Question

The slope of the tangentto the curve ${\displaystyle x=t^2+3t-8,y=2t^2-2t-5}$ at the point ${\displaystyle (2,-1)}$ is

पर्याय

• ${\displaystyle \frac{2^2}{7}}$

• ${\displaystyle \frac{6}{7}}$

• ${\displaystyle \frac{7}{6}}$

• ${\displaystyle \frac{-6}{7}}$

Answer 1

${\displaystyle \frac{6}{7}}$

Explanation:

The curves is = ${\displaystyle t^2+3t-8,y=2t^2-2t+5}$  ......(1)

Putting ${\displaystyle x=2,2=t^2+3t-8}$

∴ ${\displaystyle t^2+3t-10}$ = 0

or ${\displaystyle (t+5)(t-2)}$ = 0

Putting ${\displaystyle t=2}$ in ${\displaystyle y=2t^2-2t-5=8-4-5=-1}$

At ${\displaystyle x=2,y=-1,t=2}$

Differentiating (1)

${\displaystyle \frac{dx}{dt}=2t+3,\frac{dy}{dt}=4t-2}$

${\displaystyle \frac{dy}{dx}=\frac{dy}{dt}+\frac{dx}{dt}=\frac{4t-2}{2t+3}}$

At ${\displaystyle (t=2)\frac{dy}{dx}=\frac{4\times 2-2}{2\times 2+3}=\frac{6}{7}}$