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

Question

Elijah Abram · Accepted Answer

#(-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]}

Paisley Johnson · Answer

undefined To find the coordinates of the points on the curve where the tangent line is horizontal, we need to determine the values of x and y that satisfy the given condition.

First, we differentiate the equation with respect to x to find the slope of the tangent line:

d/dx (x^2 - xy + y^2) = 2x - y - xy'

Next, we set the derivative equal to zero to find the points where the tangent line is horizontal:

2x - y - xy' = 0

Since we are looking for horizontal tangent lines, the slope (y') should be zero. Therefore, we have:

2x - y = 0

Rearranging this equation, we get:

y = 2x

Substituting this value of y into the original equation, we have:

x^2 - x(2x) + (2x)^2 = 9

Simplifying further:

x^2 - 2x^2 + 4x^2 = 9

3x^2 = 9

x^2 = 3

Taking the square root of both sides, we find:

x = ±√3

Substituting these values of x back into the equation y = 2x, we get:

y = 2(√3) and y = 2(-√3)

Therefore, the coordinates of the points on the curve where the tangent line is horizontal are:

(√3, 2√3) and (-√3, -2√3)