Question

Let $O(0,0)$ and $A(0,1)$ be two fixed points. Then the locus of a point $P$ such that the perimeter of $\Delta AOP$ is $4$, is :

• A. $8x^2-9y^2+9y=18$
• B. $9x^2+8y^2-8y=16$
• C. $8x^2+9y^2-9y=18$
• D. $9x^2-8y^2+8y=16$

Answer 1

Answer: (B)

$AP+OP+AO=4$
$\sqrt{h^2+{\left(k-1\right)}^2}+\sqrt{h^2+k^2}+1=4$
$\sqrt{h^2+{\left(k-1\right)}^2}+\sqrt{h^2+k^2}=3$
$h^2+{\left(k-1\right)}^2=9+h^2+k^2-6\sqrt{h^2+k^2}$
$-2k-8=-6\sqrt{h^2+k^2}$
$k+4=3\sqrt{h^2+k^2}$
$k^2+16+8k=9\left(h^2+k^2\right)$
$9h^2+8k^2-8k-16=0$
Locus of P is $9x^2+8y^2-8y-16=0$