Question

consider the parametric curve x = f(t) = t^2 + t , y = g(t) = e^t. Find all t's such that the tangent line to the curve at the point (f(t), g(t)) intersects the x-axis at (4,0)

Answer 1

I did some funky voodoo and got this as the solution equation:
\[t^2-t-5=0\]
It involved getting the slope for the tangent line, dy/dx by combining dy/dt and dx/dt.
Not sure if everything I did was kosher, but if anyone is interested, I'll show the steps.

Answer 2

x' = 2t+1; y'=e^t

since the slope of the line at any point of "t" is defined as: \(\large \frac{e^t}{2t+1}~:~t\ne -\frac12\) (should we assume t is >= 0 )?

the slope intercept form of the tangent line would amount to:\[y= \frac{e^t(x-4)}{2t+1}\]\[e^t= e^t\frac{t^2+t-4}{2t+1}\]
which I would say amounts to the fraction part equating to 1; or top equals bottom\[t^2+t-4=2t+1\]\[t^2-t-5=0\]which seems to be the same as Cliffs

Answer 3

It seems to work fine for the negative value of t. I graphed it from -3<t<4 and both tangent lines look legit.