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

Question

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?

Can someone please help me with this question?

Answer 1

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

Answer 2

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).