Question

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

Answer 1

$\textbf{Given:}$

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

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

$\textbf{To find:}$

$\text{The locus of P}$

$\textbf{Solution:}$

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

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

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

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

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

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

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

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

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

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

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

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

$\text{Divide by 2}$

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

$\textbf{Answer:}$

$\textbf{The locus of P is}$

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

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Answer 2

Q):-Find the equation of locus of a point P such that PA^2+PB^2=2C^2,Where A (a,0), B (-a,0) and o <11<|c|.

Answer:-

x^2+y^2+a^2-c^2=0.

Explanation:-

Let P (x,y) be a point on the locus

given A (a,0),B (-a,0)

given condition is PA^2+PB^2=2C^2

(x-a)^2+(y-0)^2+(x+a)^2+(y-0)^2=2C^2

x^2+a^2-2ax+y^2+x^2+a^2+2ax+y^2=2C^2

2x^2+2y^2+2a^2-2C^2=0

2 (x^2+y^2+a^2-c^2)=0

x^2+y^2+a^2-C^2=0.

Therefore Equation of locus of P is x^2+y^2+a^2-C^2=0.