The graph of the equation is x^2+xy+y^2=9

Question

a) What is the equation of the right most vertical tangent?

b) That tangent touches the ellipse where y= what?

I've calculated the derivative to y'=(-y-2x)/(2y+x) and I found the horizontal tangents. How do I do this part?

Answer 1

x^2 + xy + y^2 = 9
2x + x dy/dx + y + 2y dy/dx = 0
dy/dx(x + 2y)= -2x - y)
dy/dx = (-2x-y)/(x+2y) <---- you had that, good!

to have a vertical tangent, the denominator of the above has to be zero, that is, dy/dx must be undefined.

x+2y = 0
x = -2y
sub into the original:
(-2y)^2 + (-2y)(y) + y^2 = 9
4y^2 - 2y^2 + y^2 = 9
3y^2 = 9
y^2 = 3
y = ± ?3
when y = ?3, x = -2?3
when y = -?3 , x = 2?3

so it touches at the right most point of (2?3, -?3) ,appr (3.45, -1.73)

verification by Wolfram:

http://www.wolframalpha.com/input/?i=plot+x%5E2%2Bxy%2By%5E2%3D9

Answer 2

horizontal tangents occur when dy/dx = 0

vertical tangents occur when dx/dy = 0.

dx/dy = -(x+2y)/(2x+y), =0 when x = -2y

and you get Reiny's answer.