Stuckonproblems
New member
- Joined
- Apr 10, 2012
- Messages
- 7
Find the point(s) on the curve at which the tangent line is(are) vertical given the curve x2+xy+y2=3
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Find vertical tangent
- Thread starter Stuckonproblems
- Start date
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.
\(\displaystyle y'=\frac{-(2x+y)}{x+2y}\)
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
New member
- Joined
- Apr 10, 2012
- Messages
- 7
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?
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
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
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.