Find the tangent line to the curve (y-x)^2= 2y+4 at the point (2,6)
To find the tangent line to a curve at a given point, we need to find the derivative of the curve and then substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line.
Let's start by finding the derivative of the equation (y-x)^2 = 2y + 4.
To do this, we need to rewrite the equation to make it easier to differentiate. Expanding the equation, we get:
y^2 - 2xy + x^2 = 2y + 4
Rearranging, we get:
y^2 - 2y - 2xy + x^2 - 4 = 0
Now, let's differentiate both sides with respect to x.
d/dx (y^2 - 2y - 2xy + x^2 - 4) = d/dx (0)
Differentiating each term separately, we get:
2y * dy/dx - 2 * dy/dx * x - 2y - 2x + 2x = 0
Simplifying, we get:
2y * dy/dx - 2 * dy/dx * x - 2y = 0
Now, let's solve for dy/dx or the slope of the tangent line.
Factor out dy/dx:
dy/dx * (2y - 2x) = 2y
Divide both sides by (2y - 2x):
dy/dx = 2y / (2y - 2x)
At the point (2,6), we need to substitute x = 2 and y = 6 into dy/dx:
dy/dx = 2 * 6 / (2 * 6 - 2 * 2)
dy/dx = 12 / (12 - 4)
dy/dx = 12 / 8
dy/dx = 3/2
So, the slope of the tangent line to the curve at the point (2,6) is 3/2.
Now, we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point.
Substituting in the values we found, we get:
y - 6 = (3/2)(x - 2)
Simplifying, we get:
y - 6 = (3/2)x - 3
Adding 6 to both sides, we get:
y = (3/2)x + 3
Therefore, the equation of the tangent line to the curve (y-x)^2 = 2y + 4 at the point (2,6) is y = (3/2)x + 3.