The number of chords drawn from point $(a, a)$ on the circle , $x^2+y^2=2a^2$which are bisected by the parabola $y^2=4ax$ , is 

Let $P\left(a{t}^2,2at\right)$ be a point on the parabola $y^2=4ax$ such that the chord drawn from $(a, a)$is bisected at P. Then, the equation of the chord is 

$\left(a{t}^2\right)x+\left(2at\right)y=a^2{t}^4+4a^2{t}^2$ It passes through $(a, a)$ .  

$\therefore t^2+2t=t^4+4t^2\Rightarrow t\left(t^3+3t-2\right)=0\Rightarrow t=0,t^3+3t-2=0$ 

Let $f\left(t\right)=t^3+3t-2$ .Then $f^{\mathrm{\prime }}\left(t\right)=3t^2+3>0$ for all t

So, $f\left(t\right)$is increasing. Also, $f\left(0\right) = - 2 < 0$ and $f\left(1\right) > 0.$ 

So, $f\left(t\right)$has exactly one real root.

Hence, there are two real values of t, correspondingly there are two chords.