Thread: 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.

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 and for which means . Solve for on this interval.

Originally Posted by ThePerfectHacker

. Solve for on this interval.

I must say I’m stupid. Thank you for this line. Problem solved