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 :
 

  • Option 1)

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

  • Option 2)

    8x^{2}-9y^{2}+9y=18

  • Option 3)

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

     

  • Option 4)

    8x^{2}+9y^{2}-9y=18

 

Answers (1)
S solutionqc

0(0,0)\; \; A(0,1)\; \; \; \; Let \; P = (x,y)

then perimeter = OA+AP+PO

=OA=1                                          

\\AP=\sqrt{x^{2}+(y-1)^{2}}\\\; \; \; \; \\PO=\sqrt{x^{2}+y^{2}}

=OA+AP+PO=4\; \; \; \; \; \; (given)

= 1+\sqrt{x^{2}+(y-1)^{2}}+\sqrt{(x^{2}+y^{2})}=4

\sqrt{x^{2}+(y-1)^{2}}+\sqrt{x^{2}+y^{2}}=3

\left ( \sqrt{x^{2}+(y-1)^{2}} \right )^{2}=\left ( 3-\sqrt{x^{2}+y^{2}} \right )^{2}

-2y+1=9-6\sqrt{x^{2}+y^{2}}

-2y-8=-6\sqrt{(x^{2}+y^{2})}

=y+4=3\sqrt{(x^{2}+y^{2})}

Squaring both sides =y^{2}+8y+16=9(x^{2}+y^{2})

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

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

 

 

 

 


Option 1)

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

Option 2)

8x^{2}-9y^{2}+9y=18

Option 3)

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

 

Option 4)

8x^{2}+9y^{2}-9y=18

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