Question

The parametric equations of a curve are x = 4t and y = 4 − t2.
Find the equations of the normals to the curve at the points where
the curve meets the x-axis. Hence, find the point of intersection of these normals.

Answer 1

I bet you, that second equation is y = 4-t^2

form the first: x^2 = 16t^2 --> t^2 = x^2/16
from the 2nd: t^2 = 4 - y

so x^2/16 = 4-y
x^2 = 64 - 16y , which is a parabola opening downwards.

derivative: 2x = -16 dy/dx
dy/dx = -x/8

x-intercepts of the parabola, let y = 0
x = ±8
so we have intercepts at (8,0) and (-8,0)

at (8,0), slope of tangent = -8/8 = -1
so the normal has a slope of +1
equation of normal: y = 1(x-8)

at (-8,0), slope of tangent = 1
so slope of normal = -1
equation of normal : y = -(x+8)

solving these two normals:
x-8 = -x-8
2x = 0
x = 0 , obvious from the symmetry
then y = -8

the normals intersect at (0,-8)