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?
 A: 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

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

A: $$x^2 + xy - 2y^2 = 0$$
$$2x+(y+xy^\prime)-4yy^\prime=0$$
$$2x+y=4yy^\prime-xy^\prime$$
$$2x+y=y^\prime(4y-x)$$
$$\dfrac{2x+y}{4y-x}=y^\prime$$
There's the expression differentiated. I'm assuming that's what you had a problem with. The next step would be evaluating this expression a (4,4). The find the gradient of the normal line ($m_{normal}\cdot m=-1$).
Once you have the gradient of the normal line (denoted 'm'), solve for c in y=mx+c at point (4,4). This will yield an equation of a line and the equation this equation of  line to the original expression and solve for x and y.
I leave the details of the solution for you to work out. This is a strategy as to how to go about solving it.
Good luck!
A: A key to
easily solving this
is lab bhattacharjee's observation that
$x^2+xy-2y^2=(x+2y)(x-y)
$.
Therefore,
the curve is the union
of the lines
$L_1: x+2y = 0$
and
$L_2: x-y = 0$.
Since $(4, 4)$ is on
$L_2$, and not $L_1$,
and the slope of $L_2$
is $1$,
the slope of the normal is $-1$.
The equation of the line through
$(4, 4)$
and slope $-1$ is
$\frac{y-4}{x-4} = -1
$
or
$y-4 = 4-x$
or
$y=-x+ 8$.
This intersects $L_1$
when
$0 
=x+2y
= x+2(-x+8)
= -x+16
$
or
$x = 16$.
Then
$y = -x+8
= -8
$,
so the point is
$(16, -8)$.
