If I tilt a parabola by $\pi/180$ radians in the Cartesian plane, which vertical line will cross the graph only once?

Question

This question is spurred from this question I asked a while back (by https://math.stackexchange.com/users/70134/jeppe-stig-nielsen)

If I tilt the parabola $f(x)=x^2$ from its vertex by $\pi/180$ radians in the Cartesian plane, which vertical line will cross the graph only once?

If I had a nice equation that represented the tilted parabola, maybe $f(x,y)$, then the question would become setting the discriminant of $f(x, \alpha)$ equal to zero and solving for $\alpha$ correct? If not, is there another approach?

Answer 1

For a given counterclockwise angle of rotation $\theta \in (0, \pi)$ the parametrized curve $$(x,y) = (t, t^2), \quad t \in \mathbb R$$ under the transformation matrix $$M = \begin{bmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ has the parametrization $$(u,v) = M(x,y) = (t \cos \theta - t^2 \sin \theta, t^2 \cos \theta + t \sin \theta).$$ This curve will have a vertical tangent when $du/dt = 0$; i.e., $$0 = \frac{du}{dt} = \cos \theta - 2t \sin \theta \implies t = \frac{1}{2} \cot \theta.$$ Hence the point at which the rotated parabola has a vertical tangent corresponds to $$(u,v) = \left(\frac{1}{4} \cos \theta \cot \theta, \frac{1}{4} (\cos \theta + \cot \theta \csc \theta) \right).$$ For $\theta = \pi/180$, we have $u = \frac{\cos \pi/180 \cot \pi/180}{4}\approx 14.3203$, which is the equation of the desired vertical tangent line; if the rotation is clockwise, the sign is simply negated.

Answer 2

Instead of tilting the parabola $y=x^2$ by $\pi/180$, one may also tilt vertical lines by $-\pi/180$ relative to the origin.

Vertical lines become straight lines of slope $\cot \frac\pi{180}$, and in particular the $y$-axis becomes:

$$\begin{align*} y &= x\cot\frac\pi{180}\\ x\cot\frac\pi{180} - y &= 0 \end{align*}$$

One line in this family of tilted lines touches the parabola $y=x^2$, where the slope of the parabola matches $\cot\frac\pi{180}$:

$$\begin{align*} \frac{dx^2}{dx} &= \cot\frac\pi{180}\\ 2x &= \cot\frac\pi{180}\\ x &= \frac12\cot\frac\pi{180}\\ y &= \left(\frac12\cot\frac\pi{180}\right)^2 = \frac14\cot^2\frac\pi{180} \end{align*}$$

The distance between the tilted $y$-axis and the above tangential point, using the distance formula, is:

$$\begin{align*} \text{Distance} &= \operatorname{distance}\left(x\cot\frac\pi{180} - y = 0, \left(\frac12\cot\frac\pi{180}, \frac14\cot^2\frac\pi{180}\right)\right)\\ &=\frac{\frac12\cot\frac\pi{180} \cdot \cot\frac\pi{180} - \frac14\cot^2\frac\pi{180}}{\sqrt{\left(\cot\frac\pi{180}\right)^2+1^2}}\\ &= \frac{\frac14\cot^2\frac\pi{180}}{\csc\frac\pi{180}}\\ &= \frac14\cos\frac\pi{180}\cot\frac\pi{180}\\ &\approx 14.3203\ldots \end{align*}$$

The required vertical line has the same $x$-intercept as the above distance:

$$\begin{align*} x &= \frac14\cos\frac\pi{180}\cot\frac\pi{180}\\ x &\approx 14.3203\ldots \end{align*}$$

Answer 3

EDIT: Due to @DavidK 's critique on the slope I have chosen.

Consider $f(x)=x^2$ and the line $y=\tan(\pi/180)x$. This question is actually the same as first finding the line $\perp$ to $y$, call it $y_\perp$, such that $y_\perp$ crosses graph$f$ only once. Second, finding the distance between the intersection point of $y$ and $y_\perp$ call it $(x_0,y_0)$ and the origin.

We know the slope of $y_\perp$ will be $-\cot(\pi/180)$. Now let $y_\perp -y_0 = -\cot(\pi/180)(x-x_0)$ where $(x_0, y_0)$ is the point where $y$ crosses $y_\perp$. Then \begin{align*}-\cot(\pi/180)(x-x_0)+y_0 =x^2 \\ \implies x^2+\cot(\pi/180)x-(y_0+\cot(\pi/180)x_0)=0 \end{align*} Hence if $y_\perp$ only crosses $f$ once we must have $$\cot^2(\pi/180)+4(y_0+\cot(\pi/180)x_0)=0$$ $$\implies \cot^2(\pi/180)+4(\tan(\pi/180)x_0+\cot(\pi/180)x_0)=0$$ $$\implies x_0=-(3\cot(\pi/180)+\cos(\pi/180)\csc(\pi/180))/16$$ $$\implies y_0 =-1/4$$ Finally $d(x_0, y_0) =\sqrt{x_0^2+y_0^2} =1/(4\sin(\pi/180))$. Hence, the vertical line $$x=-1/(4\sin(\pi/180))\approx -14.325$$ will cross the tilted parabola only once.