Question

$A(2,3)$ is a fixed point and $Q(3\cos \theta ,2\sin \theta )$ is a variable point. If ${}^{\prime }{P}^{\prime }$ divides $AQ$ internally in the ratio $3:1,$ find the locus of $P$.

• A. ${\left(\frac{4x+2}{9}\right)}^2+{\left(\frac{4y-3}{6}\right)}^2=1$
• B. ${\left(\frac{4x-2}{9}\right)}^2+{\left(\frac{4y-3}{6}\right)}^2=1$
• C. ${\left(\frac{4x-2}{6}\right)}^2+{\left(\frac{4y+3}{9}\right)}^2=1$
• D. ${\left(\frac{4x+2}{6}\right)}^2+{\left(\frac{4y+3}{9}\right)}^2=1$

Answer 1

Answer: (B)

${\left(\frac{4x-2}{9}\right)}^2+{\left(\frac{4y-3}{6}\right)}^2=1$

By using section formula we find point P

Here $m:n=3:1$

$\Rightarrow P=(\frac{3.3\cos \theta +1.2}{4},\frac{3.2\sin \theta +1.3}{4})$

$\Rightarrow P=(\frac{9\cos \theta +2}{4},\frac{6\sin \theta +3}{4})$

Let $P=(x,y)$

$\Rightarrow x=\frac{9\cos \theta +2}{4}$

$\Rightarrow y=\frac{6\sin \theta +3}{4}$

$\Rightarrow \frac{4x-2}{9}=\cos \theta$

$\Rightarrow \frac{4y-3}{6}=\sin \theta$

Squaring both equations ;

$\Rightarrow {\left(\frac{4x-2}{9}\right)}^2+{\left(\frac{4y-3}{6}\right)}^2=1$ is the locus of point $P$