Question

Find the equation of locus of a point such that the sum of its distances from the points $(2,0)$ and $(-2,0)$ is $6$.

Answer 1

Let the point $(x_1,y_1)$ be such that the sum of its

distance from the points $(2,0)$ and $(-2,0)$ is 6.

$\Rightarrow \sqrt{(x_1-2{)}^2+(y_1-0{)}^2}+\sqrt{(x_1+2{)}^2+(y_1-0{)}^2}=6$

$\Rightarrow (x_1-2{)}^2+y_1^2=36+(x_1+2{)}^2+y_1^2-2\times 6\times \sqrt{(x_1+2{)}^2+y_1^2}$

$\Rightarrow x^2-4x_1+4=36+x_1^2+4x_1+4-12\sqrt{(x_1+2{)}^2+y_1^2}$

$\Rightarrow 12\sqrt{(x_1+2{)}^2+y_1^2}=36+8x_1$

$\Rightarrow 3\sqrt{(x_1+2{)}^2+y_1^2}=9+2x_1$

$\Rightarrow 9\left(\right(x_1+2{)}^2+y_1^2)=4x_1^2+36x_1+81$

$\Rightarrow 9x_1^2+36+36x_1+9y_1^2=4x_1^2+36x_1+81$

$\Rightarrow 5x_1^2+9y_1^2=45$

$\Rightarrow \frac{x_1^2}{9}+\frac{y_1^2}{5}=1$

$\therefore$ The equation of locus of the point is,

$\frac{x^2}{2}+\frac{y^2}{5}=1$