Question

The coordinates of the points $A$ and $B$ are $(a,0)$ and $(-a,0)$, respectively. If a point $P$ moves so that $P{A}^2-P{B}^2=2k^2$, when $k$ is constant, then find the equation to the locus of the point $P$.

• A. $2ax+k^2=0$
• B. $2ax-k^2=0$
• C. $ax+2k^2=0$
• D. $ax-2k^2=0$

Answer 1

The correct option is A $2ax+k^2=0$

Given Point $A=(a,0),B=(-a,0)$ for a moving point $P$

${PA}^2-{PB}^2=2k^2$

for $k$ is constant

also given to find out equation to the locus of point $P$

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

$P(x,y)$ $A(a,0)$

$PA=\sqrt{{(a-x)}^2+{(0-y)}^2}$

by squaring on both sides for eqution (1) we get

${\left(PA\right)}^2={\left[\sqrt{{(a-x)}^2+{(-y)}^2}\right]}^2$

${\left(PA\right)}^2=a^2+x^2-2ax+y^2$

${\left(PA\right)}^2=x^2+y^2-2ax+a^2$

$x^2+y^2+a(a-2x)....\left(2\right)$

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

$P(x,y);B(-a,0)$

$PB=\sqrt{{(-a-x)}^2+{(0-y)}^2}...\left(3\right)$

by squaring on both side of (3) we get

${\left(PB\right)}^2={\left[\sqrt{{(-a-x)}^2+{(-y)}^2}\right]}^2$

${\left(PB\right)}^2=x^2+2ax+a^2+y^2$

$x^2+y^2+a(a+2x)....\left(4\right)$

Let us substitute ${\left(PA\right)}^2$ and ${\left(PB\right)}^2$ in given equation

${\left(PA\right)}^2-{\left(PB\right)}^2=2k^2$

$x^2+y^2+a^2-2ax-(x^2+y^2+a^2+2ax)=2k^2$

$x^2+y^2+a^2-2ax-x^2-y^2-a^2-2ax=2k^2$

$-4ax=2k^2$

$k^2=-2ax\Rightarrow$ $k^2+2ax=0$