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