You visited us 1 times! Enjoying our articles? Unlock Full Access!

Intercept Made by Circle on Axes

The radical c...

Question

The radical centre of the circles drawn on the focal chords of the parabola $y^{2} = 4 a x$ as diameters is

(- a, 0)

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

(a, 0)

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

(0, 0)

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

(a, a)

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

Open in App

Solution

The correct option is B $(0, 0)$
Given equation of parabola is
$y^{2} = 4 a x$
Vertex is at $(0, 0)$
We know that "The radical axis of the two circles described on any two focal chords as diameters, passes through the vertex of the parabola"
So, the radical centre is $(0, 0)$ .

Suggest Corrections

Similar questions

y^{2} = 4 a x

t_{1}, t_{3}

y^{2} = 4 a x

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

A (a t_{1}^{2}, 2 a t_{1})

B (a t_{2}^{2}, 2 a t_{2})

y^{2} = 4 a x,

t_{1} t_{2} = - 1

y^{2} = 4 a x

y^{2} = 4 a x

t_{1}, t_{3}

y^{2} = 4 x + y + 4

16 x + k = 0,

k

Join BYJU'S Learning Program