Find an equation of all tangent lines for the curve x^2-xy+y^2=16 x=-2
2x-(xy'+y(1))+2yy'=0
2x-xy'+y+2yy'=0
y'(-x+2y)=-2x-y
y'=-2x-y/-x+2y
y'=-2(0)-y/-(0)+2y
y'=-y/2y
Not sure what to do now?
You have y' = (2x-y)/(x-2y)
at x = -2, y = -1±√13, so there will be two lines tangent to the ellipse at x = -2.
for y = -1+√13,
y' = (-4+1-√13)/(-2-2(-1+√13))
= (-3-√13)/(-2√13)
= (13+3√13)/26
So, the line is
y = (13+3√13)/26 (x+2) + -1+√13
You can see the plots at
http://www.wolframalpha.com/input/?i=plot+x^2-xy%2By^2%3D16%2C+y+%3D+%2813%2B3%E2%88%9A13%29%2F26+%28x%2B2%29+%2B+%28-1%2B%E2%88%9A13%29
Now do the same for y = -1-√13
To find the equation of all tangent lines for the curve x^2 - xy + y^2 = 16 at the point x = -2, you've correctly started by differentiating the equation implicitly. However, it seems like there is a mistake in your differentiation.
Let's differentiate the equation correctly:
Differentiating both sides of the equation with respect to x:
d/dx(x^2 - xy + y^2) = d/dx(16)
2x - (x(dy/dx) + y) + 2y(dy/dx) = 0
Now, we need to substitute the given value x = -2 into the equation. Plugging in x = -2:
2(-2) - (-2)(dy/dx) + y + 2y(dy/dx) = 0
-4 + 2(dy/dx) + y + 2y(dy/dx) = 0
Let's combine like terms:
(2 + 2y)(dy/dx) = 4 - y
(dy/dx) = (4 - y) / (2 + 2y)
Now, we have the derivative (dy/dx) in terms of y and x, but to find the equation of the tangent line, we need to find the slope and the point of tangency at x = -2.
First, let's find the value of y when x = -2 by substituting x = -2 into the original equation:
(-2)^2 - (-2)y + y^2 = 16
4 + 2y + y^2 = 16
y^2 + 2y - 12 = 0
Now, we can solve this quadratic equation for y:
(y + 4)(y - 3) = 0
y = -4 or y = 3
So, the curve intersects the x-axis at y = -4 and y = 3 when x = -2. We will consider these as the points of tangency.
Now, we have two equations of tangent lines:
1) For y = -4:
(dy/dx) = (4 - (-4)) / (2 + 2(-4))
(dy/dx) = 8 / (-6)
(dy/dx) = -4/3
2) For y = 3:
(dy/dx) = (4 - 3) / (2 + 2(3))
(dy/dx) = 1 / 8
(dy/dx) = 1/8
Now we have two slopes of the tangent lines: -4/3 and 1/8.
To find the equations of the tangent lines, we need to use the point-slope form of the equation of a line:
y - y1 = m(x - x1)
For y = -4, x = -2, and m = -4/3:
y - (-4) = (-4/3)(x - (-2))
Simplifying this equation will give you the equation of the first tangent line.
For y = 3, x = -2, and m = 1/8:
y - 3 = (1/8)(x - (-2))
Simplifying this equation will give you the equation of the second tangent line.
I hope this helps you find the equations of the tangent lines for the given curve!