NIMCET 2025 Paper

To find the normal slope at $\left(2,-1\right)$, first find the value of $\mathit{t}$ that gives this point.
Solve:
${\mathit{t}}^2+3\mathit{t}-8=2\Rightarrow {\mathit{t}}^2+3\mathit{t}-10=0$
This factors:

Now compute derivatives:
$\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=2\mathit{t}+3,\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}=4\mathit{t}-2.$
At $\mathit{t}=2$ :
$\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=7,\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}=6$
So the slope of the tangent is:
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{6}{7}$
Thus, the slope of the normal is the negative reciprocal: