Question

The line joining $(1,0)$ to $(2\cos \theta ,2\sin \theta )$ is divided internally in the ratio $1:2$ at $P$, then the locus of $P$ is

• A. ${\left(3x-1\right)}^2+9y^2=4$
• B. $x+y-3=0$
• C. $(x-3{)}^2+y^2=16$
• D. $x^2=y-3$

Answer 1

The correct option is A ${\left(3x-1\right)}^2+9y^2=4$

Let the coordinate of $P$ is $(x,y)$.

Then by the given question,

$\begin{array}{c}x=\left(\frac{1+2\cos \theta }{\left(1+2\right)}\right)=\frac{1+2\cos \theta }{3}\\ \Rightarrow 3x=1+2\cos \theta \\ \Rightarrow 2\cos \theta =3x-1\\ \Rightarrow \cos \theta =\frac{\left(3x-1\right)}{2}and\\ y=\frac{\left(0+2\sin \theta \right)}{\left(1+2\right)}=\frac{2\sin \theta }{3}\\ \Rightarrow 3y=2\sin \theta \\ \Rightarrow \sin \theta =\frac{3y}{2}\\ Weknowthat,\\ {\sin }^2\theta +{\cos }^2\theta =1\\ Now,\\ \Rightarrow {\left(\frac{3y}{2}\right)}^2+{\frac{\left(3x-1\right)}{2^2}}^2=1\\ \Rightarrow {\frac{9y}{4}}^2+{\frac{\left(3x-1\right)}{4}}^2=1\\ \Rightarrow {\left(3x-1\right)}^2+9y^2=4\end{array}$

Hence, this is the answer.