Finding the distance between the tangent and normal line of a function at y=0

Question

It doesn't seem difficult at first but I think the reason why I can't seem to figure this problem out is because I am used to figuring out tangents and normals at a certain point while in this problem it doesn't state at which point the tangent and normal line intersect.

Here is the assignment: Prove that the length of the segment that is cut off from the x-axis by the normal and tangent line of the following function is 2a (a=const). The function is: $$ x=2a\left(\ln\left(\sin t\right)-\sin ^2 t\right)$$

$$\:y=a \sin 2t $$

Answer 1

Let $(x_0,y_0)$ be any random point on the curve $y=f(x)$ where tangent and normal are drawn.

$X_t=$ point where tangent cuts $X$ axis.

$X_n=$ point where normal cuts $X$ axis.

Check that

$X_t=x_0-{y_0\over y'(x_0)}$ and $X_n=x_0+y_0\cdot y'(x_0)\implies X_n-X_t=y_0\left(y'(x_0)+{1\over y'(x_0)}\right)$

Now $$y'(x)={dy\over dx}={dy\over dt}\cdot{dt\over dx}={2a\cos(2t)\over 2a\left({\cos(t)\over \sin(t)}-2\sin(t)\cos(t)\right)}=\tan(t)$$

$$\therefore X_N-X_t=2a\sin(2t_0)\left(\tan(t_0)+{1\over \tan(t_0)}\right)=2a$$ BINGO!