Solving a Cartesian and parametric equation at a intersection.

Question

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

Answer 1

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 $t$.

Answer 2

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 $B=(4\cos 2t_2, 3\sin t_2)=\left(\frac{367}{512},\frac{-123}{64}\right)$.