Find the equation of the locus of a point whose distance from $\left(a, 0\right)$ is always four times its distance from the $Y$ axis.
Let $\;$ $A = \left(a, 0\right)$
Let $P \left(x, y\right)$ be a point on the required locus.
Distance $\;$ $PA = \sqrt{\left(x - a\right)^2 + \left(y - 0\right)^2}$
i.e. $\;$ $PA = \sqrt{x^2 + y^2 + a^2 - 2ax}$
$\therefore \;$ $\left(PA\right)^2 = x^2 + y^2 + a^2 - 2ax$
Distance of point $P \left(x, y\right)$ from the $Y$ axis is $= x$
As per question, $\;$ $PA = 4 \times x$
i.e. $\;$ $\left(PA\right)^2 = 16 \times x^2$
i.e. $\;$ $x^2 + y^2 + a^2 - 2ax = 16x^2$
i.e. $\;$ $15x^2 - y^2 + 2ax - a^2 = 0$ $\;\;\;$ is the equation of the required locus.