For a particle executing SHM, x= displacement from mean position, v= velocity and a= acceleration at any instant, then

$v=\omega \sqrt{A^2-x^2}$

$\therefore \frac{v^2}{A^2{\omega }^2}+\frac{x^2}{(A{)}^2}=1$

i.e. v-x graph is an ellipse.

$a=-{\omega }^{2}x$

i.e. a-x graph is a straight line passing through origin with negative slope.