# The line that is normal to the curve x^2=2xy-3y^2=0 at(1,1) intersects the curve at what other point?

Please help. Thanks in advance.

We have x^{2}=2xy - 3y^{2} = 0

Are there supposed to be 2 equal signs in this expression or is it

x^{2} + 2xy - 3y^{2} = 0 ?

I'll suppose it's the second one.

You need to differentiate this implicitly to find dy/dx. Then find an equation for the normal line and set it equal to the curve.

The implicit derivative looks like

2x + 2x*dy/dx + 2y -6y*dy/dx = 0

Solve for dy/dx and use the negative reciprocal of dy/dx at the point (1,1) as the slope of the normal.

I'm not sure if you know what this second degree curve is, but it's an ellipse. The normal line at any point should intersect the ellipse in two points.

It appears your function is

x^{2} + 2xy = 3y^{2}

This is not an ellipse, but a pair of lines that pass through the origin, at least according to what I got in my graphics application.

Differentitate this to get

2x + 2x*dy/dx + 2y = 6y*dy/dx so

2(x+y) = 2*dy/dx(3y -x) and

dy/dx = (x+y)/(3y -x)

At (1,1) dy/dx = 2/2 = 1 so the normal has slope = -1

The equation of the normal is

y-1 = -1(x-1) or

y = -x + 2

Check the points (1,1) and (3,-1)

Sorry it is x^2+2xy-3y^2=0

Thanks for the answers.

How did you get the point (3,-1)? Thanks.

Your function factors as (x-y)(x+3y)=0

These are the lines y=x and y=(-1/3)x

The point (1,1) is on the line y=x with slope 1. The normal has slope -1 and goes through that point. It intersects the other line at (3,-1).

(2x-y)(3y+x)