How do you find horizontal and vertical tangent lines after using implicit differentiation of #x^2+xy+y^2=27#?

Question

Isaiah Allen · Accepted Answer

#y = pm 6#
#x = pm6# Given #f(x,y)=x^2+xy+y^2-27=0# #df=f_x dx + f_y dy = 0# so #dy/dx = - f_x/(f_y) = (2x+y)/(2y+x)# The horizontal tangent lines have #f_x = 0->x = -y/2# and the vertical tangent lines have #f_y = 0->x = -2y# So for horizontals #f(-y/2,y) = y^2/4-2y^2+y^2-27=0->y=pm6# and for verticals #f(x,-x/2) = x^2-x^2/2+x^2/4 - 27=0->x=pm 6#

Paisley Johnson · Answer

undefined To find the horizontal and vertical tangent lines after using implicit differentiation of the equation x^2+xy+y^2=27, we need to solve for dy/dx and set it equal to zero for horizontal tangents, and find the values of x and y that make dy/dx undefined for vertical tangents.

First, we differentiate the equation implicitly with respect to x:
2x + (x(dy/dx) + y) + 2y(dy/dx) = 0

Next, we isolate dy/dx by moving the terms involving dy/dx to one side:
x(dy/dx) + y(dy/dx) + 2y - 2x = 0

Factoring out dy/dx:
(dy/dx)(x + y) = 2x - 2y

Finally, we solve for dy/dx:
dy/dx = (2x - 2y) / (x + y)

To find horizontal tangents, we set dy/dx equal to zero:
(2x - 2y) / (x + y) = 0

Simplifying the equation:
2x - 2y = 0

This implies that 2x = 2y, or x = y. Therefore, the equation of the horizontal tangent line is y = x.

To find vertical tangents, we need to find the values of x and y that make dy/dx undefined. This occurs when the denominator of dy/dx is equal to zero:
x + y = 0

This implies that y = -x. Therefore, the equation of the vertical tangent line is y = -x.