For the curve given by 4x^2+y^2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.

Answer 1

if the line is horizontal, then dx/dy==0

8xdx+2ydy=2xdy+2ydx

dy/dx (2y-2x)=2y-8x

or 2y=8x or y=4x
but if this is true at x=2, then y must be 8.
In the original equation
4(4)+8^2 =? 48+2(2*8)
80=? 80, which is true, so at x=2 dy/dx is zero at the point of tangency

Answer 2

For the curve given by 4x^2 + y^2 = 48 + 2xy, find the positive y-coordinate given that the x-coordinate is 2.