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

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Aicharya · Accepted Answer

Let $P\left(x_1, y_1ight)$ be any point, then $\begin{aligned} \sqrt{\left(x_1-0ight)^2+} & \left(y_1-2ight)^2 \ & +\sqrt{\left(x_1-0ight)^2+\left(y_1+2ight)^2}=6 \end{aligned}$ $\begin{aligned} & \Rightarrow \sqrt{x_1^2+\left(y_1-2ight)^2}=6-\sqrt{x_1^2+\left(y_1+2ight)^2} \ & \Rightarrow x_1^2+\left(y_1-2ight)^2=36+\left(x_1^2+\left(y_1+2ight)^2ight) \ & \quad-12 \sqrt{x_1^2+\left(y_1+2ight)^2} \ & \Rightarrow-8 y_1=36-12 \sqrt{x_1^2+\left(y_1+2ight)^2} \ & \Rightarrow 3 \sqrt{x_1^2+\left(y_1+2ight)^2}=2 y_1+9 \ & \Rightarrow 9\left(x_1^2+\left(y_1+2ight)^2ight)=4 y_1^2+81+36 y_1 \ & \Rightarrow \quad 9 x_1^2+5 y_1^2=45\end{aligned}$ Hence, locus of a point is $9 x^2+5 y^2=45$

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

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