Question

Find dy/dx 4x^2+y^2-8x+4y+4=0, find points on graph of the equation where there is a vertical or horizontal tangent line

Answer 1

This can be solved without taking derivatives.

4x^2 +y^2 -8x +4y +4 = 0
4(x^2 -2x +1) +(y^2 +4y +4) = 8
4(x-1)^2 + (y+2)^2 = 8
That is the equation of an ellipse with center at (1,-2)

There are vertical tangents at y = -2, where x-1 = + or - sqrt2
x = 1 +/-sqrt2

There are horizontal tangents at x =1, where y+2 = +/1 sqrt8
y = 2 +/-sqrt8

If you want an equation for dy/dx, try implicit differentiation

8x + 2y dy/dx -8 +4 dy/dx = 0
dy/dx (2y +4) = (-8x +8)
dy/dx = (-8x +8)/(2y +4)
This clearly shows that the derivative is zero at x=1 (horizontal slope) and infinity at y= -2 (vertical slope), in agreement with the other method. The y(x) equation can be used to solve for the other coordinates, when x=1 or y = -2.