Question

If $A=(5,0)$ and $B=(0,4)$, then the locus of moving point $P$ such that $|PA{|}^2-|PB{|}^2=9$ is

• A. $5x-4y=0$
• B. $4x-5y=0$
• C. $5x+4y=0$
• D. $4x+5y=0$

Answer 1

The correct option is A $5x-4y=0$

Let $P\equiv (x,y)$ be an arbitrary point satisfying the condition.Then we have $|PA{|}^2-|PB{|}^2=9$
$\left(\right(x-5{)}^2+y^2)-(x^2+(y-4{)}^2)=9$
$\Rightarrow -10x+25+8y-16=9$
$\Rightarrow 5x-4y=0$
Hence, the equation of the locus of $P$ is $5x-4y=0$