A parabola with a twist

Instead of finding a formula for the rotated parabola, we can just rotate the x and y axes instead.

One of the axes when rotated by $\frac\pi5$ will have a slope of $\tan\frac\pi5$ (it doesn't matter whether this is positive or negative as the parabola is symmetrical). For simplicity, let this be the constant $t=\tan\frac\pi5$ . The other axis, which is perpendicular, will have a slope of $-\frac1t$ .

Now we need to find the lines with these slopes that are tangent to the parabola. To do this, we will differentiate the parabola to get $y'=\frac x8$ , and substitute in the required slopes:

$\begin{aligned}t&=\frac{x_1}8&-\frac1t&=\frac{x_2}8\\ x_1&=8t&x_2&=-\frac8t\\ y_1&=\frac{(8t)^2}{16}=4t^2&y_2&=\frac{\left(-\frac8t\right)^2}{16}=\frac4{t^2}\\\\ y-y_1&=m(x-x_1)&y-y_2&=m(x-x_2)\\ y-4t^2&=t(x-8t)&y-\frac4{t^2}&=-\frac1t\left(x+\frac8t\right)\\ y-4t^2&=tx-8t^2&y-\frac4{t^2}&=-\frac1tx-\frac8{t^2}\\ y&=tx-4t^2&y&=-\frac xt-\frac4{t^2}\end{aligned}$

We already know that these lines intersect the parabola at $x_1=8t$ and $x_2=-\frac8t$ respectively. Lastly, we need their intersection point:

$\begin{aligned}tx-4t^2&=-\frac xt-\frac4{t^2}\\ xt+\frac xt&=4t^2-\frac4{t^2}\\ x\left(t+\frac1t\right)&=4\left(t+\frac1t\right)\left(t-\frac1t\right)\\ x&=4\left(t-\frac1t\right)\end{aligned}$

Now we can use these x-coordinates as the terminals to integrals between the parabola and the lines to find the total area:

$\text{Area}=\int_{-\frac8t}^{4\left(t-\frac1t\right)} \frac{x^2}{16}-\left(-\frac xt-\frac4{t^2}\right)dx+\int_{4\left(t-\frac1t\right)}^{8t} \frac{x^2}{16}-\left(tx-4t^2\right)dx$

Evaluating these integrals gives $\text{Area}\approx\boxed{24.8}$ .

Retaining the original $x$ - and $y$ -axes of the parabola $y=ax^2$ , the graphs are as shown above. Instead of the parabola tilts $\frac \pi 5$ clockwise, it is equivalent to the new $x'$ -axis (orange line) tilts $\frac \pi 5$ anticlockwise. Instead of the parabola translates up and to the right, the $x'$ -axis translates down and to the left until it is tangent to the parabola.

Therefore, $x'$ -axis has a gradient of $-\cot \frac \pi 5$ and so is the point on the parabola it is tangent to. Since the gradient of the parabola is $\dfrac {dy}{dx} = 2ax = \dfrac x8 = - \cot \frac \pi 5$ , $\implies x = - 8\cot \frac \pi 5$ and $y = ax^2 = 4\cot^2 \frac \pi 5$ . Therefore, the point of tangent is $\left(-8\cot \frac \pi 5, 4 \cot^2 \frac \pi 5\right)$ . From $\dfrac {y-4\cot^2 \frac \pi5}{x+8\cot \frac \pi5} = - \cot \frac \pi 5$ , the equation of $x'$ -axis is $y = - x\cot \frac \pi 5 - 4\cot^2 \frac \pi 5$ .

The $y'$ -axis (grey line) is perpendicular to $x'$ -axis, hence it has a gradient of $\tan \frac \pi 5$ . The point of tangent is $\left(8\tan \frac \pi 5, 4 \tan^2 \frac \pi 5\right)$ and the equation of $y'$ -axis is $y = x\tan \frac \pi 5 - 4\tan^2 \frac \pi 5$ .

From $- x\cot \frac \pi 5 - 4\cot^2 \frac \pi 5 = x\tan \frac \pi 5 - 4\tan^2 \frac \pi 5$ , we find that $x'$ -axis and $y'$ -axis intersect at $x=4\left(\tan \frac \pi 5 - \cot \frac \pi 5\right)$ .

