Question

A(0,0)& B(-a,0) are two points find the locus of p such angle APB =π/2​

Answer 1

$\textbf{Given:}$

$\text{The points A(0,0), B(-a,0) and \angle{APB}=\frac{\pi}{2} }$

$\textbf{To find:}$

$\text{The locus of P}$

$\text{Let the coordinates of P be (h,k)}$

$\text{Since \angle{APB}=\frac{\pi}{2} ,}$

$\textbf{Slope of AP}{\times}\textbf{Slope of PB}=-1$

$\dfrac{k-0}{h-0}{\times}\dfrac{k-0}{h+a}=-1$

$\dfrac{k}{h}{\times}\dfrac{k}{h+a}=-1$

$\dfrac{k^2}{h^2+ah}=-1$

$k^2=-h^2-ah$

$\implies\,h^2+k^2+ah=0$

$\therefore\textbf{The locus of P is \bf\,x^2+y^2+ax=0 }$

Find more:

B and c are fixed points having coordinates (3,0) and (-3,0) respectively. If the vertical angle bac is 90 then locus

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If A =(-2,3) and B=(4,1) are given points find the equation of locus of point P, such that PA=2PB.

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Answer 2

Answer:

The locus is x^2+y^2+ax=0

Step-by-step explanation:

Hope it helps you