{a t^{2}}{2}, a t)$
Let locus of R is $(x, y)$
$∴ x = \frac{a t^{2}}{2}, y = a t$
Elimination $t$ we get required locus $y^{2} = 2 a x$

Suggest Corrections

Similar questions

y^{2} = 4 a x

y^{2} = 2 a x .

y^{2} = 4 a x

y^{2} = 2 a x

y^{2} = 4 a x

y^{2} = 4 a x

y^{2} = 4 a x

Join BYJU'S Learning Program