How to solve for a specific gradient in an implicit relationship, when no points are known?

Question

The problem

Consider the following relation: $$x^2-3xy+y^2=7$$

I'm struggling with what is essentially the following task:

Find all coordinates of all points where the gradient of the tangent of the curve is ${2\over3}$.

Using implicit differentiation, I arrive at the following derivative of $y$ wrt $x$: $${\text dy\over\text dx}=\frac{3y-2x}{2y-3x}$$

This is somewhat of a mindbender for me, since it looks like the derivative itself depends on both $x$ and $y$. How can this be? How does one work with this?

Even more daunting is the task of solving $\frac{\text dy}{\text dx}=\frac{2}{3}$. The equation cannot be solved as we have a single, two-variable equation: $$2(2y-3x)=3(3y-2x)$$ Sadder yet is that by my logic, there is an infinite number of solutions ($xy$ pairs) that satisfy this. Unfortunately, graphing this relation gives the solution set $(x,0)$ where $x\inℝ$.

The correct answer is $(\sqrt7,0)\cup(-\sqrt7,0)$.

My questions

1. How do I work with a derivative which depends on both $x$ and $y$, and conceptually, how is that even possible?
2. Why is it that graphing the solutions of $2(2y-3x)=3(3y-2x)$ did not work and gave erroneous solutions?
3. How is the answer $(\sqrt7,0)\cup(-\sqrt7,0)$ obtained?

Thank you very much :)

Answer 1

Note: As per comment everywhere where you read ellipse you should read hyperbola.

In your case the curve is an ellipse. If you go to a value $x$, what is the derivative of the function? Well, it is not clearly defined. First you need to find $y$. You can have 0, 1, or 2 $y$ values corresponding to a single $x$. So where do you calculate the derivative? You need to choose $y$ to be on the ellipsis. Think for example that you have a circle of radius $1$, centered at the origin. And you want to calculate the derivative at $x=0.5$. You notice that you have two intersections. At $y=\sqrt 3/2$ you have a negative derivative, but at $y=-\sqrt 3/2$ the derivative is positive.

So you have an equation for the derivative that involve $x$ and $y$. But you also know that $(x,y)$ is on the given curve. That's the second equation.

What you need to do is to write $y(x)$ from the derivative equation, plug it into the equation for the curve, find $x$, then find $y(x)$.

Let me know how these steps work for you.

Answer 2

When you differentiated you obtained the derivative for any of the curves $$x^{2}-3xy+y^{2}=c. \tag{1}$$ Too see that these curves are different you can let $c=0$ then $(0,0)$ is on the curve, but if $c=1$ then $(0,0)$ is not a point on the curve, more generally $(\sqrt{c},0)$ is unique on each curve in $(1)$.

Simplifying $2(2y-3x)=3(3y-2x)$ you will obtain $y=0$ and pluggin that into $(1)$ gives $$x^{2}=\sqrt{c}$$ which as you can see has a solution for any $c\geq 0$. Here you can see how the solutions to your problem came about, but more importantly the family of hyperbels in $(1)$ where $c\geq 0$ all have the slope $\frac{2}{3}$ at two points. Since there are infinitley many choices of $c$ and each curve in the family is different, there must be an infinite number of points on the derivative to account for every $c$ and therefore it is a curve.