# Problems with finding horizontal and vertical tangents for this equation $3(x^2+y^2)^2=100xy$

Our professor gave us this function to differentiate $$3(x^2+y^2)^2=100xy$$ and I did differentiate it $$\frac{dy}{dx}=\frac{3x^4+3xy^2-25y}{25x-3x^2y+3y^3}$$ But I'm having trouble finding the points that have a vertical or horizontal tangent. I am aware the numerator needs to =0 for the tangent to be horizontal and denominator =0 for tangent to be vertical

I tried using the quadratic formula to get $$y=\frac{25\pm\sqrt{25^2-36x^5}}{6x}$$ and I know we are supposed to replace y into the original function to get the points, but seeing as the professor doesn't allow us to use any sort of calculator, I have feeling there should be a much simpler way to do this?

• Your expression for $dy/dx$ is not quite right. It should be $\frac{3x^3+3xy^2-25y}{25x-3x^2y-3y^3}$
– smcc
Oct 16, 2018 at 18:29
• I might try solving for $x^2+y^2$ instead of $y$ and substituting that back into the original equation.
– amd
Oct 16, 2018 at 19:02

Suppose your equation implicitly defines $$y$$ as a function of $$x$$:

$$3(x^2+y(x)^2)^2=100xy(x)$$

Differentiating with respect to $$x$$:

$$6(x^2+y^2)[2x+2yy'(x)]=100[y+xy'(x)] \tag{1}$$

Substitute $$y'(x)=0$$ into $$(1)$$:

$$12(x^2+y^2)x=100y \iff 3(x^2+y^2)x=25y$$

This can be used with the original equation to obtain the point(s) where $$y'(x)=0$$.

The other point(s) can be found by the symmetry of the problem in $$x$$ and $$y$$. (You could go through the same working as above but consider the equation as defining $$x$$ as an implicit function of $$y$$.)

Alternatively:

Let

$$F(x,y)=3(x^2+y^2)^2-100xy$$

By the implicit function theorem

\begin{align*}y'(x)&=-\frac{F_x}{F_y}\quad \text{ if F_y\neq 0}\\ x'(y)&=-\frac{F_y}{F_x} \quad\text{ if F_x\neq 0}\end{align*}

$$F_x=12x(x^2+y^2)-100y\qquad \text{ and } \qquad F_y=12y(x^2+y^2)-100x$$

The points you require are where $$F_x=0$$ or $$F_y=0$$ (but not both).

• The point about symmetry is a good one, but the equation that you’ve derived is the same one that the OP has already solved for $y$.
– amd
Oct 16, 2018 at 18:56
• Not quite, and I had left the equation as it as in order to hint that it is probably better not to solve for $y$ as they have done.
– smcc
Oct 16, 2018 at 19:03
• It’s probably better to make that hint less subtle, then.
– amd
Oct 16, 2018 at 19:04

Hint: You must use both equations $$3(x^2+y^2)^2=100xy$$ $$3x^4+3xy^2-25y=0$$

• So should I just discard my calculations or..? There's gotta be more hints than that. Oct 16, 2018 at 18:35