$P$ is a variable point such that the distance of $P$ from $A(4,0)$ is twice the distance of $P$ from $B(-4…
$P$ is a variable point such that the distance of $P$ from $A(4,0)$ is twice the distance of $P$ from $B(-4,0)$. If the line $3 y-3 x-20=0$ intersects the locus of P at the points C and D , then the distance between C and D is
8
$\frac{8 \sqrt{2}}{3}$
$\frac{32}{3}$
$\frac{8}{3}$
Solution
$\begin{aligned}
& \text { } \sqrt{(x-4)^2+y^2}=2 \sqrt{(x+4)^2+y^2} \\
& \Rightarrow 3 x^2+3 y^2+40 x+48=0 \\
& \text { Centre } \equiv\left(\frac{-20}{3}, 0\right) ; \text { Radius }=\sqrt{\left(\frac{20}{3}\right)^2-16}=\frac{16}{3}
\end{aligned}$ Given line is $3 y-3 x-20=0$
It passes through $\left(\frac{-20}{3}, 0\right)$ i.e. the centre.
Distance between C \& D = Diameter $=\frac{32}{3}$