Question

The y - intercept of tangent drawn to the curve $x=t^2+3t-8$ and $y=2t^2-2t-5$ at the point $(2,-1)$ is

• A. $\frac{12}{7}$
• B. $\frac{15}{7}$
• C. $\frac{-11}{7}$
• D. $\frac{-19}{7}$

Answer 1

The correct option is D $\frac{-19}{7}$

$t^2+3t-8=2,2t^2-2t-5=-1$
$\Rightarrow t=2,-5$ $\Rightarrow t=-1,2$
So, $t=2$
${\begin{array}{c}\frac{dy}{dx}\end{array}|}_{t=2}={\begin{array}{c}\frac{dy/dt}{dx/dt}\end{array}|}_{t=2}={\begin{array}{c}\frac{4t-2}{2t+3}\end{array}|}_{t=2}={\begin{array}{c}\frac{6}{7}\end{array}|}_{t=2}$
Equation of tangent is
$(y+1)=\frac{6}{7}(x-2)$
At $x=0,y=\frac{-12}{7}-1=\frac{-19}{7}$