The number on points on the curve $\mathrm{y}=2{\mathrm{t}}^2+3\mathrm{t}-5\text{ and }\mathrm{x}={\mathrm{t}}^3-4{\mathrm{t}}^2-3\mathrm{t}$ such that the normals drawn at them on the curve are parallel to X-axis is 

$\mathrm{y}=2{\mathrm{t}}^2+3\mathrm{t}-5\text{ and }\mathrm{x}={\mathrm{t}}^3-4{\mathrm{t}}^2-3\mathrm{t}$

$\frac{dy}{dt}=4t+3$ and $\frac{dx}{dt}=3t^2-8t-3$

$\frac{dx}{dt}=\frac{3t^2-8t-3}{4t+3}$

Parallel to x-axis so,slope=0

$3t^2-8t-3=0$

will get two values of t ,

then will saticefy two points