Thread: Find all points where the tangent line is horizontal.

Find all points where the tangent line is horizontal.
The equation is x^(2) + xy + y^(2) = 1

Okay so from a previous thread I learned that the horizontal tangent line is when a single point touches the equation and in this case the cirlce/oval.

I took the derivative and i got:
2x + xy' + y + 2y^(2) y' = 0

I'm looking at the graph of this and I think there are only two horizontal tangents but I dont know how to compute them. Please show me how. Thanks.

2x + xy' + y + 2y^(2) y' = 0

y'(x+2y^2)= -(2x+y)

y'= -(2x+y)/(x+2y^2)

so where y'=0 there's a horizontal tangent so:

-(2x+y)=0
y+2x=0 divided by -1 but 0/-1 =0 so


where y=-2x there will be a tangent line i think you can sub this back into your original so

x^(2) + x*(-2x)+ (-2x)^(2) = 1

so x^2-2x^2+4x^2=1

so 3x^2=1

x= (1/3)^(1/2)


i did this on the computer while typing it and my maths a bit rusty but there


***********mistake it should be x= + or - (1/3)^(1/2)

Originally Posted by moonman

Find all points where the tangent line is horizontal.
The equation is x^(2) + xy + y^(2) = 1

Okay so from a previous thread I learned that the horizontal tangent line is when a single point touches the equation and in this case the cirlce/oval.

I took the derivative and i got:
2x + xy' + y + 2y^(2) y' = 0 Mr F says: Wrong. It's 2x + xy' + y + 2y y' = 0. Now solve this equation for y'.

I'm looking at the graph of this and I think there are only two horizontal tangents but I dont know how to compute them. Please show me how. Thanks.

.

.

So you have the following equations:

.... (1)

.... (2)

Solve equations (1) and (2) simultaneously: . Therefore .

Okay i didn't bother checking your other math, nice catch Mr. Fantastic,

my works is still fine, just change my denominator even though I get the same answer

Oh yeah forgot to find the way but as mr. fantastic said "..." -you can do that yourself