Search any question & find its solution
Question:
Answered & Verified by Expert
The locus of a point such that the sum of its distances from the points $(0,2)$ and $(0,-2)$ is 6 , is
Options:
Solution:
1940 Upvotes
Verified Answer
The correct answer is:
$9 x^2+5 y^2=45$
Let $P\left(x_1, y_1\right)$ be any point, then
$\begin{aligned}
\sqrt{\left(x_1-0\right)^2+} & \left(y_1-2\right)^2 \\
& +\sqrt{\left(x_1-0\right)^2+\left(y_1+2\right)^2}=6
\end{aligned}$
$\begin{aligned} & \Rightarrow \sqrt{x_1^2+\left(y_1-2\right)^2}=6-\sqrt{x_1^2+\left(y_1+2\right)^2} \\ & \Rightarrow x_1^2+\left(y_1-2\right)^2=36+\left(x_1^2+\left(y_1+2\right)^2\right) \\ & \quad-12 \sqrt{x_1^2+\left(y_1+2\right)^2} \\ & \Rightarrow-8 y_1=36-12 \sqrt{x_1^2+\left(y_1+2\right)^2} \\ & \Rightarrow 3 \sqrt{x_1^2+\left(y_1+2\right)^2}=2 y_1+9 \\ & \Rightarrow 9\left(x_1^2+\left(y_1+2\right)^2\right)=4 y_1^2+81+36 y_1 \\ & \Rightarrow \quad 9 x_1^2+5 y_1^2=45\end{aligned}$
Hence, locus of a point is
$9 x^2+5 y^2=45$
$\begin{aligned}
\sqrt{\left(x_1-0\right)^2+} & \left(y_1-2\right)^2 \\
& +\sqrt{\left(x_1-0\right)^2+\left(y_1+2\right)^2}=6
\end{aligned}$
$\begin{aligned} & \Rightarrow \sqrt{x_1^2+\left(y_1-2\right)^2}=6-\sqrt{x_1^2+\left(y_1+2\right)^2} \\ & \Rightarrow x_1^2+\left(y_1-2\right)^2=36+\left(x_1^2+\left(y_1+2\right)^2\right) \\ & \quad-12 \sqrt{x_1^2+\left(y_1+2\right)^2} \\ & \Rightarrow-8 y_1=36-12 \sqrt{x_1^2+\left(y_1+2\right)^2} \\ & \Rightarrow 3 \sqrt{x_1^2+\left(y_1+2\right)^2}=2 y_1+9 \\ & \Rightarrow 9\left(x_1^2+\left(y_1+2\right)^2\right)=4 y_1^2+81+36 y_1 \\ & \Rightarrow \quad 9 x_1^2+5 y_1^2=45\end{aligned}$
Hence, locus of a point is
$9 x^2+5 y^2=45$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.