# Thread: Normal line to a curve

1. ## Normal line to a curve

x^2 + 2xy - 3y^2 = 0

The line that is normal to this curve at (1,1) intersects the curve at what other point?

I got the slope to be 1. Can someone tell me how to do this? Please?

2. Originally Posted by Morgan82
x^2 + 2xy - 3y^2 = 0

The line that is normal to this curve at (1,1) intersects the curve at what other point?

I got the slope to be 1. Can someone tell me how to do this? Please?
normal slope is -1

$y-1 = -1(x-1)$

$y = -x + 2$

substitute (-x+2) in for y in the original equation ... solve for x, then determine y

3. If I plug it back in, am I supposed to get (-16+- sqrt(192)) / 8?

Im still confused.

4. Originally Posted by Morgan82
If I plug it back in, am I supposed to get (-16+- sqrt(192)) / 8?

Im still confused.
no ... you should get x = two very nice integer solutions.

5. Which equation do I plug it into?

6. Originally Posted by skeeter

substitute (-x+2) in for y in the original equation ...
..

7. oh I got it... a dumb mistake in the solving. Thankyou!