Sure, let's use implicit differentiation to find the points where the parabola has horizontal and vertical tangent lines.
To find horizontal tangent lines, we differentiate the given equation with respect to x. Let's differentiate term by term:
d/dx (x^{2}-2xy+y^{2}-6x+2y+17) = 0
2x - 2y(dy/dx) + 2y(dy/dx) - 6 + 2(dy/dx) = 0
Simplifying this equation, we get:
2x - 6 + 2(dy/dx) = 0
Now, to find the points where the parabola has horizontal tangent lines, we are looking for the values of x that make the derivative (dy/dx) equal to zero. So, let's solve for (dy/dx):
2(dy/dx) = 6 - 2x
dy/dx = (6 - 2x) / 2
Simplifying this further, we find:
dy/dx = 3 - x
Now, to find the points where the parabola has horizontal tangent lines, we set dy/dx = 0:
3 - x = 0
x = 3
Substituting this value of x back into the original equation, we can solve for y:
(3)^2 - 2(3)y + y^2 - 6(3) + 2y + 17 = 0
9 - 6y + y^2 - 18 + 2y + 17 = 0
y^2 - 4y + 8 = 0
Now we can solve this quadratic equation to find the values of y:
Using the quadratic formula, we have:
y = (-(-4) ± √((-4)^2 - 4(1)(8))) / (2(1))
y = (4 ± √(16 - 32)) / 2
y = (4 ± √(-16)) / 2
As we cannot take the square root of a negative number without involving complex numbers, it shows that the parabola does not have any real points where it has horizontal tangent lines.
Now, let's find the points where the parabola has vertical tangent lines. To do this, we differentiate the equation with respect to y. Let's differentiate term by term:
d/dy (x^{2}-2xy+y^{2}-6x+2y+17) = 0
-2x + 2y(dx/dy) + 2xdx/dy + 2 + 2(dx/dy) = 0
Simplifying this equation, we get:
-2x + 4y(dx/dy) + 2(dx/dy) + 2 = 0
Now, to find the points where the parabola has vertical tangent lines, we are looking for the values of y that make the derivative (dx/dy) equal to zero. So, let's solve for (dx/dy):
4y(dx/dy) + 2(dx/dy) = -2 + 2x
(4y + 2)(dx/dy) = -2 + 2x
dx/dy = (-2 + 2x) / (4y + 2)
Simplifying this further, we find:
dx/dy = (2x - 2) / (4y + 2)
Now, to find the points where the parabola has vertical tangent lines, we set dx/dy = 0:
(2x - 2) / (4y + 2) = 0
2x - 2 = 0
2x = 2
x = 1
Substituting this value of x back into the original equation, we can solve for y:
(1)^2 - 2(1)y + y^2 - 6(1) + 2y + 17 = 0
1 - 2y + y^2 - 6 + 2y + 17 = 0
y^2 = -12
As we cannot take the square root of a negative number without involving complex numbers, it shows that the parabola does not have any real points where it has vertical tangent lines.
In conclusion, the parabola defined by the equation x^{2}-2xy+y^{2}-6x+2y+17 = 0 does not have any real points where it has either horizontal or vertical tangent lines. Well, isn't that parabola being quite picky? It doesn't want to have any special friends!