Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the curve defined by the parametric equations $x=t^2 +t-1$ and $y=te^{2t}$

i) Show that $dy/dx =e^{2t}$

ii) Hence show that the tangent to the curve at the point on the curve where $t= -1$ passes through the origin.

I'm sorry to bug you guys, but I'm clueless and would help me if someone could help me, so I get questions that are similar to this. Thanks!

share|cite|improve this question
for the first part, use the fact that dy/dx = (dy/dt)/(dx/dt). – symplectomorphic Jul 8 '13 at 6:40
up vote 3 down vote accepted


$$\frac{dx}{dt}=2t+1 $$


At $t=-1,x=-1, y=-e^{-2}$

So, the equation of the tangent will be $$\frac{y-(-e^{-2})}{x-(-1)}=\frac{dy}{dx}_{\text{(at }t=-1)}=e^{-2}\implies y=x\cdot e^{-2}$$ which clearly passes through the origin $(0,0)$

share|cite|improve this answer
THANKS, your a genius! – Red Queen10101 Jul 8 '13 at 7:19
@RogerShan, my pleasure. – lab bhattacharjee Jul 8 '13 at 7:40
Excuse me, I would like to ask how do you mark a question to be answered? sorry Im new with this site. – Red Queen10101 Jul 8 '13 at 9:17
hahaha dw got it – Red Queen10101 Jul 8 '13 at 9:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.