Question

A point P lying inside the curve $y=\sqrt{2ax-x^2}$ is moving such that its shortest distance from the curve at any position is greater than its distance from x-axis. The point P enclose a region whose area is equal to

• A. $\frac{\pi a^2}{2}$
• B. $\frac{a^2}{3}$
• C. $\frac{2a^2}{3}$
• D. $\left(\frac{3\pi -4}{6}\right)a^2$

Answer 1

The correct option is C $\frac{2a^2}{3}$

$y=\sqrt{2ax-x^2}\Rightarrow (x-a{)}^2+y^2=a^2$
Let P(h, k) be a point then BP > PN
For the boundary condition BP = PN = k
Now $AP=a-k=\sqrt{(h-a{)}^2+k^2}\Rightarrow k=h-\frac{h^2}{2a}$
$\therefore$ boundary of the region is $y=x-\frac{x^2}{2a}$
Required area $2{\int }_0^a(x-\frac{x^2}{2a})dx=\frac{2a^2}{3}$