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  • Let and be two fixed points, then the locus of a point P such that the perimeter of <img alt="AOP\Delta"

Let and A(1,2)be two fixed points, then the locus of a point P such that the perimeter of AOP\Delta is 4, is 

 

Option: 1

8 x^2-24 x+16 y^2-64 y+56=32 \sqrt{(2)}


Option: 2

8 x^2-24 x-16 y^2-64 y-56=32 \sqrt{(2)}


Option: 3

10 x^2-44 x-26 y^2-94 y-56=32 \sqrt{(2)}


Option: 4

10 x^2+44 x+26 y^2+94 y+56=-32 \sqrt{(2)}


Answers (1)

best_answer

Let p(x,y) be any point on the locus.

Then, using the distance formula, we have:

\begin{aligned} & \left.A P=\sqrt{(}(x-1)^2+(y-2)^2\right) \\ & \left.O P=\sqrt{(}(x-1)^2+y^2\right) \\ & O A=\sqrt{\left(1^2+1^2\right)}=\operatorname{sqrt}(2) \\ & \end{aligned}

The perimeter of triangle AOP is:


\begin{aligned} & A P+O P+O A \\ = & \left.\left.\sqrt{(}(x-1)^2+(y-2)^2\right)+\sqrt{(}(x-1)^2+y^2\right)+\sqrt{(2)} \\ = & \left.\left.\sqrt{(}(x-1)^2+(y-2)^2\right)+\sqrt{(}(x-1)^2+y^2\right)+\sqrt{(2)}=4 \\ = & \left.\left.\sqrt{(}(x-1)^2+(y-2)^2\right)+\sqrt{(}(x-1)^2+y^2\right)=4-\sqrt{(2)} \\ = & \left.\left((x-1)^2+(y-2)^2\right)+2 \sqrt{(}(x-1)^2+(y-2)^2\right)\left((x-1)^2+y^2\right)+\left((x-1)^2+y^2\right)= \\ ( & \left(4-\sqrt{(2))^2}\right. \\ = & \left.2 \sqrt{(}(x-1)^2+(y-2)^2\right]\left[(x-1)^2+y^2\right)=14-8 \sqrt{(2)} \\ = & 4\left((x-1)^2+(y-2)^2\right)\left((x-1)^2+y^2\right)=\left(14-8 \sqrt{(2))^2}\right. \\ = & 8 x^2-24 x+16 y^2-64 y+56=32 \sqrt{(2)} \end{aligned}

Therefore, the locus of the point P is the set of all points (x,y)  that satisfy the equation:
\Rightarrow 8 x^2-24 x+16 y^2-64 y+56=32 \sqrt{(2)}

 

 

Posted by

Rishabh

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