Then the area bounded by the parabola, $x'$ -axis, and $y'$ -axis is given by:

$\begin{aligned} A & = \int_{4\left(\tan \frac \pi 5 - \cot \frac \pi 5\right)}^{8\tan \frac \pi 5} \left(ax^2 - \left(x\tan \frac \pi 5 - 4\tan^2 \frac \pi 5\right)\right) dx + \int^{4\left(\tan \frac \pi 5 - \cot \frac \pi 5\right)}_{-8\cot \frac \pi 5} \left(ax^2 - \left(-x\cot \frac \pi 5 - 4\cot^2 \frac \pi 5\right)\right) dx \\ & = \int_{-8\cot \frac \pi5}^{8\tan \frac \pi 5} \frac {x^2}{16} dx - \int_{4\left(\tan \frac \pi 5 - \cot \frac \pi 5\right)}^{8\tan \frac \pi 5} \left(x\tan \frac \pi 5 - 4\tan^2 \frac \pi 5\right) dx + \int^{4\left(\tan \frac \pi 5 - \cot \frac \pi 5\right)}_{-8\cot \frac \pi 5} \left(x\cot \frac \pi 5 + 4\cot^2 \frac \pi 5\right) dx \\ & = \frac {x^3}{48} \bigg|_{-8\cot \frac \pi5}^{8\tan \frac \pi 5} - \left[\frac {x^2}2 \tan \frac \pi 5 - 4x\tan^2 \frac \pi 5\right]_{4\left(\tan \frac \pi 5 - \cot \frac \pi 5\right)}^{8\tan \frac \pi 5} + \left[\frac {x^2}2 \cot \frac \pi 5 + 4x\cot^2 \frac \pi 5\right]^{4\left(\tan \frac \pi 5 - \cot \frac \pi 5\right)}_{-8\cot \frac \pi 5} \\ & \approx \boxed{24.799} \end{aligned}$

The parametric equation of the parabola is,

$\mathbf{r}(t) = \mathbf{d} + \mathbf{R} [t, a t^2]^T \hspace{12pt} (1)$

where

$\mathbf{R} = \mathbf{R}(\theta) = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} \hspace{12pt} (2)$

(our angle of rotation $\theta = -\dfrac{\pi}{5}$ ) and $\mathbf{d}=[d_x, d_y]^T$ is the coordinate vector of the vertex. The parameter is $t \in \mathbb{R}$ .

The tangent to the parabola is obtained by differentiating $\mathbf{r}(t)$ with respect to $t$ ,

$\mathbf{r'}(t) = \mathbf{R} [1, 2 a t]^T \hspace{12pt} (3)$

Expanding the right hand side of (3), yields,

$\mathbf{r'}(t) = [ c - 2a t s , s + 2a t c ]^T \hspace{12pt} (4)$

where $c = \cos \theta$ , and $s = \sin \theta$ .

Setting the y-coordinate of $\mathbf{r'}(t)$ to 0 (horizontal tangent), yields

$t_1 = - \dfrac{1}{2a} \tan \theta \gt 0 \hspace{12pt} (5)$ ,

and setting the x-coordinate of $\mathbf{r'}(t)$ to 0 (vertical tangent), yields

$t_2 = \dfrac{1}{2a} \cot \theta \lt 0 \hspace{12pt} (6)$

Further, at $t_1$ , the y-coordinate is zero, so

$d_y + [0, 1] \mathbf{R} [t_1, at_1^2]^T = 0 \hspace{12pt} (7)$

and at $t_2$ , the x-coordinate is zero, so that,

$d_x + [1, 0] \mathbf{R} [t_2 , at_2^2]^T = 0 \hspace{12pt} (8)$

hence, we have found the vertex $\mathbf{d}$ . (You can check that $d_x = 4.454065458$ ,and $d_y =1.70820393$ )

The area bounded by a parametric curve $\mathbf{p}(t) = [x(t), y(t)]^T$ over the interval $[t_1, t_2]$ and the line segments connecting the origin with $\mathbf{p}(t_1)$ and the origin with $\mathbf{p}(t_2)$ is given by the famous formula,

$A = \dfrac{1}{2} \left| \displaystyle \int_{t_1}^{t_2} \left( x(t) y'(t) - x'(t) y(t) \right) dt \right| \hspace{24pt} (9)$

For our curve,

$x(t) = d_x + c t - a s t^2$

$y(t) = d_y + s t + a c t^2$

And, the derivatives are,

$x'(t) = c - 2 a s t$

$y'(t) = s + 2 a c t$

Plugging this into the above formula gives,

$A = \dfrac{1}{2} \left| \displaystyle \int_{t_1}^{t_2} \left( (d_x + c t - a s t^2 )(s + 2 a c t ) - (d_y + s t + ac t^2)(c - 2 a s t ) \right) dt \right|$

Expanding, simplifying and re-arranging, this becomes,

$A = \dfrac{1}{2} \left| \displaystyle \int_{t_1}^{t_2} \left( (s dx - c dy ) + 2a t ( c dx + s dy ) + a t^2 ) \right) dt \right|$

Now this is straightforward to integrate, and it evaluates to $\boxed{ \approx 24.8 }$

Any area bounded by the $x$ -axis, $y$ -axis, and a parabola in the form of $y = ax^2$ that is rotated by $\theta$ and translated so that it becomes tangent to the $x$ and $y$ axes is always $\frac{1}{6}$ the area of the rectangle defined by origin and the two tangent points, no matter what the value of $a$ or $\theta$ is. The proof is as follows:

Instead of rotating the parabola $y = ax^2$ , rotate the $x$ - and $y$ - axes by $\theta$ instead, and draw rectangle $ADBE$ defined by the origin and the two tangent points, and draw parallelogram $ABFG$ so that $FG$ is tangent to the parabola at $C$ and $BF$ and $AG$ are parallel to the $y$ -axis.

