Find the point(s) (if any) of horizontal tangent lines for the equation #x^2+xy+y^2=6#. If none exist, why?

Question

Cameron Alexander · Accepted Answer

graph{(x^2+xy+y^2-6)((x-sqrt2)^2+(y+2sqrt2)^2-0.01)((x+sqrt2)^2+(y-2sqrt2)^2-0.01)=0 [-8.89, 8.89, -4.444, 4.445]}

#P_1=(sqrt2, -2sqrt2)#
#P_2=(-sqrt2, +2sqrt2)# Implicitly differentiating over x the equation we get (using y'=#dy/dx#) #2x+y+xy'+2yy'=0# So #2x+y=-(x+2y)y'# #y'=-(2x+y)/(x+2y)# So #y'=0# only when #y=-2x# (excluding #y=0# when #y'=-2#) Now intersect with the equation #{(y=-2x),(x^2+xy+y^2=6):}# #{(y=-2x),(x^2-2x^2+4x^2=6):}# #{(y=-2x),(3x^2=6):}# #{(y=-2x),(x=+-sqrt2):}#

Charles Anctil · Answer

undefined To find the point(s) of horizontal tangent lines for the equation x^2+xy+y^2=6, we need to find the values of x and y where the derivative of the equation with respect to y equals zero.

Taking the derivative of the equation with respect to y, we get: 2x + x(dy/dy) + 2y(dy/dy) = 0

Simplifying, we have: 2x + x + 2y(dy/dy) = 0

Combining like terms, we get: 3x + 2y(dy/dy) = 0

Since we are looking for horizontal tangent lines, the derivative dy/dy is zero. Therefore, the equation becomes: 3x = 0

Solving for x, we find that x = 0.

Substituting x = 0 back into the original equation, we get: 0^2 + 0y + y^2 = 6

Simplifying, we have: y^2 = 6

Taking the square root of both sides, we find that y = ±√6.

Therefore, the points of horizontal tangent lines for the equation x^2+xy+y^2=6 are (0, √6) and (0, -√6).