the question 5x^2-6xy+5y^2=16 represent an ellipse

a-)) Determine dy/dx at(1.-1)

b )) DETERMINE TWO POINTS ON THE ELLIPSE AT WHICH THE TANGENT IS HORIZONTAL.

implicit differentiation yields, 10x-6y-6xy'+10yy'=0,dy/dx=(6y-10x)/(10y-6x),at(1,-1) dy/dx=-16/-16 =1

if tangent is horizontal,dy/dx=0,y≠(6x/10)and y=(10x/6),substituting in the ellipse equation

5x^2-10x^2+(125x²/9)=16,80x^2 =144, x=±(3/√5),y=±√5, the two points are (3/√5,√5) and (-3/√5,-√5)

Approved