To which degree must I rotate a parabola for it to be no longer the graph of a function?

Question

To which degree must I rotate a parabola for it to be no longer the graph of a function?

I have no problem with narrowing the question down by only concerning the standard parabola: $$f(x)=x^2.$$

I am looking for a specific angle measure. One such measure must exist as the reflection of $f$ over the line $y=x$ is certainly no longer well-defined. I realize that preferentially I should ask the question on this site with a bit of work put into it but, alas, I have no intuition for where to start. I suppose I know immediately that it must be less than $45^\circ$ as such a rotation will cross the y-axis at $(0,0)$ and $(0,\sqrt{2})$.

Any insight on how to proceed?

Answer 1

Rotating the parabola even by the smallest angle will cause it to no longer be well defined.

Intuitively, you can prove this for yourself by considering the fact that the derivative of a parabola is unbounded. This means that the parabola becomes arbitrarily "steep" for large (or small) values of $x$, i.e. its angle being closer and closer to $90^\circ$, and rotating it by even a little will tip it over the $90$ degrees.

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For a formal proof, first, we need to explain exactly what a rotation of a parabola is. In general, a rotation in $\mathbb R^2$ is multiplication with a rotation matrix, which has, for a rotation by $\phi$, the form $$\begin{bmatrix}\cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{bmatrix}$$

In other words, if we start with a parabola $P= \{(x,y)|x\in\mathbb R\land y=x^2\}$, then the parabola, rotated by an angle of $\phi$, is

$$\begin{align}P_\phi &= \left.\left\{\begin{bmatrix}\cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{bmatrix}\cdot\begin{bmatrix}x\\y\end{bmatrix}\right| x\in\mathbb R, y=x^2\right\}\\ &=\{(x\cos\phi - y\sin\phi, x\sin\phi + y\cos\phi)|x\in\mathbb R, y=x^2\}\\ &= \{(x\cos\phi-x^2\sin\phi, x\sin\phi + x^2\cos\phi)| x\in\mathbb R\}\end{align}.$$

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The question now is which values of $\phi$ construct a well defined parabola $P_\phi$, where by "well defined", we mean "it is a graph of a function", i.e., for each $\overline x\in\mathbb R$, there exists exactly one value $\overline y$ such that $(\overline x,\overline y)\in P_\phi$.

Clearly, if $\phi = 0$, we have $P_0=\{(x, x^2)|x\in\mathbb R\}$ which is well defined, because for every $\overline x$, the value $\overline y=\overline x^2$ is the unique value required for $(\overline x,\overline y)$ to be in $P_0$. Also, if $\phi=\pi$, then $P_\pi = \{(-x, -x^2)|x\in\mathbb R\}$ is also well defined because if $(\overline x,\overline y)\in P_\pi$, then $\overline y=-\overline x^2$.

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Now, observe what happens if $\phi\notin\{0,\pi\}$. For now, let's assume that $\phi\in(0,\frac\pi2)$. In that case, $\sin\phi\neq 0$, which means that the equation $$x\cos\phi-x^2\sin\phi=0$$ has two solutions. One solution is $x=0$, the other is $x=\frac{\cos\phi}{\sin\phi} = \cot\phi$.

This means that, if we take $\overline x=0$, there are two values of $x$ that can create a point $(\overline x, \overline y)$, and we have two possible values of $\overline y$ as well. One is $\overline y_1 = 0$, the other is $$\overline y_2 = x\sin\phi + x^2\cos\phi = \frac{\cos\phi}{\sin\phi} \sin\phi + \left(\frac{\cos\phi}{\sin\phi}\right)^2\cos\phi =\cos\phi + \frac{\cos^3\phi}{\sin\phi}$$

and, because $\phi\in(0,\frac\pi2)$, we know that $\overline y_2>0$, which means $\overline y_2\neq \overline y_1$, and therefore, $P_\phi$ is not a graph of a function.

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Note that the options when $\phi$ is in one of the other three quadrants can be solved similarly as the one above, or, you can use symmetry to translate all of the other three cases to the one already solved above.

