Consider the curve defined by x^2 + 4y^2 = 7 + 3xy
Show that there is a point P with x-coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P
dy/dx = (3y - 2x)/ (8y-3x)
I think you forgot to tell us what the x-coordinate is. For now I'll just call it .
So the derivative is . The horizontal tangent line occurs when the derivative is zero, namely, when . Since we are given the x-coordinate (which I'm calling ), we have that . Solving for gives us so the point is .
So whatever the x-coordinate actually is, just plug it in for and that will give you the answer.