The equation x^2+xy+y^2=9 has two horizontal tangent lines. I implicitly differentiated it to come up with (dy/dx)=(-2x-y)/(x+2y). I'm not quite sure how to proceed to find the two horizontal and two vertical tangents, though. Since it's a fraction, when I set the derivative to zero, I end up with y=-2x, which seems to be bisecting it, rather than tangent. Any direction?

Question

The equation x^2+xy+y^2=9 has two horizontal tangent lines.  I implicitly differentiated it to come up with (dy/dx)=(-2x-y)/(x+2y).  I'm not quite sure how to proceed to find the two horizontal and two vertical tangents, though.  Since it's a fraction, when I set the derivative to zero, I end up with y=-2x, which seems to be bisecting it, rather than tangent.  Any direction?

shubhamsrg · Answer

2x = -y is the condition fot horizontal tangency..
put y = -2x in your original eqn.

x^2 - 2x^2 + 4x^2 =9
x^2 = 3
x= sqrt(3), -sqrt(3)
y=-sqrt(3)/2, sqrt(3)/2

these are the co-ordinates from which your tangent passes,,
let eqn of the tangent be
y+sqrt3/2 = m(x-sqrt3)

you here know that m=0

thus eqn of horizontal tangent will be y=-sqrt3 /2 and y = sqrt3 /2

shubhamsrg · Answer

hope that'd help..

shubhamsrg · Answer

you may do the same for vertical tangents/./

anonymous · Answer

Brilliant.  That's exactly what I needed.

anonymous · Answer

Well, I thought I followed your explanation pretty well, but that's not lining up as tangent to the ellipse.  I'm not sure where I've gone wrong, but it looks like it ought to be somewhere around 3.5.

shubhamsrg · Answer

didnt get your query? come again..

anonymous · Answer

I thought I understood what you were doing by plugging -2x back into the original equation, but that's not the right answer.  

y=sqrt(3)/2 is not tangent to x^2+xy+y^2=9

anonymous · Answer

Oh, I see.  It's 2*sqrt(3), not sqrt(3)/2.  Thanks again.

shubhamsrg · Answer

ohh yes,,my mistake!! sorry! :P