Use implicit differentiation to find the points where the parabola defined by
x2−2xy+y2−4y+4=0
has horizontal and vertical tangent lines.
The parabola has horizontal tangent lines at the point(s)
(x-y)/x-y+2
.
The parabola has vertical tangent lines at the point(s)
Well, you got that y' = (x-y)/(x-y+2)
If x=y, then y' = 0/2 = 0 so horizontal tangent
If x-y = -2 then y' = -2/0 so vertical tangent
So, where on the parabola is x=y? When
x^2−2x^2+x^2−4x+4=0
x = 1
So, at (1,1) the tangent is horizontal.
similarly, at (0,2) the tangent is vertical
To find the points at which the parabola has horizontal and vertical tangent lines, we can use implicit differentiation.
To begin, let's differentiate the equation of the parabola with respect to x:
d/dx (x^2 - 2xy + y^2 - 4y + 4) = 0
Applying the chain rule, we get:
2x - 2y(dy/dx) - 2y + 2y(dy/dx) - 4(dy/dx) = 0
Simplifying, we find:
2x - 2y(dy/dx) - 2y - 4(dy/dx) = 0
Combining like terms, we have:
2x - 2y - 6(dy/dx) = 0
Now, to find the points where the parabola has horizontal tangent lines, we set dy/dx equal to zero:
2x - 2y - 6(0) = 0
2x - 2y = 0
Solving this equation, we can rearrange it to solve for y in terms of x:
2y = 2x
y = x
Therefore, the parabola has horizontal tangent lines at the points where y = x.
Next, to find the points where the parabola has vertical tangent lines, we need to find where the derivative is undefined, which occurs when the denominator of dy/dx is equal to zero.
From the previous equation, we have:
2x - 2y - 6(dy/dx) = 0
For vertical tangent lines, the slope (dy/dx) would be undefined, so we set the denominator equal to zero:
2x - 2y - 6(0) = 0
2x - 2y = 0
Solving this equation, we can rearrange it to solve for x in terms of y:
2x = 2y
x = y
Therefore, the parabola has vertical tangent lines at the points where x = y.
In summary, the parabola defined by x^2 - 2xy + y^2 - 4y + 4 = 0 has horizontal tangent lines at the points where y = x and vertical tangent lines at the points where x = y.