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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)
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