Let O(0,0) and A(0,1) be two fixed points.

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 $	riangle AOP$ is 4, is:

Accepted Answer

Let point C(h, k)

Given, AB + BC + AC = 4

From graph, AB = 1

$$ \therefore $$ BC + AC = 3

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

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

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

$$ \Rightarrow $$ $$9\left( {{h^2} + {k^2}} \right) = {k^2} + 8k + 16$$

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

$$ \therefore $$ Locus of point C will be,

9x2 + 8y2 -- 8y = 16

Solutions · 1