The line that is normal to the curve x^2 + 2xy -3y^2 = 0 at (6,6) intersects the curve at what other point.

Using the derivatives of this I was able to determine that the normal line is y = -x+12. I am not sure what to do next to figure out the other point of the equation.

Question

The line that is normal to the curve x^2 + 2xy -3y^2 = 0 at (6,6) intersects the curve at what other point.  

Using the derivatives of this I was able to determine that the normal line is y = -x+12.  I am not sure what to do next to figure out the other point of the equation.

ganeshie8 · Answer

you want to find where this normal line intersects the curve again

andu1854 · Answer

yes

ganeshie8 · Answer

one dumb way is to solve the system :

x^2 + 2xy -3y^2 = 0
y = -x+12

ganeshie8 · Answer

plugin y = -x+2 into the first equation
you will get a quadratic in x which you can solve by factoring

andu1854 · Answer

Ok that makes sense and by doing this it will give me two sets of coordinates (6,6) and whatever the other point might be?

ganeshie8 · Answer

or if you're clever ennough you will see that the given curve is just a pair of intersecting lines and canc be factored into form : 
$$(y-m_1x)(y-m_2x)=0$$

ganeshie8 · Answer

you're right!

andu1854 · Answer

Cool, thank you very much, sometimes I just need a little help with being pointed in the right direction...

ganeshie8 · Answer

you're welcome :)

anonymous · Answer

Okay, I'm surprised the straight forward approach wasn't pointed out here...

anonymous · Answer

Differentiate with respect to $x$, then set $(x,y)=(6,6)$ and solve for $y'$.
This gives the tangent slope.  The normal line is perpendicular, so it's slope will be $-1/y'$.

ganeshie8 · Answer

the op already has a normal @wio