Answer 2

HINT:

The convex region $\{(x,y)\ | \ y \ge x^2\}$ has $\{0\} \times [0, \infty)$ as recession cone.

$\bf{Added:}$ We can ask a related question: given a convex function $f\colon \mathbb{R}\to \mathbb{R}$, ($f(0)=0$ to keep it simple), what are the cones with vertex at $(0,0)$ that fit inside the epigraph of $f$ ? Here is the answer: the derivative of $f$ is increasing, and its range is an interval $[m, M]$ (which could be infinite). The the slopes of the cone are $m$, $M$ ( the terminal slopes of $f$).

In our case for $f(x) = x^2$, $m=-\infty$, $M= \infty$, so the cone becomes a half-line.

Let's take another example:

$$f(x) = \log (\cosh x)$$

We have $f(0)= 0$, $f'(x) = \tanh x$. One checks that the larges cone that fits in has branches $y = \pm x$. ( that is the recession cone).

Question: what is the largest angle of rotation to still get the graph of a function? (left or right will be the same in this case).

Answer 3

Let $P_\theta$ be the parabola obtained from the parabola $P$ of equation $y=x^2$ by a rotation of angle $\theta\in(0,2\pi)$.

A rotation of $\pi$ is simply a symmetry about the origin so $P_\pi$ is the graph of $y=-x^2$, the graph of a function. Moreover, if $\theta=\frac{3\pi}{2}$, then $P_{\theta}$ is the graph of $x=y^2$, not the graph of a function. Similarly, if $\theta=\frac{\pi}{2}$, $P_{\theta}$ is not the graph of a function. From now on, we assume $\theta\notin\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$.

$P_\theta$ is the graph of a function if and only if every vertical line intersects its graph in at most one point.

Assume that $P_\theta$ is the graph of a function. Then the line $L$ of equation $x=0$ intersects $P_{\theta}$ in one point, the origin. Now, rotate $P_\theta$ and $L$ by $-\theta$. The image of $L$ by this rotation is the line $L_\theta$ through the origin with inclination $\frac{\pi}{2}-\theta$. $L_\theta$ should intersects the parabola $P$ in exactly one point.

Since $\theta\notin\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$, $L_\theta$ is neither horizontal nor vertical. It then intersects $P$ in two points, a contradiction.

As a conclusion, $P_\theta$ is the graph of a function if and only if $\theta=\pi$ (assuming $0<\theta<2\pi$).

Answer 4

You write (emphasis added):

I am looking for a specific angle measure. One such measure must exist as the reflection of $f$ over the line $y=x$ is certainly no longer well-defined.

It sounds like you are looking for $\theta_\text{min}$, where a rotation through an angle $\theta \geq \theta_\text{min}$ produces a graph that is no longer the graph of a function, but a rotation through $\theta < \theta_\text{min}$ still produces the graph of a function.

But, in this case, the equality needs to be on the other side: a rotation through $\theta > \theta_\text{min}$ produces a graph that is no longer the graph of a function, but a rotation through $\theta \leq \theta_\text{min}$ still produces the graph of a function.

In fact, as we will show, $\theta_\text{min} = 0$: any rotation (except for a half-revolution) produces a graph that is no longer the graph of a function.

You then write:

I suppose I know immediately that it must be less than $45^\circ$ as such a rotation will cross the y-axis at $(0,0)$ and $(0,1)$.

Actually, it crosses the $y$-axis at $(0,0)$ and $(0,\sqrt{2})$. But the important point is that it crosses a vertical line at multiple points.

So now we want to rotate the parabola through some other angle $\theta$ and check whether it crosses a vertical line at multiple points. 5xum’s answer shows how to do this, and that is great. But for all the talk about it being “elementary”, it requires transformation matrices and trigonometry. A truly elementary solution would require no knowledge beyond lines, parabolas and basic algebra: such a solution is presented below.

