1. ## Find vertical tangent

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

2. Differentiate implicitly:

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

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.

3. 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?
y

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

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