The graph of the equation is x^2+xy+y^2=9
a) What is the equation of the right most vertical tangent?
b) That tangent touches the ellipse where y= what?
I've calculated the derivative to y'=(-y-2x)/(2y+x) and I found the horizontal tangents. How do I do this part?
x^2 + xy + y^2 = 9
2x + x dy/dx + y + 2y dy/dx = 0
dy/dx(x + 2y)= -2x - y)
dy/dx = (-2x-y)/(x+2y) <---- you had that, good!
to have a vertical tangent, the denominator of the above has to be zero, that is, dy/dx must be undefined.
x+2y = 0
x = -2y
sub into the original:
(-2y)^2 + (-2y)(y) + y^2 = 9
4y^2 - 2y^2 + y^2 = 9
3y^2 = 9
y^2 = 3
y = ± ?3
when y = ?3, x = -2?3
when y = -?3 , x = 2?3
so it touches at the right most point of (2?3, -?3) ,appr (3.45, -1.73)
verification by Wolfram:
http://www.wolframalpha.com/input/?i=plot+x%5E2%2Bxy%2By%5E2%3D9
horizontal tangents occur when dy/dx = 0
vertical tangents occur when dx/dy = 0.
dx/dy = -(x+2y)/(2x+y), =0 when x = -2y
and you get Reiny's answer.
To find the equation of the rightmost vertical tangent, we need to find the point at which it touches the graph of the equation x^2 + xy + y^2 = 9.
a) To find the equation of the rightmost vertical tangent, we need to find the x-coordinate of the point of tangency. This can be done by setting y' = 0 and solving for x.
From the equation you provided, y' = (-y - 2x) / (2y + x).
Setting y' to 0, we have:
0 = (-y - 2x) / (2y + x)
Multiply both sides by (2y + x):
0 = -y - 2x
Now, for the rightmost vertical tangent, the slope of the tangent line should be approaching infinity, which means the denominator (2y + x) should be equal to zero.
Therefore, we have:
2y + x = 0
Solving this equation for y, we get:
y = -x/2
So, the equation of the rightmost vertical tangent is y = -x/2.
b) To find the value of y where the rightmost vertical tangent touches the ellipse, we substitute the value of x from part (a) into the equation x^2 + xy + y^2 = 9.
Replacing x with -x/2, we have:
(-x/2)^2 + (-x/2)y + y^2 = 9
x^2/4 - xy/2 + y^2 = 9
(x^2 - 2xy + 4y^2)/4 = 9
Multiplying both sides by 4 to eliminate the fraction, we have:
x^2 - 2xy + 4y^2 = 36
Now, substitute the value of x from part (a):
(-x/2)^2 - 2(-x/2)y + 4y^2 = 36
(x^2/4) + xy - 2y^2 = 36
x^2 + 4xy - 8y^2 = 144
Since we know that x = -2y (from part a), we can substitute -2y for x:
(-2y)^2 + 4(-2y)y - 8y^2 = 144
4y^2 - 8y^2 - 8y^2 = 144
-12y^2 = 144
Dividing both sides by -12, we get:
y^2 = -12
However, this equation has no real solutions, which means the rightmost vertical tangent does not touch the ellipse.
Therefore, there is no value of y where the rightmost vertical tangent touches the ellipse.
To find the equation of the rightmost vertical tangent, you need to determine the x-coordinate of the point at which the tangent line intersects the graph. Here's how you can do it:
Step 1: Rewrite the given equation in terms of y:
x^2 + xy + y^2 = 9
y^2 + xy + x^2 - 9 = 0
Step 2: Differentiate implicitly with respect to x:
Taking the derivative of both sides of the equation, you have to apply the chain rule for terms involving y:
2y * dy/dx + (x + y) + 2x = 0
Simplifying this expression, you get:
2y * dy/dx = -x - y - 2x
2y * dy/dx = -3x - y
Step 3: Solve for dy/dx:
Divide both sides by 2y:
dy/dx = (-3x - y)/(2y)
Step 4: Find the equation of the rightmost vertical tangent:
For a vertical tangent, dy/dx approaches infinity. This occurs when the denominator (2y) equals zero.
Set 2y = 0 and solve for y:
2y = 0
y = 0
Therefore, the vertical tangent touches the graph when y = 0.
Hence, the answers to your questions are:
a) The equation of the rightmost vertical tangent is x = its x-coordinate.
b) The tangent touches the ellipse where y = 0.