Graph of $v^2-x^2$ for a particle executing SHM is

$$\begin{gathered} velocity of oscillator=v=\omega \sqrt{A^2-x^2} ; here \omega =angular velocity; A=amplitude; x= displacement \\ square on both sides \\ \Rightarrow v^2={\omega }^2{A}^2-{\omega }^2{x}^2 \end{gathered}$$

Let  $v^2=Y$ and  $x^2=X$
So,  $Y={\omega }^2{A}^2-{\omega }^{2}X$
This is of the form Y = mX + c  (equation of a straight line), where  $c={\omega }^2{A}^2$ and  slope, $m=-{\omega }^2$ .