Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

# The slope of tangent o the curve $x=t^2+3t-8,y=2t^2-2t-5$ at the point (2,-1) is :$(A)\;\frac{22}{7}\quad(B)\;\frac{6}{7}(C)\;{-6}{7}\quad(D)\;-6$

Can you answer this question?

## 1 Answer

0 votes
Toolbox:
• Slope of a line is $ax+by+c=0$ is $- \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
• If $y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p =$ slope of the normal to $y = f(x)$ at point $p$
Step 1
The parameters of the curve are
$x = t^2+3t-8$
differentiating w.r.t $t$ we get,
$\large\frac{dx}{dt}=2t+3$
and $y=2t^2-2t-5$
differentiating w.r.t $t$ we get,
$\large\frac{dy}{dt}=4t-2$
$\therefore \large\frac{dy}{dt}=\large\frac{dy}{dt} \times \large\frac{dt}{dx}$
$= \large\frac{4t-2}{2t+3}$
Step 2
when $x=2$
then $t$ is
$2=(t^2+3t-8)$
$\Rightarrow t^2+3t-10=0$
$\Rightarrow (t+5)(t-2)=0$
$\Rightarrow t=-5\: or \: t=2$
when $t=2$
$\large\frac{dy}{dx} = \large\frac{4(2)-2}{2(2)+3}$
$= \large\frac{6}{7}$
Hence the correct option is B
answered Aug 11, 2013

0 votes
0 answers

0 votes
1 answer

0 votes
1 answer