The equation `5x^2 - 6xy + 5y^2 = 16` represents an ellipse. Determine two points on the ellipse at which the tangent is horizontal.
`5x^2 - 6xy + 5y^2 = 16`
`` To find the points we need to find the derivative by using implicit differentiation and then find the derivatives zero.
==> `10x - 6y - 6xy' + 10yy' `
`==gt 10yy'- 6xy' = 6y-10x `
`==gt y' (10y - 6x) = 6y-10x `
`=gt 6y - 10x = 0 `
`==gt y= 10x/6 = (5x)/3`
`` ==> Now we will substitute `y= (5x)/3` into the equation.
==> `5x^2 - 6xy + 5y^2 = 16 `
`==gt 5x^2 - 6x(5x/3) + 5(5x/3)^2 = 16 `
`==gt 5x^2 - 10x^2 + 125x^2 /9 = 16`
`` ==> Multiply by 9.
==> `45x^2 - 90x^2 + 125x^2 = 144 `
`==gt 80x^2= 144 `
`==gt x^2 = 144/80 = 9/5 `
`==gt x = +- 3/sqrt5 = 3sqrt5/5 `
`==gt y= 5x/3 ==gt y= +-sqrt5`
`` Then the points such that the tangent is horizontal are:
`((3sqrt5)/5 , sqrt5 ) and ((-3sqrt5)/5 , -sqrt5)`