Meaning of (apparantly meaningless) roots from eliptical curve points of interaction When answering the question 'Find the equations of the tangents with gradient 1 to the curve $x^2 + 2y^2 = 6$' .
$$\frac{dy}{dx} = \frac{-2x}{2\sqrt{2}\sqrt{6-x^2}}$$
solving for $\frac{dy}{dx} = 1$ reduces to $\frac{-x}{\sqrt{2}} = \sqrt{6 - x^2}$
This give solutions $2, -2, \sqrt{12}$.
$2$ and $-2$ are correct and useful solutions but the two solutions from $\sqrt{12}$ seem meaningless, looking at the graph of the elipse there is no value of y at $x = \pm\sqrt{12}$. Do these roots have any meaning?
 A: So by my eye, your current issues arise when finding the points on the curve with the specified derivative.
There's a few issues here.

Firstly, you took the derivative incorrectly, because what you did was solve for $y$. This isn't good enough: you can't explicitly solve for $y$ here and naively treat its derivative, because $y$ is not a function of $x$. (The ellipse has two $y$ values corresponding to almost every $x$, so it can't be a function.)
What you should do is implicit differentiation. Rewrite the relation as so:
$$x^2 + 2 \Big( y(x) \Big)^2 = 6$$

Note: Using the notation $y(x)$ here is a little misleading, considering what I just said, but we can write the graph as multiple functions of $x$, and treat them all at the same time like this. Your issue is that, by solving for $y$, you're assuming a particular form of $y$. For instance, in $x^2+y^2=1$, it is equally justifiable to say
$$y = \sqrt{1-x^2} \qquad y = -\sqrt{1-x^2}$$
in the solving process, and each can lead to an incomplete view.

Differentiation with the chain rule gives us
$$2x+4yy'=0$$
If we want gradient $1$, then $y'=1$, so
$$2x+4y=0 \implies y = -\frac 1 2 x$$
Thus, points of the form $(x,-x/2)$ on the ellipse are possible points.
Of course, it becomes easy to see that, plugging in $y=-x/2$ into the original equation, that
$$x^2 + 2 \left( - \frac x 2 \right)^2 = 6 \implies \frac 3 2 x^2 = 6 \implies x^2 = 4 \implies x = \pm 2$$
Notice how no $\sqrt{12}$ terms are popping out. You will need to compare with the graph to figure out the corresponding $y$ values with positive tangent slope.

Now, let us also note that $\sqrt{12}$ is an extraneous solution to your original equation in the first place.
In fact, $2$ is also an extraneous solution.
The "original equation" I'm referring to is
$$ - \frac{x}{\sqrt 2} = \sqrt{6-x^2}$$
If you plug in $x=2$, then you get
$$ - \frac{2}{\sqrt 2} = \sqrt{6-2^2} \implies - \sqrt 2 = \sqrt 2$$
an obvious problem. $\sqrt{12}$ has a similar but even more fundamental issue:
$$\sqrt{6 - \left( \sqrt{12} \right)^2} = \sqrt{6-12} = \sqrt{-6} \not \in \mathbb{R}$$
This is not a real number, so applying it to a graph in the $x,y$-plane is kind of silly (especially since the left-hand side is not imaginary).
This is the sort of information loss you arrive at when you solve for $y$ in this scenario (where $y$ cannot be expressed as a single function of $x$), and also arises in the solving process (since you had to square both sides to get it).
You could get $x=2$ as a legitimate solution by using instead the "different" solution for $y$, wherein
$$y = -\frac{1}{\sqrt{2}} \cdot \sqrt{6-x^2}$$
analogous to a previous example I gave.

In summary:

*

*When squaring equations, you may "lose" information, or "gain" extra solutions, which are not solutions to your original equation. (This is because $x^2$ is not an "injective" function: it can send multiple values to the same value. You can't totally undo it.)


*When dealing with equations that represent a graph, and that graph is not a function of $x$, just solving for $y$ will cause you to lose information and somewhat narrow your view of the problem on the whole. For calculus, it will be better to use the chain rule and implicit differentiation.
