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 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
$(16, -8)$.
