If a line is tangent to one point and normal at another point on the curve $$x=4t^2 + 3 , y=8t^3 - 1$$, then possible slopes of such a line is?
what I tried: I got the slope of a tangent at any point on the curve which is $y' $at $t_1 = 3t_1$ and hence slope of normal at the other point ,$y'$ at$ t_2= {1\over-3t_2}$. After this I formed equations of tangent and normal and compared these to get a relation between $t_1$ and $t_2$. But I seem to get nowhere. Is there any better approach to this. Please help.
You have already presumably found the equation of the tangent at the point with parameter value $t$, which is $$y-8t^3+1=3t(x-4t^2-3)$$
Let this line intersect with the curve at a point with parameter value $s$ where $s\neq t$, in which case $s$ satisfies the equation $$(8s^3-1)-8t^3+1=3t((4s^2+3)-4t^2-3)$$
This can be simplified and the resulting cubic in $s$ must have a factor $(s-t)^2$ corresponding to the tangent point. This factor can be removed and we end up with $s=-\frac 12 t$ as the only other solution.
For this tangent to be the same as the normal at point $s$, we require the gradient to be $$-\frac{1}{3s}=3t$$
Solving these two equations for $t$ we end up with $$t=\pm\frac{\sqrt{2}}{3}$$
So the possible gradients of the tangents satisfying these conditions are $$m=\pm\sqrt{2}$$