Question

A=a,0 B=-a,0 p is a moving point such that PAB-PBA=π/2,then locus of P is​

Answer 1

Answer:

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Answer 2

Answer:

The locus of the point P is $x^{2} +y^{2} =a^{2}$

Step-by-step explanation:

Given the coordinates, A = (a, 0) and B = (-a, 0)

Let the point P be (x, y)

And let the $\angle PAB=\theta_{1}$ and $\angle PBA=\theta_{2}$.

Then according to the given condition,

$\angle PAB-\angle PBA=\theta_{1}-\theta_{2}=\dfrac{\pi}{2}$

Taking tan on both sides,

$tan\big(\theta_{1}-\theta_{2}\big)=tan\dfrac{\pi}{2}$

Using the trigonometric relation,

$tan(A-B)= \dfrac{tan A-tanB}{1+tanAtanB}$ , we get

$\dfrac{tan\theta_{1}-tan\theta_{2}}{1+tan\theta_{1}tan\theta_{2}} =tan\dfrac{\pi}{2}$

Since $tan\dfrac{\pi}{2}$ cannot be defined, we can write

$1+tan\theta_{1}tan\theta_{2}=0$                              ...(1)

Using the slope formula,

$tan\theta=\dfrac{y-y_0}{x-x_0}$

From the given coordinates,

$tan\theta_{1}=\dfrac{y-0}{x-a} =\dfrac{y}{x-a}$

$tan\theta_{2}=\dfrac{y-0}{x-(-a)} =\dfrac{y}{x+a}$

Substituting these in equation (1),

$\implies 1+\dfrac{y}{x-a}\times \dfrac{y}{x+a}=0$

$\implies1+\dfrac{ y^{2}}{x^{2} -a^{2}} =0$

$\implies x^{2} -a^{2}+y^{2} =0\implies x^{2} +y^{2} =a^{2}$

Hence, the locus of the point P is $x^{2} +y^{2} =a^{2}$