Calculus: Point of Normal Line

Question

Find all point on the graph of x^2/3 + y^2/3 = 8 at which the normal line is parallel to the line y −x = 10.

Answer 1

A normal line is perpendicular to the tangent line at the point of tangency. We want the normal line to be parallel to y = x + 10 which has a slope of 1.

That means we want to find points on the original relation where the slope of the tangent line is -1 (perpendicular to the normal).

To determine the first derivative:

2/3 x^-1/3 + 2/3 y^-1/3y' = 0 (implicit differentiation)

x^-1/3 + y^-1/3y' = 0 (divide every term by 2/3)

y'/y^1/3 = -1/x^1/3

y' = - (y^1/3/x^1/3) = -(y/x)^1/3

When does y' = -1 ?

-(y/x)^1/3 = -1

(y/x)^1/3 = 1

y/x = 1

y = x

So y' only equals -1 when y = x. From the original relation substituting x for y:

x^2/3 + x^2/3 = 8

2x^2/3 = 8

x^2/3 = 4

x = 4^3/2 = 8

Since y = x, y = 8.

The point (8,8) has a tangent line with a slope of -1, so a normal line with a slope of 1.

The equations are:

Tangent:

y - 8 = -1(x - 8)

y = -x + 16

Normal:

y - 8 = 1(x - 8)

y = x

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