A (−a,0)and B (a,0) are two fixed points. The locus of a point P such as $∠ A P B$ is a right angle, is

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

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

None of these

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

Open in App

Solution

The correct option is A + =
Let P (x, y) be the moving point whose locus is required. $L e t A \equiv (- a, 0) a n d B \equiv (a, 0), G i v e n ∠ A P B = \frac{π}{2} ∴ A P^{2} + P B^{2} = A B^{2}$

$\Rightarrow (x + a)^{2} + y^{2} + (x - a)^{2} + y^{2} = (2 a)^{2} \Rightarrow 2 x^{2} + 2 y^{2} + 2 a^{2} = 4 a^{2} \Rightarrow x^{2} + y^{2} = a^{2}$
Which is the required locus.
Hence, (a) is the correct answer.

Suggest Corrections

Similar questions

∠ A P B

A (- a, 0)

B (a, 0)

P

∠ A P B

Two fixed points are A(a, 0) and B(-a, 0). If $∠ A - ∠ B = θ$ , then the locus of point C of triangle ABC will be

Let Two fixed points are A(a, 0) and B(-a, 0). If $∠ A - ∠ B = θ$ , then the locus of point C of triangle ABC will be

A (- a, 0)

B (a, 0)

P

∠ A P B

Join BYJU'S Learning Program