Question

How do you find the equations for the normal line to `x^2+y^2=9` through (0,3)?

Answer 1

`x= 0`

Explanation:

We find the derivative.

`2x + 2y(dy/dx) = 0`

`2y(dy/dx) = -2x`

`dy/dx =(-2x)/(2y)`

`dy/dx = -x/y`

We now determine the slope of the tangent.

`m_"tangent" = -0/3`

`m_"tangent" = 0`

The normal line is perpendicular to the tangent, so the slope of the normal line will be `-1/k` of that of the tangent, where `k` is the slope of the tangent.

`m_"normal" = O/`

Hence, the normal line will be vertical at `(0, 3)` and will be of the form `x = a`.

Since the point of intersection is `x = 0`, the normal line will have equation `x= 0`.

Hopefully this helps!

Answer 2

Here is a solution using geometry.

Explanation:

`x^2+y^2=9` is a circle centered at `(0,0)` with radius `3`.

The point `(0,3)` is on the `y`-axis.

The line segment joining `(0,0)` and `(0,3)` is a radius of the circle and the radius is perpendicular to the tangent.

So the `y` axis is the normal line at that point.

Answer 3

Here is a solution using calculus, but without implicit differentiation.

Explanation:

`x^2+y^2=9`

`y = +-sqrt(9-x^2)`

The point `(0,3)` is a solution to `y = sqrt(9-x^2)` so we are interested in the function `f(x) = sqrt(9-x^2)`.

Differentiate using the chain rule to get

`f'(x) = 1/(2sqrt(9-x^2)) * (-2x) = (-x)/sqrt(9-x^2)`

At `(0,3)`, the tangent line has slope `f'(0) = 0`. So the tangent line is a horizontal line and the normal line is vertical.

The vertical line through `(0,3)` has equation `x=0`.