Solving a Cartesian and parametric equation at a intersection. A curve C has parametric equations:
$x=4cos(2t)$ and $y=3sin(t)$ $-\frac{\pi}{2} < t < \frac{\pi}{2}$
The normal of a point A$(2,1.5)$ on curve C has the equation $6y-16x+23=0$
The curve and the normal intersect again at point B. What is the Y and X values at point B?
I think that at point B, the x and y values of the normal and the curve C are identical and hence I tried to substitute the parametric equations into the normal equation in a attempt to determine T, which would allow me to find the X and Y values. 
However, after subsitiution, I got $18sin(t) -64cos(2t) +23 = 0$
I don't know how to solve this and it feels like i'm not on the right path.
Can somebody clarify as to the best way to approach this problem?
Many thanks
 A: It seems the following.
You proceeded OK, and it rests to solve the equality $18\sin t  - 64\cos 2t +23 = 0$. Do it.  We have $\cos 2t=1- 2\sin^2 t$. So 
$18\sin t  - 64(1- 2\sin^2 t) +23 = 0$
$128\sin^2 t+18\sin t  - 41 = 0$
$\sin t=\frac{-9\pm 73}{128}$
$\sin t_1=\frac{1}{2}$ - this is for the first intersection point
$\sin t_2=\frac{-41}{64}$ 
$\cos 2t_2=1- 2\sin^2 t_2=\frac{367}{2048}$ - this is for second intersection point
Then $B=(4\cos 2t_2, 3\sin t_2)=\left(\frac{367}{512},\frac{-123}{64}\right)$.
A: You are on the right path.  If you solve that trigonometric equation for $t$, that will give you the values (plural) of the parameter for which the normal line intersects the parametric curve.  The Cartesian coordinates then correspond to plugging in that value of $t$ into the parametric equations.
So to get you started, write $$\begin{align*}  0 &= 18 \sin t - 64 \cos 2t + 23 \\ &= 18 \sin t - 64 (1 - 2\sin^2 t) + 23 \\ &= 128 \sin^2 t + 18 \sin t - 41. \end{align*}$$  Then let $u = \sin t$ and solve the resulting quadratic in $u$; then take the inverse sine to recover $t$.
