question_answer

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 : [JEE Main 8-4-2019 Morning]

A) \[8{{x}^{2}}9{{y}^{2}}+9y=18\]

B) \[9{{x}^{2}}+8{{y}^{2}}8y=16\]

C) \[8{{x}^{2}}+9{{y}^{2}}9y=18\]

D) \[9{{x}^{2}}8{{y}^{2}}+8y=16\]

Correct Answer: B

Solution :

\[AP+OP+AO=4\] \[\sqrt{{{h}^{2}}+{{(k-1)}^{2}}}+\sqrt{{{h}^{2}}+{{k}^{2}}}+1=4\] \[\sqrt{{{h}^{2}}+{{(k-1)}^{2}}}+\sqrt{{{h}^{2}}+{{k}^{2}}}=3\] \[{{h}^{2}}+{{(k-1)}^{2}}=9+{{h}^{2}}+{{k}^{2}}-6\sqrt{{{h}^{2}}+{{k}^{2}}}\] \[-2k-8=-6\sqrt{{{h}^{2}}+{{k}^{2}}}\] \[k+4=3\sqrt{{{h}^{2}}+{{k}^{2}}}\] \[{{k}^{2}}+16+8k=9({{h}^{2}}+{{k}^{2}})\] \[9{{h}^{2}}+8{{k}^{2}}-8k-16=0\] Locus of P is \[9{{x}^{2}}+8{{y}^{2}}8y16=0\]