Question

The slope of tangent at $(1,1)$ to the curve $x=t^2+4t+1$ and $y=2t^2+3t+1$ is

• A. $\frac{-1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{-3}{4}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{3}{4}$

Given $x=1,y=1$
$\Rightarrow x=t^2+4t+1=1$
$\Rightarrow t^2+4t=0t=0,-4$
When $y=1$
$\Rightarrow y=2t^2+3t+1=1$
$\Rightarrow 2t^2+3t=0$
$\Rightarrow t=0,-\frac{3}{2}$
$\therefore$ Value of $t=0$
$\Rightarrow {\frac{dy}{dx}|}_{t=0}=\frac{4t+3}{2t+4}=\frac{3}{4}$