The equation of the normal when the gradient of the tangent = 0

Question

Q. Find the equation of the tangent and normal to $x^2-xy+y^2 = 3$ at $(1,2)$.

I have done the first part. I found:

$$\displaystyle\frac{dy}{dx} = \displaystyle\frac{y-2x}{2y+x}$$

and substituted $x = 1$ and $y = 2$.

I found the gradient to be 0.

I found the equation of the tangent line to be:

$m=0, x_1 = 1, y_1 = 2$

$$y-y_1 = m(x-x_1)$$

$$y-2 = 0(x-1)$$

$$y = 2$$

I now want to find the equation of the normal. I put down that it is undefined. My professor said the normal itself isn't undefined, it's just a line with an undefined gradient.

So he still wants the equation of the normal.

As it is going to be a vertical line off a horizontal line, is the answer that the equation of the normal is $x = 1$?

Answer 1

The slope of the tangent line at $(1,2)$ being $0$ implies it is the horizontal line $y=2$.

The normal line is perpendicular to the tangent line at $(1,2)$, hence it is the vertical line $x=1$.

Refer to the graph:

$\hspace{1cm}$

Answer 2

The slope of normal at that point is $\frac{-1}{dy/dx}$ clearly since slope of tangent here is zero therefore slope of normal at that point must be undefined from above formula, this implies a vertical line is the normal i.e. x=a but since this has to pass the above point, x=1 is your normal at that point.