Write the equation of the tangent at (2,2) to the curve x^2-2xy+y^2+2x+y-6=0
Can anyone please give me some ideas to do it?Thanks!
first you have to differentiate implicitly with respect to x
2x - 2xdy/dx - 2y + 2ydy/dx + 2 + dy/dx = 0
now solve this for dy/dx by using common factors
dy/dx(2y - 2x + 1) = -2x + 2y + 2
dy/dx = (-2x+2y+2)/(2y-2x+1)
now sub in x = 2 from your point (2,2) to get the slope of the tangent.
from there it should be easy, let me know what you got.
Is it y=2x-2?
correct
To find the equation of the tangent at any given point on a curve, we need to first find the derivative of the curve and then substitute the coordinates of the given point into the derivative equation.
Let's start by rewriting the given equation in a standard form:
x^2 - 2xy + y^2 + 2x + y - 6 = 0
Rearranging the equation, we have:
x^2 + y^2 - 2xy + 2x + y = 6
Now, we need to find the derivative with respect to x. Differentiating both sides of the equation, we get:
2x + 2yy' - 2(xy' + y) + 2 + y' = 0
Simplifying the equation, we find:
2x + 2yy' - 2xy' - 2y + 2 + y' = 0
Grouping the terms, we have:
(2x - 2y + 2) + (2y - 2x + y') = 0
Simplifying further, we get:
(2x - 2y + y') + (2y - 2x) = -2
Now, let's substitute the coordinates (2,2) into the derivative equation:
(2(2) - 2(2) + y') + (2(2) - 2(2)) = -2
Simplifying, we find:
-2 + y' = -2
y' = 0
Since y' represents the slope of the curve at the given point, (2,2), we can conclude that the slope is 0.
Therefore, the tangent line at (2,2) is horizontal, meaning it has the equation y = 2.
To summarize, the equation of the tangent line at (2,2) to the curve x^2 - 2xy + y^2 + 2x + y - 6 = 0 is y = 2.