Finding slope of line so that its normal at one point and tangent at another I am given a curve $$x= 4t^2 +1$$ and $$y=8t^3 -1$$ now by applying chain rule I get $dy/dx$ as $3t$. Now writing equation of line as 
$$(y- 8t^3 +1)=3t(x- 4t^2 -1)$$ now how to ensure that its tangent at one point and normal at other ?
 A: Hints:


*

*You know that the tangent at  $(4t_1^2+1,8t_1^3-1)$ has the form  $(y−8t_1^3 +1)=3t_1(x−4t_1^2 −1)$.  

*You want to know where else this intersects the curve so you need to solve $(8t_2^3-1−8t_1^3 +1)=3t_1(4t_2^2+1−4t_1^2 −1)$ for $t_2$ in terms of $t_1$ remembering that $t_1=t_2$ will be a solution but not a useful one.

*You want to find $t_1$ and $t_2$ such that the tangents at those two points are perpendicular, i.e. so $3t_2=-\frac{1}{3t_1}$.  

*$3t_1$ for that specific $t_1$ will then be the slope of a line which is both tangent and normal to the curve.
A: I believe the problem you're trying to solve is this: given the parametric curve $x = 4t^2+1$, $y = 8t^3 - 1$, find a line $L$ that meets the curve at two points $P$ and $Q$ and is tangent to the curve at $P$ and normal to the curve at $Q$. 
Here's how I'd solve it: as you have observed, at parameter value $t$, the slope is $3t$. You want to find two parameter values $s$ and $t$ with these properties:


*

*The slope at $P = (x(t), y(t))$ is the negative reciprocal of the slope at $Q = (x(s), y(s))$.

*The slope at $P$ is the same as the slope of the line from $P$ to $Q$.
The first of these tells you that 
$$
3t = \frac{-1}{3s}
$$
so $t = -1/(9s)$ or $s = -1/(9t)$. Let's use the second form. We now know that
\begin{align}
P &= (4t^2 + 1, 8t^3 - 1)\\
S &= (4s^2 + 1, 8s^3 - 1)\\
&= (\frac{4}{81t^2} + 1, \frac{-8}{729t^3} - 1)
\end{align}
The second condition above says that $3t$ is the same as the slope of the line from $P$ to $Q$, which says
\begin{align}
3t &= \frac{(\frac{-8}{729t^3} - 1) - (8t^3 - 1)}{\frac{4}{81t^2} + 1- (4t^2 + 1)}\\
&= \frac{(\frac{-8}{729} - t^3) - (8t^6 - t^3)}{\frac{4t}{81} + t^3- 4t^5 - t^3}\\
\end{align}
That's now a hideous mess that you can solve for $t$, but I can't bring myself to do so. 
Once you find $t$, you can find $s$ from it, and then use these to compute the points $P$ and $Q$.
By the way, playing with this on paper, it looks as if I probably made a mistake in the algebra somewhere, because the resulting equation ends up having three double roots, which doesn't make intuitive sense to me. But perhaps my algebra on paper (or my intuition) is wrong, too.   
