How do you find the coordinates of the points on the curve #x^2-xy+y^2=9# where the tangent line is horizontal?

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Answer 1 · 2017-01-10T23:56:52

#(-sqrt(3),-2sqrt(3))# and #(sqrt(3),2sqrt(3))#

# x^2 - xy +y^2 = 9 #

Differentiating Implicitly (and applying the Product rule) gives:

# 2x - { (x)(dy/dx) + (1)(y) } + 2ydy/dx = 0 #
# :. 2x - xdy/dx - y + 2ydy/dx = 0 #

The tangent is horizontal at a turning point ( ie #dy/dx = 0#)

# => 2x - 0 - y + 0 = 0 #
# :. y=2x #

Subs #y=2x# into original equation:

# => x^2 - x(2x) +(2x)^2 = 9 #
# :. x^2 - 2x^2+4x^2=9 #
# :. 3x^2 =9 #
# :. x^2 =3 #
# :. x =+-sqrt(3) #

Hence the coordinates are #(-sqrt(3),-2sqrt(3))# and #(sqrt(3),2sqrt(3))#

graph{x^2 - xy +y^2 = 9 [-10, 10, -5, 5]}