The normal line intersects a curve at two points. What is the other point? The line that is normal to the curve $\displaystyle x^2 + xy - 2y^2 = 0 $ at $\displaystyle (4,4)$ intersects the curve at what other point?
I can not find an example of how to do this equation. Can someone help me out?

Given the curve $\mathcal{C}$ as

$$
x^2 + x y - 2 y^2 = 0. \tag 1
$$

Solve the curve $\mathcal{C}$
$$
x^2 + x y - 2 y^2 = 0.\\
\Downarrow\\
4 x^2 + 4 x y - 8 y^2 = 0.\\
\Downarrow\\
\Big( 2 x + y \Big)^2 - 9 y^2 = 0.\\
\Downarrow\\
\Big( 2 x + y \Big)^2 = \Big( 9 y \Big)^2.\\
\Downarrow\\
2 x + y = \pm 3 y.\\
\Downarrow\\
2 x = \Big( \pm 3 - 1 \Big) y.\\
\Downarrow\\
y = \frac{2}{ \pm 3 - 1 } x.\\
\Downarrow\\
y = \frac{2}{2} x \vee y = \frac{2}{-4} x.\\
$$
So the curve $\mathcal{C}$ are two crossing lines,
given by

$$
y = x \vee y = - \frac{1}{2} x. \tag{2}
$$

The normal through point $(4,4)$
The normal through point $(4,4)$ is given by

$$
y = 8 - x. \tag 3
$$

Intersection
We need to find the intersection between
$$
y = - \frac{1}{2} x
$$
and
$$
y = 8 - x.
$$
So we get

$$
y = - \frac{1}{2} x = 8 - x.\tag 4
$$

Solving intersection
As
$$
- \frac{1}{2} x = 8 - x,
$$
we get
$$
\frac{1}{2} x = 8,
$$
so

$$
x = 16.
$$

And we have
$$
y = - \frac{1}{2} x,
$$
so

$$
y = - 8.
$$

Solution
The solution is given by

$$
\bbox[16px,border:2px solid #800000] { (x,y) = (16,-8). } \tag 5
$$

