Question:
The slope of the tangent to the curve $x=t^{2}+3 t-8, y=2 t^{2}-2 t-5$ at the point $(2,-1)$ is
A. $\frac{22}{7}$
B. $\frac{6}{7}$
C. $\frac{7}{6}$
D. $-\frac{6}{7}$
Solution:
Given that $x=t^{2}+3 t-8, y=2 t^{2}-2 t-5$
Differentiating w.r.t. $t$,
$\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t}+3, \frac{\mathrm{dy}}{\mathrm{dt}}=4 \mathrm{t}-2$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4 \mathrm{t}-2}{2 \mathrm{t}+3}$
For $(2,-1)$,
The given point is $(2,-1)$
$2=t^{2}+3 t-8,-1=2 t^{2}-2 t-5$
On solving we get,
$\mathrm{t}=2$ or $-5$ and $\mathrm{t}=2$ or $-1$
$\because \mathrm{t}=2$ is the common solution
$\mathrm{SO}, \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{8-2}{4+3}=\frac{6}{7}$