Instead of rotating the parabola, then checking if it crosses a vertical line at multiple points, we can rotate the axes, then check if the parabola crosses a line that is “vertical” relative to the rotated axes at multiple points. This is the idea behind Taladris’s answer.

As stated in that answer, it always does (again, excluding half-revolutions). But how can we prove this?

This “rotated vertical” line could (relative to the original axes) be anything other than a vertical line. That is, it could be a line with gradient $m$, for any value of $m$. Now, for any such value, we can find a corresponding line that crosses the parabola at multiple points. To see this, note that for $m = 0$, we can simply use any line above the origin, and for any other value of $m$, we can use the line through the origin, which crosses the parabola there and at a second point given by:

\begin{align} mx &= x^2\\ x &= m\\ y &= m^2. \end{align}

Answer 5

My attempt to phrase Brian Drake's answer even more simply:

The original problem can be phrased geometrically as, "What is the smallest angle $\theta$ by which we can rotate the standard parabola $P$ around the origin in order to make its rotated version $R$ intersect a given vertical test line $v$ at two points?"

There are two keys to solving the problem.

Key #1: In this problem, position is relative! The absolute positions of the rotated parabola $R$ and vertical line $v$ on the $xy$-plane do not change whether or not $R$ describes a function.

Key #2: Rotations are also always relative!

Let's now bring these two ideas together.

According to Key #1, we can choose the $y$-axis for our vertical line $v$. Why this choice? Some reasons:

• Lines passing through the origin have an easily written function
• The $y$-axis passes through the origin and so has an easy function
• Standard parabola $P$ also passes through the origin
• Rotated parabola $R$ was generated from rotating $P$ around the origin and so also passes through the origin
• So, the $y$-axis passes through both $R$ and $P$

Then, the relativity of rotation from Key #2 mean it's completely OK to think of $\theta$ as a rotating $v$ around the origin while keeping the standard parabola $P$ fixed.

And I think that makes it obvious that any non-zero $\theta$ will make the point that any diagonal line passing through the origin will intersect the standard parabola at two points.

Of course, if you're not a visual/intuitive person, then you can formally justify this result in many ways.

• 5xum does it above using rotations. He finds that $x_{intersection} = tan(\theta)$.
• Brian Drake does it above by noting that a diagonal line passing through the origin will have an equation $y = mx$ for some slope $m$. He then finds the (other) intersection of this line with the standard parabola by setting $y_{intersection} = mx_{intersection} = x_{intersection}^2 \implies x_{intersection} = m$.
• These two solutions are related by the fact that for all lines, $tan(\theta) = \frac{sin(\theta)}{cos(\theta)} = \frac{\Delta y}{\Delta x} = slope = m$.

Answer 6

A part of the question is: how much do we have to rotate $y = x^2$ around the origin such that it hits the $y$-axis in a second point, in addition to the origin? To simplify this, instead of rotating the parabola (by $\varphi$) we can rotate the $y$-axis (by $-\varphi$). Now any non-vertical line through the origin has an equation $y=ax$ and $$x^2=ax$$ has a solution at $x=a$, so the line hits the parabola at $(a, a^2)$. We see that we cannot rotate the parabola even a tiny bit without losing the property that it is the graph of a function.

(In the above you may have noticed that I glossed over two cases: For $\varphi=\pi$ the $y$-axis is taken to itself, so it is vertical, and of course the parabola is mapped to $y=-x^2$, the graph of a function. And in the case $a=0$, corresponding to $\varphi=\pi/2$ or $\varphi=3\pi/2$, the point $(a,a^2)=(0,0)$ is the origin, so we would have to use a different line to show that the rotated parabola is not the graph of a function. But this is obvious.)

