1. ## Differentiation

The curve C has parametric equations

x = 4 cos 2t, y = 3 sin t, -pi/2 < t < pi/2

A is the point (2, 3/2), and lies on C.

a) Show that an equation of the normal to C at A is 6y – 16x + 23 = 0.

The normal at A cuts C again at the point B.
b) Find the y-coordinate of the point B.

I’ve done a, but how can I find b. Please help me.

2. Originally Posted by geton
The curve C has parametric equations

x = 4 cos 2t, y = 3 sin t, -pi/2 < t < pi/2

A is the point (2, 3/2), and lies on C.

a) Show that an equation of the normal to C at A is 6y – 16x + 23 = 0.

The normal at A cuts C again at the point B.
b) Find the y-coordinate of the point B.
To do the second part if the parametrized curve intersects the line then $\displaystyle x=4\cos 2t$ and $\displaystyle y=3\sin t$ for $\displaystyle t\in (-\pi/2,\pi/2)$ which means $\displaystyle 18\sin t - 64 \cos 2t +23 = 0$. Solve for $\displaystyle t$ on this interval.

3. Originally Posted by ThePerfectHacker
$\displaystyle 18\sin t - 64 \cos 2t +23 = 0$. Solve for $\displaystyle t$ on this interval.
I must say I’m stupid. Thank you for this line. Problem solved