Question

The point $(p,p+1)$ lies on the locus of the point which moves such that its distance from the point $(1,0)$ is twice the distance from $(0,1)$. The value of $\frac{1}{2p^2}+\frac{1}{2p^4}$ is equal to

Answer 1

Let $P(x,y)$ be the point on the locus such that its distance from the point $A(1,0)$ is twice the distance from $B(0,1)$
i.e $PA=2PB$
$\Rightarrow P{A}^2=4P{B}^2$
$\Rightarrow (x-1{)}^2+y^2=4(x^2+(y-1{)}^2)$
but $(p,p+1)$ lies on the locus.
$\Rightarrow (p-1{)}^2+(p+1{)}^2=4(p^2+p^2)$
$\therefore p^2=\frac{1}{3}$
Now,$\frac{1}{2p^2}+\frac{1}{2p^4}$ = $\frac{1}{2}3+\frac{1}{2}9=6$
$\therefore \frac{1}{2p^2}+\frac{1}{2p^4}=6$