Find the y-values of each point on the curve 2x^2 + y^2 - xy = 3 where the tangent line is vertical
To find the y-values of each point on the curve where the tangent line is vertical, we need to find the points where the derivative of y with respect to x is undefined.
First, we differentiate the equation of the curve with respect to x:
d/dx (2x^2 + y^2 - xy) = d/dx (3)
4x + 2y * dy/dx - y - x * dy/dx = 0
Next, we solve for dy/dx:
dy/dx * (2y - x) = x - 4x + y
dy/dx = (x - 4x + y) / (2y - x)
For the tangent line to be vertical, the derivative dy/dx must be undefined. This occurs when the denominator is equal to zero:
2y - x = 0
Rearranging the equation:
2y = x
y = x/2
Substituting y = x/2 into the equation of the curve:
2x^2 + (x/2)^2 - x(x/2) = 3
2x^2 + x^2/4 - x^2/2 = 3
8x^2 + x^2 - 2x^2 = 12
7x^2 = 12
x^2 = 12/7
Taking the square root:
x = ±√(12/7)
Therefore, the x-values of the points on the curve where the tangent line is vertical are ±√(12/7).
To find the corresponding y-values, substitute these x-values into the equation y = x/2:
When x = √(12/7):
y = √(12/7) / 2 = √3 / √7
When x = -√(12/7):
y = -√(12/7) / 2 = -√3 / √7
So the y-values of the points on the curve where the tangent line is vertical are √3 / √7 and -√3 / √7.