Question

Given the eqn 2y^2 - 2xy - 4y + x^2 = 0 , how to get the x coordinates of each point at which the curve has a horizontal tangent?

Answer 1

Use Calculus

Explanation:

Implicitly differentiate both sides of the equation, rearrange to get
`dy/dx = (y-x)/(2y-x-2)`
Horizontal tangent: gradient =0
`0=(y-x)/(2y-x-2)`
`0=y-x`
`y=x`

So horizontal tangents occur when y=x