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 x2=2xy - 3y2 = 0
Are there supposed to be 2 equal signs in this expression or is it
x2 + 2xy - 3y2 = 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
x2 + 2xy = 3y2
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)

This might help: You already know how to find the equation for the normal line is y=-x+2. Plug this equation into the original equation of the curve in place of y:

x^2+(2x)(-x+2)-(3)(-x+2)^2=0
x^2-2x^2+4x-3x^2+12x-12=0
-4x^2+16x-12=0
(-4)(x^2-4x+3)=0
x^2-4x+3=0
(x-1)(x-3)=0

Now we know that x=1 and x=3 are the only two places where the graphs can cross. We already know when x=1 then y=1, so we only need to pay attention to x=3. Plug x=3 into the line for the normal curve to determine the y-value at that point:

y=-x+2
y=-3+2
y=-1

So we know the second point is (3, -1). I posted this response so belatedly because I had a tough time figuring this out from the responses I found (here and on other sites), so I hope this helps other students who are struggling with the concept.

Awesome!

omg a question asked in 2006!!!

Apologies for the confusion. In that case, your function factors as (2x-y)(3y+x)=0. Therefore, the curve is a pair of lines: y=2x and y=-x/3.

Since the line normal to the curve passes through the point (1,1) and intersects the curve at another point, we need to find the equation of the normal line and see where it intersects the lines of the curve.

The normal line will have a slope that is the negative reciprocal of the derivative at the point (1,1).

Let's differentiate the curve implicitly:
2x - y + 2y(dy/dx) + 3y + x(dy/dx) = 0

Solving for dy/dx:
dy/dx = (y - 2x) / (2y + x)

At the point (1,1), dy/dx = (1 - 2*1) / (2*1 + 1) = -1/3.

Therefore, the slope of the normal line is 3.

Using the point-slope form of a line (y - y1) = m(x - x1) with the point (1,1) and slope 3:
(y - 1) = 3(x - 1)

Expanding and simplifying:
y = 3x - 2

To find where this line intersects the curve, substitute y = 3x - 2 into the equation of the curve:
x^2 + 2xy - 3y^2 = 0
(x^2 + 6x - 2) - 3(3x - 2)^2 = 0

Solving this equation will give you the other point where the line intersects the curve.

To find the point at which the normal line intersects the curve, we first need to find the equation of the normal line.

The equation of the normal line at any point on the curve is given by the equation:
(y - y1) = m(x - x1)

Where (x1, y1) is the given point on the curve and m is the slope of the normal line.

First, let's find the derivative of the curve equation x^2 + 2xy - 3y^2 = 0 with respect to x:

d/dx (x^2 + 2xy - 3y^2) = d/dx (0)
2x + 2y * (dy/dx) - 6y * (dy/dx) = 0

Now, let's solve for dy/dx:
2x + 2y * (dy/dx) - 6y * (dy/dx) = 0
2x - 4y * (dy/dx) = 0
dy/dx = (2x) / (4y)
dy/dx = x / (2y)

To find the slope of the normal line at the point (1, 1), substitute x = 1 and y = 1 into the equation for dy/dx:
dy/dx = (1) / (2 * 1) = 1/2

Since the normal line has a slope that is the negative reciprocal of the curve's derivative at the point of contact, the slope of the normal line at (1, 1) is -2.

Now we can use the point-slope form of the line equation to find the equation of the normal line:
(y - 1) = -2(x - 1)
y - 1 = -2x + 2
y = -2x + 3

To find the point of intersection, we need to solve the system of equations:
x^2 + 2xy - 3y^2 = 0
y = -2x + 3

Substitute the equation for y from the line into the curve equation:
x^2 + 2x(-2x + 3) - 3(-2x + 3)^2 = 0
Simplify and solve for x.

After solving for x, substitute the value of x back into the equation for the line to find the corresponding value of y.

These calculations will give the coordinates of the other point where the normal line intersects the curve.