Show that no point on the graph of X^2-3XY+Y^2=1 has a horizontal tangent line.

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anonymous · Answer

Are you still online? I just got an answer to your question and want to know you're there before I spend several minutes typing it out. :P

anonymous · Answer

I would greatly appreciate it! =D

anonymous · Answer

Use implicit differentiation (or chain rule)$$2x-3(y+x(dy/dx))+2y(dy/dx)=0$$Expand brackets$$2x-3y-3x(dy/dx)+2y(dy/dx)=0$$Factorise and rearrange$$(2y-3x)(dy/dx)=3y-2x$$And finally$$dy/dx=(3y-2x)/(2y-3x)$$We have horizontal tangents when the derivate is equal to zero, so$$(3y-2x)/(2y-3x)=0$$$$3y-2x=0$$As the bottom part doesn't effect whether it's zero or not

(continuing on next post)

anonymous · Answer

Therefore$$3y=2x$$$$y=2x/3$$Substitute this back into the original equation to get$$x^2-3x(2x/3)+(2x/3)^2=1$$$$x^2-2x^2+(4x^2)/9=1$$$$(-5/9)x^2=1$$$$x^2=-9/5$$$$x=\pm \sqrt{-9/5}$$As you may no, there are no real solutions to the square root of a negative value. Therefore there are no real x values for the derivative to be zero. Hence there are no horizontal tangents.

anonymous · Answer

know :P

anonymous · Answer

thanks so much, I was so lost lol