Find vertical tangent

[Stuckonproblems]

Find the point(s) on the curve at which the tangent line is(are) vertical given the curve x²+xy+y²​=3

 

[galactus]

Differentiate implicitly:

$\displaystyle y'=\frac{-(2x+y)}{x+2y}$

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.

 

[Stuckonproblems]

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 x²+xy+y²=3 to find the x and y points for the tangent points right?

[FONT=MathJax_Math]y[/FONT]​

 

[HallsofIvy]

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