Prove the following


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

  1. (1) $8 x^{2}-9 y^{2}+9 y=18$

  2. (2) $9 x^{2}-8 y^{2}+8 y=16$

  3. (3) $9 x^{2}+8 y^{2}-8 y=16$

  4. (4) $8 x^{2}+9 y^{2}-9 y=18$

Correct Option: , 3


Let point $P(h, k)$

$O A=1$

So, $O P+A P=3$

$\Rightarrow \sqrt{h^{2}+k^{2}}+\sqrt{h^{2}+(k-1)^{2}}=3$

$\Rightarrow h^{2}+(k-1)^{2}=9+h^{2}+k^{2}-6 \sqrt{h^{2}+k^{2}}$

$\Rightarrow 6 \sqrt{h^{2}+k^{2}}=2 k+8$

$\Rightarrow 9 h^{2}+8 k^{2}-8 k-16=0$

Hence, locus of point $\mathrm{P}$ is

$9 x^{2}+8 y^{2}-8 y-16=0$

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