Use implicit differentiation to find the points where the parabola defined by
x^{2}-2xy+y^{2}-8x+4y+20 = 0
has horizontal and vertical tangent lines.
To find the points where the given parabola has horizontal and vertical tangent lines, we can use implicit differentiation.
Starting with the given equation:
x^2 - 2xy + y^2 - 8x + 4y + 20 = 0
We differentiate both sides of the equation with respect to x implicitly:
d/dx(x^2) - d/dx(2xy) + d/dx(y^2) - d/dx(8x) + d/dx(4y) + d/dx(20) = 0
Simplifying each term:
2x - 2y(dy/dx) + 2y(dy/dx) - 8 + 4(dy/dx) = 0
Combine like terms:
2x - 8 + 4(dy/dx) = 0
Rearranging the equation:
dy/dx = (8 - 2x) / 4
To find the points where the parabola has horizontal tangent lines, we set dy/dx = 0:
(8 - 2x) / 4 = 0
Solving for x:
8 - 2x = 0
-2x = -8
x = 4
Now substitute x = 4 back into the original equation to find the corresponding y-coordinate:
4^2 - 2(4)y + y^2 - 8(4) + 4y + 20 = 0
16 - 8y + y^2 - 32 + 4y + 20 = 0
y^2 - 4y + 4 = 0
(y - 2)^2 = 0
y - 2 = 0
y = 2
So, when x = 4, y = 2. Therefore, the point (4, 2) is the point where the parabola has a horizontal tangent line.
To find the points where the parabola has vertical tangent lines, we focus on the equation dy/dx = (8 - 2x) / 4. For a vertical tangent line, dy/dx is undefined, so we set the denominator equal to zero:
4 = 0
Since this is not possible, there are no points where the parabola has vertical tangent lines.
Therefore, the point (4, 2) is the only point where the given parabola has a horizontal tangent line.
To find the points where the given parabola has horizontal and vertical tangent lines, we need to use implicit differentiation.
Let's start by differentiating the equation of the parabola with respect to x.
d/dx (x^2 - 2xy + y^2 - 8x + 4y + 20) = 0
To differentiate each term, we treat y as a function of x and apply the chain rule.
2x - 2y(dy/dx) + 2y(dy/dx) - 8 + 4(dy/dx) = 0
Simplifying, we get:
2x - 8 + 4(dy/dx) = 0
Now, let's find the slope of the tangent line when it is horizontal, which means dy/dx = 0.
Substituting dy/dx = 0 into the equation above, we have:
2x - 8 + 4(0) = 0
2x - 8 = 0
2x = 8
x = 4
So, the parabola has a horizontal tangent line at x = 4.
Now let's find the points where the parabola has a vertical tangent line. A vertical tangent line occurs when dx/dy = 0.
Taking the derivative of the parabola equation with respect to y, we get:
d/dy (x^2 - 2xy + y^2 - 8x + 4y + 20) = 0
Applying the chain rule, we have:
-2x(dy/dx) + 2y - 8(dy/dx) + 4 = 0
Simplifying, we get:
-2x(dy/dx) - 8(dy/dx) + 2y + 4 = 0
Factoring out dy/dx, we have:
(-2x - 8)(dy/dx) + 2y + 4 = 0
Since we are looking for dx/dy = 0, we set dy/dx = 0 and solve for y.
(-2x - 8)(0) + 2y + 4 = 0
2y + 4 = 0
2y = -4
y = -2
So, the parabola has a vertical tangent line at y = -2.
Therefore, the points where the parabola has horizontal and vertical tangent lines are (4, -2) and (4, -2), respectively.
x^2 -2xy + y^2 - 8x + 4y + 20 = 0
2x - 2y - 2xy' + 2yy' - 8 + 4y' = 0
(-2x+2y+4)y' = -2x+2y+8
y' = (x-y-4)/(x-y-2)
So, y'=0 when x-y-4 = 0
y' is undefined when x-y-2 = 0
To find the points, intersect the lines with the parabola