Question

If A(a,0), B(-a,0) then the locus of the point P such that PA2+PB2=2c2 is :

Answer 1

$\textbf{Given:}$

$\text{P is the moving point and}$

$PA^2+PB^2=2c^2$

$\textbf{To find:}$

$\text{The locus of P}$

$\textbf{Solution:}$

$\text{Let P(h,k) be the moving point}$

$PA=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$PA=\sqrt{(h-a)^2+(k-0)^2}$

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

$PB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$PB=\sqrt{(h+a)^2+(k-0)^2}$

$PB=\sqrt{(h+a)^2+k^2}$

$\text{Condition:}\;PA^2+PB^2=2c^2$

$\implies\,(h-a)^2+k^2+(h+a)^2+k^2=2c^2$

$\implies\,h^2+a^2-2ah+k^2+h^2+a^2+2ah+k^2=2c^2$

$\implies\,h^2+a^2+k^2+h^2+a^2+k^2=2c^2$

$\implies\,2h^2+2k^2+2a^2-2c^2=0$

$\text{Divide by 2}$

$\implies\,h^2+k^2+a^2-c^2=0$

$\textbf{Answer:}$

$\textbf{The locus of P is}$

$\bf\,x^2+y^2+a^2-c2=0$

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

Step-by-step explanation:

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Maths

 Let A=(1,0) , B=(−1,0),C=(2,0), the locus of a point P such that PB2+PC2=2PA2 is

A

a straight line parallel to x-axis

B

a straight line parallel to y-axis

Correct Answer

C

pair of straight line

D

combined equation of coordiante axes

Answer

Let P=(x,y)

Then, PA=(x−1)2+y2 ...(i)

PB=(x+1)2+y2 ...(ii)

PC=(x−2)2+y2 ...(iii)

Hence, PC2+PB2=2PA2

⇒[(x−2)2+y2]+[(x+1)2+y2]=