I found part B, but stuck on part A.
Use implicit differentiation to find the points where the parabola defined by x^2-2xy+y^2+4x-8y+24= 0 has horizontal and vertical tangent lines.
A) The parabola has horizontal tangent lines at the point(s):
B)The parabola has vertical tangent lines at the point(s): (2,6)
The derivative of the function given is (x-y+2)/(x-y+4).
To find the horizontal tangent line you should set (x-y+2)/(x-y+4)= 0 ,but when I set it to 0 the x and y cancel each other out...
I agree with your derivative, and I a agree with your idea of
(x-y+2)/(x-y+4)= 0 for the horizontal tangent.
Now consider where the zero could come from.
It certainly can't come from the denominator, since we can't divide by zero (this will be considered in the next paragraph)
So it must be x-y + 2 = 0
or y = x+2
sub that back into the original x^2-2xy+y^2+4x-8y+24= 0
x^2 - 2x(x+2) + (x+2)^2 + 4x - 8(x+2) + 24= 0
4x + 12 = 0 , check my algebra.
x = -3
y = -1
So you have a horizontal tangent at (-3,-1)
for the vertical tangent, the slope must be undefined.
This can only happen if the denominator is zero.
so set x-y+4 = 0 ---> y = x+4
and proceed like did for the horizontal
x^2-2xy+y^2+4x-8y+24=0
2x-2y-2xy'+2yy'+4-8y'=0
y' = -(2x-2y+4)/(-2x+2y-8) = (x-y+2)/(x-y+4)
so far, so good. For y'=0, we have
y=x+2
x^2-2x(x+2)+(x+2)^2+4x-8(x+2)+24=0
x=3 so, y=5
Looks like you missed a minus sign somewhere there. A peek at the graph at
http://www.wolframalpha.com/input/?i=x%5E2-2xy%2By%5E2%2B4x-8y%2B24%3D+0
confirms this, with the horizontal tangent at (3,5).
I'll let you work out the algebra, but it appears that the vertical tangent is at (2,6)
To find the points where the parabola has horizontal tangent lines, you need to set the derivative equal to zero. However, in this case, the derivative you calculated simplifies to 0/0, which is an indeterminate form. This means we need to use implicit differentiation again to solve for the derivative.
Let's go through the steps of implicit differentiation for the given equation:
1. Start with the equation: x^2 - 2xy + y^2 + 4x - 8y + 24 = 0.
2. Find the derivative of each term with respect to x. The derivative of x^2 is 2x, the derivative of -2xy is -2y - 2xy', the derivative of y^2 is 2yy', the derivative of 4x is 4, the derivative of -8y is -8y', and the derivative of 24 is 0.
3. Combine the derivatives and set the result equal to zero: 2x - 2y - 2xy' + 2yy' + 4 - 8y' = 0.
4. Simplify the equation by combining like terms: -2xy' + 2yy' - 8y' = -2x + 2y - 4.
5. Factor out y' from the left side: y'(-2x + 2y - 8) = -2x + 2y - 4.
6. Divide both sides of the equation by (-2x + 2y - 8) to isolate y': y' = (-2x + 2y - 4)/(-2x + 2y - 8).
7. Simplify the expression for y'.
Now we have the derivative in terms of x and y, but it is still not easy to set it equal to zero and solve for x and y. To simplify further, we can try substituting y with a different variable, like t, to make the equation more manageable. Let's substitute y with t:
y’ = (-2x + 2t - 4)/(-2x + 2t - 8).
Now, to find the horizontal tangent lines, we should set the derivative equal to zero. However, since we have the expression in terms of t, we need to consider the value of t when y' = 0.
Setting y' = 0, we have:
0 = (-2x + 2t - 4)/(-2x + 2t - 8).
To solve for x, we can simply set the numerator of the expression equal to zero:
-2x + 2t - 4 = 0.
Rearranging the equation to isolate x:
-2x = -2t + 4,
x = t - 2.
Thus, the parabola has horizontal tangent lines at the points where x = t - 2. To find the corresponding y-values, we can substitute this value of x back into the original equation x^2-2xy+y^2+4x-8y+24= 0 and solve for y.