Question #95baf

2 Answers
Mar 10, 2017

We have the points:

P_1 = (-1,-1)P1=(1,1) and P_2 = (1,1)P2=(1,1)

Explanation:

The slope of the tangent line is given by dy/dxdydx, which we can calculate by implicit differentiation:

d/dx(x^2y^2+xy) = 0ddx(x2y2+xy)=0

2x^2ydy/dx+2xy^2 +y +xdy/dx=02x2ydydx+2xy2+y+xdydx=0

(x^2y+x)dy/dx= -xy^2 -y(x2y+x)dydx=xy2y

Since from the original equation we can see that xy!=0xy0, divide both sides by xyxy

(x+1/y)dy/dx = -(y+1/x)(x+1y)dydx=(y+1x)

(xy+1)/y dy/dx =- (xy+1)/xxy+1ydydx=xy+1x

We can see that xy != -1xy1, otherwise, (xy)^2+xy(xy)2+xy would be null, in contradiction with the original equation, so:

dy/dx =-y/xdydx=yx

And we have:

-y/x = -1yx=1

which means:

x=yx=y

Substituting this in the original equation we find the values of xx for which dy/dx = -1dydx=1

{(x^2y^2+xy=2),(x=y):}

x^4+x^2 = 2

x^4+x^2-2=0

Solving this as a second degree equation with unknown x^2:

x^2 = (-1+-sqrt(1+8))/2

And choosing only the positive solution since x^2 cannot be negative for x in RR:

x^2 = 1

x=+-1

Thus we have the points:

P_1 = (-1,-1) and P_2 = (1,1)