Answer 7

Without any loss of generality, assume the center of rotation is the origin, so that the first coordinate of the image of any point $(x,y)$ under a rotation by $\theta$ equals $x\cos\theta - y\sin\theta.$

Because

$$\cot\theta \cos\theta - \cot^2\theta \sin\theta = \frac{\cos\theta}{\sin\theta}\cos\theta - \frac{\cos^2\theta}{\sin^2\theta}\sin\theta = 0,$$

the point $(\cot\theta,\cot^2\theta)$ on the parabola is mapped by a rotation of $\theta$ to a point with coordinates of the form $(0,y).$

Provided $\theta$ is not an integral multiple of $\pi/2,$ such a point exists and has positive distance from the origin, showing that $(0,y)\ne(0,0).$ However, because $(0,0)$ is also on the rotated parabola, the rotated parabola does not define a function in any neighborhood of $0.$

Answer 8

A rigid parabolic arc remains a parabola for any arbitrary displacement or rotation imparted to the parabolic arc.

That is why we can describe any parabola in the plane by an intrinsic or natural equation

$$ \kappa = \dfrac{\cos^3\phi}{2f} $$

where $\kappa$ is curvature and $ \phi $ the rotation of tangent reckoned from its vertex V and $f= VF, $ its focal length.

In rectangular coordinates with any constants $(a,b,c,d)$ we have two connected arcs of the same parabola separated by a tangent parallel to x-axis (not shown):

$$ y = ax+b \pm \sqrt{cx+d}.$$

Answer 9

A little more generally ... Consider a non-degenerate (and, for the sake of this discussion, non-circular) conic with eccentricity $e>0$, focus $(f_x,f_y)$, focus-to-directrix length $d>0$, and major/transverse axis direction vector $(\cos\theta,\sin\theta)$. By the focus-directrix definition of a conic, the curve has this equation: $$\sqrt{(x-f_x)^2+(y-f_y)^2}=e \left(\; (x-f_x)\cos\theta+(y-f_x)\sin\theta-d \;\right) \tag1$$ Squaring and gathering terms gives us $$x^2 (1-e^2\cos^2\theta) - 2 x y e^2\cos\theta\sin\theta + y^2 (1-e^2\sin^2\theta) + \cdots = 0 \tag2$$ As a quadratic in $y$, the equation cannot represent a function of $x$. (Conceivably, the equation can give double-roots for every $x$, but this is clearly a degenerate case, which we're ignoring. (But see the Note below.)) Therefore, to have $y$ as a function of $x$, we need to eliminate the equation's quadratic-ness, which we do by making the coefficient of $y^2$ vanish. $$\sin\theta=\pm\frac1e \tag{$\star$}$$ We can conclude ...

• For ellipses ($e<1$), equation $(\star)$ has no solutions, confirming the fact that no (non-degenerate) ellipse is the graph of a function.
• For parabolas ($e=1$), equation $(\star)$ has the solutions $\theta=\pm90^\circ$. Thus, a (non-degenerate) parabola must have a vertical axis to be the graph of a function. (This answers OP's question.)
• For hyperbolas ($e>1$), equation $(\star)$ has exactly two solutions. We can interpret this as saying that any (non-degenerate) hyperbola has exactly two orientations that make it the graph of a function; the reader can check that these orientations correspond to making one or the other asymptote vertical.

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Note. Equation $(2)$ gives double-root $y$ value when its $y$-discriminant vanishes; that is, $$x^2 (1 - e^2) - 2 x ( f_x (1 - e^2) - e^2 d \cos\theta) + f_x^2 (1 - e^2) - 2 \cos\theta d e^2 f_x - d^2 e^2 = 0$$ For a polynomial in $x$ to vanish across a range of $x$-values, each coefficient must vanish. Thus, $1-e^2=0$, which implies $d\cos\theta=0$, and then also $d=0$. So, this situation applies only to a parabola with $d=0$, for which equation $(2)$ factors as $$\left(\;(x-f_x)\sin\theta - (y-f_y)\cos\theta\;\right)^2 = 0$$ making the parabola (a portion of) a "double-line" coinciding with its axis. (The "portion" consists of the focus and the points on the appropriate side of the focus to satisfy $(1)$.) In this case, the graph is a function except when that axis is vertical.

Answer 10

$$\lim_{n\to\infty}{\alpha_n}=\frac{\pi}{2}$$ So we cannot rotate the graph of $y=x^2$ at all.