
New Member
Find vertical tangent
Find the point(s) on the curve at which the tangent line is(are) vertical given the curve x^{2}+xy+y^{2}=3

Elite Member
Differentiate implicitly:
[tex]y'=\frac{(2x+y)}{x+2y}[/tex]
Now, vertical tangents occur where the slope is undefined. When the denominator is 0.
Set the denominator equal to 0 and solve for y. Sub this back into the original and solve for the x. y coordinates follow.

New Member
So I set the denominator equal to 0.
x+2y= 0
x=2y
y=x/2
so I plug these two back into the original equation x^{2}+xy+y^{2}=3 to find the x and y points for the tangent points right? y

Senior Member
Those are not two separate equations. Putting y= x/2 into [tex]x^2+ xy+ y^2= 3[/tex] gives [tex]x^2 x^2/2+ x^2/4=3x^2/4= 3[/tex]. Solve that for x and then use y= x/2 to find the corresponding values for y. OR put x= 2y into the equation: [tex]4y^2 2y^2+ y^2= 3y^2= 3[/tex]. Solve that for y and then use x= 2y to find the correspondinmg values for x.
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