Find all points on the curve x^2+xy+y^2=1 where the tangent line is parallel to the line y= -2x
To find the points on the curve where the tangent line is parallel to y = -2x, we need to find the gradient of the tangent line and set it equal to the gradient of the line y = -2x.
The gradient of y = -2x is -2. Let's call this gradient m.
To find the gradient of the tangent line, we need to find the derivative of the curve equation x^2 + xy + y^2 = 1 with respect to x.
Differentiating both sides with respect to x:
2x + (x*dy/dx + y) + 2y*(dy/dx) = 0
2x + xy' + y + 2yy' = 0
xy' + 2yy' = -2x - y
y'(x + 2y) = -2x - y
y' = (-2x - y) / (x + 2y)
To set y' equal to -2 (the gradient of the line y = -2x):
(-2x - y) / (x + 2y) = -2
-2(2x + 2y + 4y) = -2x - y
-4x - 4y - 8y = -2x - y
-4x - 12y = -2x - y
-2x - 11y = 0
Simplifying this equation, we get:
2x + 11y = 0
So, the points on the curve where the tangent line is parallel to y = -2x are the points that satisfy both equations:
x^2 + xy + y^2 = 1
2x + 11y = 0
These are the points we need to find. Let's solve these equations simultaneously.
From the second equation, we can solve for x:
x = -11y/2
Substituting this value of x into the first equation, we get:
(-11y/2)^2 + (-11y/2)y + y^2 = 1
(121y^2/4) - (11y^2/2) + y^2 = 1
(121y^2 - 22y^2 + 4y^2)/4 = 1
103y^2/4 = 1
103y^2 = 4
y^2 = 4/103
Taking the square root of both sides, we get:
y = ±2/sqrt(103)
Substituting these values of y back into the equation x = -11y/2, we get:
x = -11(±2/sqrt(103))/2
Simplifying further:
x = ±(-22/sqrt(103))
So, the points on the curve x^2 + xy + y^2 = 1 where the tangent line is parallel to y = -2x are:
( -22/sqrt(103), 2/sqrt(103) )
( 22/sqrt(103), -2/sqrt(103) )