Question

A(c,0) and B(-c,0) are two points. If P is a point such that PA + PB = 2a where 0 < c < a, then find the locus of P.​

Answer 1

$\textbf{Given:}$

$\textsf{A(c,0) and B(-c,0) and}$

$\textsf{P is a point such that PA+PB=2a}$

$\textbf{To find:}$

$\textsf{Locus of P}$

$\textbf{Solution:}$

$\textbf{Let the coordinates of the moving point P be (h,k)}$

$\mathsf{PA+PB=2a}$

$\mathsf{\sqrt{(h-c)^2+k^2}+\sqrt{(h+c)^2+k^2}=2a}$

$\mathsf{\sqrt{(h-c)^2+k^2}=2a-\sqrt{(h+c)^2+k^2}}$

$\textsf{Squaring on bothsides, we get}$

$\mathsf{(h-c)^2+k^2=(2a-\sqrt{(h+c)^2+k^2})^2}$

$\mathsf{(h-c)^2+k^2=4a^2+(h+c)^2+k^2-4a\sqrt{(h+c)^2+k^2}}$

$\mathsf{h^2+c^2-2hc+k^2=4a^2+h^2+c^2+2hc+k^2-4a\sqrt{(h+c)^2+k^2}}$

$\mathsf{-2hc=4a^2+2hc-4a\sqrt{(h+c)^2+k^2}}$

$\mathsf{-4hc-4a^2=-4a\sqrt{(h+c)^2+k^2}}$

$\mathsf{-4(hc+a^2)=-4a\sqrt{(h+c)^2+k^2}}$

$\mathsf{hc+a^2=a\sqrt{(h+c)^2+k^2}}$

$\textsf{Squaring again,}$

$\mathsf{(hc+a^2)^2=a^2[(h+c)^2+k^2]}$

$\mathsf{h^2c^2+a^4+2hca^2=a^2[h^2+c^2+2hc+k^2]}$

$\mathsf{h^2c^2+a^4+2hca^2=a^2h^2+a^2c^2+2hca^2+a^2k^2]}$

$\mathsf{h^2c^2+a^4=a^2h^2+a^2c^2+a^2k^2}$

$\textsf{Rearranging terms, we get}$

$\mathsf{h^2(a^2-c^2)+k^2a^2=a^4}$

$\therefore\mathsf{The\;locus\;of\;P\;is\;(a^2-c^2)x^2+a^2y^2=a^4}$

$\textbf{Find more:}$

Find the locus of the point p such that PA2+pb2=2c2 where A(a,0),B(-a,o) and 0<|a|<|c|​

https://brainly.in/question/19831924

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

https://brainly.in/question/3201572