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 $△AOP$ is $4$, is:

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

Answer 1

The correct option is B $9x^2+8y^2-8y=16$

Given,

$OA+AP+OP=4$
$\Rightarrow 1+\sqrt{x^2+(y-1{)}^2}+\sqrt{x^2+y^2}=4$
$\Rightarrow (\sqrt{x^2+y^2}{)}^2=(3-\sqrt{x^2+(y-1{)}^2}{)}^2$
$\Rightarrow x^2+y^2=9+x^2+y^2-2y+1-6\sqrt{x^2+(y-1{)}^2}$
$\Rightarrow 3\sqrt{x^2+(y-1{)}^2}=(5-y)$
$\Rightarrow 9(x^2+(y-1{)}^2)=(5-y{)}^2$
$\Rightarrow 9x^2+8y^2-8y=16$