A particle is executing SHM in a straight line. Let x, v, and a represent instantaneous  displacement, velocity and acceleration respectively. Graph of a² vs v² is

Displacement,    $x=A\sin \left(\omega t+\varphi \right)$
Velocity,     $v=A\omega \cos \left(\omega t+\varphi \right)$
Acceleration,    $a=-A{\omega }^2\sin \left(\omega t+\varphi \right)$
 $v=\omega \sqrt{A^2-x^2}\Rightarrow v^2={\omega }^2\left(A^2-x^2\right)$   ...(1)
 $a=-{\omega }^{2}x\Rightarrow a^2={\omega }^2{x}^2$     ...(2)
(2) in (1) gives,
So,  $v^2={\omega }^2\left(A^2-\frac{a^2}{{\omega }^4}\right)\Rightarrow \frac{v^2}{A^2{\omega }^2}+\frac{a^2}{A^2{\omega }^4}=1$  
It is of the form $y{K}_1+x{K}_2=1$ , where K₁ and K₂ are positive constants.
   $y=\frac{1}{K_1}-\frac{x{K}_2}{K_1}\Rightarrow y=c-mx$. It is a straight line with negative slope and positive  intercept.