Question #95baf

Question

1977 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-10T17:22:50

See below.

Answer 2 · 2017-03-10T17:23:35

We have the points:

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

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

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

$2x^2ydy/dx+2xy^2 +y +xdy/dx=0$

$(x^2y+x)dy/dx= -xy^2 -y$

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

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

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

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

$dy/dx =-y/x$

And we have:

$-y/x = -1$

which means:

$x=y$

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

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

$x^4+x^2 = 2$

$x^4+x^2-2=0$

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 $P_2 = (1,1)$