Use implicit differentiation to find the points where the parabola defined by
x^{2}-2xy+y^{2}-6x+2y+17 = 0
has horizontal and vertical tangent lines. List your answers as points in the form (a,b).
The "solving" becomes a bit easier if you change the original equation to
(x-y)^2 - 6x + 2y + 17 = 0
sub in : y = x - 3
(x - (x-3))^2 - 6x + 2(x-3) + 17 = 0
9 - 6x + 2x - 6 + 17 = 0
-4x = -20
x = 5
then y = 2
the horizontal asymptote touches at (5,2)
I will do the vertical asymptote without the simplification:
x^2 - 2x(x-2) + (x-2)^2 - 6x + 2(x-2) + 17 = 0
x^2 - 2x^2 + 4x + x^2 - 4x + 4 - 6x + 2x-4 + 17 = 0
-4x = -17
x = 17/4
then y = 17/4 - 2 = 9/4
the vertical asymptote touches at (17/4 , 9/4)
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!
To find the points where the parabola has horizontal and vertical tangent lines, we need to take the derivative of the given equation with respect to x and y.
First, let's take the derivative with respect to x. When differentiating terms that involve both x and y, we will treat y as a constant.
Differentiating the given equation with respect to x:
d/dx (x^2 - 2xy + y^2 - 6x + 2y + 17) = d/dx (0)
Using the power rule, the derivatives of the terms are:
2x - 2y(dy/dx) - 6 + 0 + 0 = 0
Simplifying the equation:
2x - 2y(dy/dx) - 6 = 0
Next, let's take the derivative with respect to y. When differentiating terms that involve both x and y, we will treat x as a constant.
Differentiating the given equation with respect to y:
d/dy (x^2 - 2xy + y^2 - 6x + 2y + 17) = d/dy (0)
Using the power rule, the derivatives of the terms are:
0 + x(dy/dy) - 2x + 2 + 0 + 0 = 0
Simplifying the equation:
x - 2x + 2 = 0
Combining the derived equations:
2x - 2y(dy/dx) - 6 = x - 2x + 2
Simplifying:
x - 2y(dy/dx) - 6 = -x + 2
Combining like terms:
3x - 2y(dy/dx) = 8
Now, this equation represents the relationship between x, y, and dy/dx. To find the points where the parabola has horizontal tangent lines, we set dy/dx = 0.
Setting dy/dx = 0:
3x - 2y(0) = 8
Simplifying:
3x = 8
x = 8/3 = 2.67
Substituting the value of x into the original equation:
(2.67)^2 - 2(2.67)y + y^2 - 6(2.67) + 2y + 17 = 0
7.1289 - 5.34y + y^2 - 16.02 + 2y + 17 = 0
Rearranging terms:
y^2 - 3.34y + 8.8929 = 0
Using the quadratic formula to solve for y:
y = (-(-3.34) ± sqrt((-3.34)^2 - 4(1)(8.8929))) / (2(1))
y = (3.34 ± sqrt(11.1556 - 35.5716)) / 2
y = (3.34 ± sqrt(-24.416)) / 2
Since the square root of a negative number is not a real number, there are no real solutions for y.
Therefore, there are no points where the parabola has horizontal tangent lines.
To find the points where the parabola has vertical tangent lines, we set dx/dy = 0.
Setting dx/dy = 0:
2x - 2y = 0
Rearranging terms:
2x = 2y
x = y
Substituting the value of y into the original equation:
x^2 - 2x(x) + (x)^2 - 6x + 2(x) + 17 = 0
Rearranging terms:
2x^2 - 4x^2 - 6x + 2x + 17 = 0
-2x^2 - 4x + 17 = 0
Using the quadratic formula to solve for x:
x = (-(-4) ± sqrt((-4)^2 - 4(-2)(17))) / (2(-2))
x = (4 ± sqrt(16 + 136)) / -4
x = (4 ± sqrt(152)) / -4
x = (4 ± 2sqrt(38)) / -4
Simplifying:
x = -1 ± 0.53
x ≈ -1.53, -0.47
Therefore, the points where the parabola has vertical tangent lines are (-1.53, -1.53) and (-0.47, -0.47).
To find the points where the parabola has horizontal and vertical tangent lines, we can use implicit differentiation. Implicit differentiation allows us to find the derivative of y with respect to x even when y is not explicitly defined as a function of x.
Let's start by differentiating both sides of the equation:
d/dx (x^2 - 2xy + y^2 - 6x + 2y + 17) = d/dx (0)
To differentiate the left side, we will use the chain rule and differentiate each term separately.
d/dx (x^2) - d/dx (2xy) + d/dx (y^2) - d/dx (6x) + d/dx (2y) + d/dx (17) = 0
Differentiating each term:
2x - 2y * (dy/dx) + 2y * (dy/dx) - 6 + 2 * (dy/dx) + 0 = 0
Now, let's simplify the equation:
2x - 2y * (dy/dx) + 2y * (dy/dx) + 2 * (dy/dx) - 6 = 0
Combining like terms:
2x + 2 * (dy/dx) - 6 = 0
To find the points where the parabola has horizontal tangent lines, we need to find when the derivative dy/dx equals 0. Since we are looking for horizontal tangents, we want the slope of the tangent line to be zero.
Setting dy/dx = 0:
2x + 2 * (dy/dx) - 6 = 0
2x + 2 * 0 - 6 = 0
2x - 6 = 0
2x = 6
x = 3
Now that we have found the x-coordinate where the slope is zero, we can substitute this value back into the original equation to find the y-coordinate. Using the original equation, we have:
x^2 - 2xy + y^2 - 6x + 2y + 17 = 0
(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
To solve this quadratic equation, we can use factoring or the quadratic formula. However, upon examination, we see that this quadratic does not have real solutions. Therefore, there are no points on the parabola where it has a horizontal tangent line.
To find the points where the parabola has vertical tangent lines, we need to find when the derivative dy/dx is undefined. Since we are looking for vertical tangents, the slope of the tangent line will be undefined.
To find when dy/dx is undefined, we need to determine when the denominator of the derivative becomes zero. Recall the derivative:
2x + 2 * (dy/dx) - 6 = 0
Setting the denominator equal to zero:
2 + 2 * (dy/dx) = 0
2 * (dy/dx) = -2
(dy/dx) = -1
Now, we can plug this value into the original equation:
x^2 - 2xy + y^2 - 6x + 2y + 17 = 0
(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
Again, we have the same quadratic equation. Since it does not have real solutions, there are no points on the parabola where it has a vertical tangent line.
In conclusion, the parabola defined by x^2 - 2xy + y^2 - 6x + 2y + 17 = 0 does not have any points with either horizontal or vertical tangent lines.
2xdx-2y dx-2xdy + 2y dy -6dx+2dy=0
for horizontal lines, dy/dx=0
2x-2y-6)/(-2x+2y+2)=0
or y=x-3
for vertical tangent, dy/dx=undifined
or -2x+2y+2=0
y=x-2
now solve the points on the parabola. for horizontal lines, substutute x-2 for y in the given equation, solve.
Then, do the same for vertical lines.