Find the unit normal to an ellipse given by an equation

Question

The equation of the ellipse is given as being: $$x^2 -xy + y^2 = 7$$We're instructed to find a unit normal to the curve at a general point $P(x,y)$, and also at point $(-1,2)$ in particular.

My intuition is that I need to parameterize this equation, but I have no idea how to go about doing that. I thought that once I had the parametric equations, I could take the derivative of that set of equations with respect to $t$, and get a general equation for the slope at any point. With that I could probably use the dot product to find the perpendicular normal.

That's my reasoning, but of course, I'm really unsure if any of that is correct.

Answer 1

We have $2x-y-xy'+2yy'=0$ so $y' =\frac {y-2x} {2y-x}$. At any point $(x,y)$ on the ellipse the slope of the normal line is $\frac {2y-x} {2x-y}$. At $(-1,2)$ it is $-5/4$. (A unit vector along the normal is $(\frac 1 {\sqrt {1+m^{2}}},\frac m {\sqrt {1+m^{2}}})$ where $m$ is the slope).

Answer 2

The general parametrization of an axis aligned ellipse is $$\pmatrix{x = a \cos(t) \\ y = b \sin(t) }$$

But this is oblique ellipse, which can be made parametric, just as above, but with a rotation.

$$\pmatrix{x = a \cos(t) \cos(\theta) - b \sin(t) \sin(\theta) \\ y = a \cos(t) \sin(\theta) + b \sin(t) \cos(\theta) }$$

Plug into the ellipse equation, and force all coefficients of $\sin(t)$ or $\cos(t)$ to be zero. With some effort, you will get to

$$ \pmatrix{ x = \sqrt{7} \cos(t) - \frac{\sqrt{7}}{\sqrt{3}} \sin(t) \\ y = \sqrt{7} \cos(t) + \frac{\sqrt{7}}{\sqrt{3}} \sin(t) }$$

which is a result of $\theta = \frac{\pi}{4}$, $a=\sqrt{2}\sqrt{7}$ and $b=\sqrt{2}\sqrt{7}/\sqrt{3}$.

Now apply the tricks of differential geometry to find the tangent, normal, curvature etc.

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Another approach is to construct the implicit derivative of $f(x,y) = x^2 - x y + y^2 -7$

$$ {\rm d}f = \frac{\partial f}{\partial x} {\rm d}x + \frac{\partial f}{\partial y}{\rm d}y $$

$$ {\rm d}f = (2x-y) {\rm d}x + (2y-x){\rm d}y = 0$$

Note that the tangent and the normal are

$$\begin{aligned} \vec{e} & = \frac{1}{\sqrt{{\rm d}x^2 + {\rm d}y^2}} \pmatrix{{\rm d}x \\ {\rm d}y} & \vec{n} & = \frac{1}{\sqrt{{\rm d}x^2 + {\rm d}y^2}} \pmatrix{-{\rm d}y \\ {\rm d}x} \end{aligned}$$

and from the implicit derivate, we can arbitrarily make ${\rm d}x = \frac{x-2 y}{2x -y} {\rm d}y$ which is used above to evaluate

$$\begin{aligned} \vec{e} & = \pmatrix{\frac{x-2y}{\sqrt{5x^2-8xy+5y^2}} \\ \frac{2x-y}{\sqrt{5x^2-8xy+5y^2}}} & \vec{n} & = \pmatrix{ \frac{y-2x}{\sqrt{5x^2-8xy+5y^2}} \\ \frac{x-2y}{\sqrt{5x^2-8xy+5y^2}} } \end{aligned}$$

Now use different values of $(x,y)$ find a solution.