Stop there, locus!

Geometry Level 3

Find the equation of the locus of a point which moves such that the ratio of its distances from $(2,0)$ and $(1,3)$ is $5:4$ .

If the equation can be written as $9{ x }^{ 2 }+{ 9y }^{ 2 }+14x-150y+\alpha =0$ , find $\alpha$ .

The answer is 186.

1 solution

Rishabh Jain
Jun 10, 2016

Using Distance formula ,

$\large\dfrac{\sqrt{(x-2)^2+y^2}}{\sqrt{(x-1)^2+(y-3)^2}}=\dfrac{5}{4}$

Cross multiplication and squaring gives:-

$16((x-2)^2+y^2)=25((x-1)^2+(y-3)^2)$

Some simplification gives :-

$9{ x }^{ 2 }+{ 9y }^{ 2 }+14x-150y+\color{#20A900}{186} =0$

$\Large \therefore\color{#20A900}{\alpha=\boxed{186}}$

Is it a circle?Can we use equation of circle to solve it? @Rishabh Cool

Anik Mandal - 5 years ago

Absolutely... We have to find locus of point $P$ such that $\dfrac{PF_1}{PF_2}=\lambda$ $(F_1\equiv(2,0);F_2\equiv(1,3);\lambda=\dfrac 54)$ . Now define two point $Q$ and $R$ which divide $F_1F_2$ in the ratio $5:4$ internally and externally respectively. Now req locus (a circle) is obtained using diametric form of circle on $Q$ and $R$ which is the same as obtained above.

Rishabh Jain - 5 years ago

Wow Nice!Thanks a lot!

Anik Mandal - 5 years ago

@Anik Mandal – No problem! :-)

Rishabh Jain - 5 years ago

Stop there, locus!

The answer is 186.

1 solution

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