The rotated $x$ -axis has a slope of $\tan \theta$ and the rotated $y$ -axis has a slope of $-\cot \theta$ . The slopes of the parabola $y = ax^2$ is defined by $m = y' = 2ax$ , so at $A$ it is $2ax = \tan \theta$ and this solves to coordinates $A(\frac{\tan \theta}{2a}, \frac{\tan^2 \theta}{4a})$ , and at $B$ it is $2ax = -\cot \theta$ and this solves to coordinates $B(-\frac{\cot \theta}{2a}, \frac{\cot^2 \theta}{4a})$ . The rotated $x$ -axis follows a line with the equation $y - \frac{\tan^2 \theta}{4a} = \tan \theta (x - \frac{\tan \theta}{2a})$ and the rotated $y$ -axis follows a line with the equation $y - \frac{\cot^2 \theta}{4a} = -\cot \theta (x - \frac{\cot \theta}{2a})$ , which intersects at $D(\frac{\tan \theta - \cot \theta}{4a}, -\frac{1}{4a})$ . Using these coordinates and the distance formula, segment $AD$ has a length of $\frac{1}{4a \sin \theta \cos^2 \theta}$ and segment $BD$ has a length of $\frac{1}{4a \sin^2 \theta \cos \theta}$ , which means rectangle $ADBE$ has an area of $A_{ADBE} = \frac{1}{16a^2 \sin^3 \theta \cos^3 \theta}$ .

Using the slope formula, the slope between $A$ and $B$ is $\frac{1}{2}(\tan \theta - \cot \theta)$ , and the equation of the line along $AB$ is $y - \frac{\tan \theta}{2a} = \frac{1}{2}(\tan \theta - \cot \theta)(x - \frac{\tan^2 \theta}{4a})$ or $y = \frac{1}{2}(\tan \theta - \cot \theta)x + \frac{1}{4a}$ . The slopes of the parabola $y = ax^2$ is defined by $m = y' = 2ax$ , so at $C$ it is $2ax = \frac{1}{2}(\tan \theta - \cot \theta)$ and this solves to coordinates $C(\frac{\tan \theta - \cot \theta}{4a}, \frac{(\tan \theta - \cot \theta)^2}{16a})$ . The equation of the line along $FG$ through $C$ is $y - \frac{(\tan \theta - \cot \theta)^2}{16a} = \frac{1}{2}(\tan \theta - \cot \theta)(x - \frac{\tan \theta - \cot \theta}{4a})$ or $y = \frac{1}{2}(\tan \theta - \cot \theta)x - \frac{(\tan \theta - \cot \theta)^2}{16a}$ . The height of parallelogram $ABFG$ is the difference of the $y$ -intercepts of the lines along $AB$ and $FG$ , which is $h_{ABFG} = \frac{1}{4a} - -\frac{(\tan \theta - \cot \theta)^2}{16a} = \frac{1}{16a \sin^2 \theta \cos^2 \theta}$ . The base of parallelogram $ABFG$ is the difference of the $x$ -coordinates of $A$ and $B$ , which is $b_{ABFG} = \frac{\tan \theta}{2a} - \frac{\cot \theta}{2a} = \frac{1}{2a \sin \theta \cos \theta}$ . The area of parallelogram $ABFG$ is $A_{ABFG} = b_{ABFG} \cdot h_{ABFG} = \frac{1}{32a^2 \sin^3 \theta \cos^3 \theta}$ , so the area of the region in the parabola enclosed by the parallelogram is $A_{para} = \frac{2}{3}A_{ABFG} = \frac{1}{48a^2 \sin^3 \theta \cos^3 \theta} = \frac{1}{3}A_{ADBE}$ .

The area of the shaded region is the difference between the area of $\triangle ADBE$ and the parabola enclosed by parallelogram $ABFG$ , or $A_{s} = \frac{1}{2}A_{ADBE} - \frac{1}{3}A_{ADBE} = \frac{1}{6}A_{ADBE} = \frac{1}{96a^2 \sin^3 \theta \cos^3 \theta}$ . Therefore, the area bounded by the $x$ -axis, $y$ -axis, and a parabola in the form of $y = ax^2$ that is rotated by $\theta$ and translated so that it becomes tangent to the $x$ and $y$ axes is always $\frac{1}{6}$ the area of the rectangle defined by origin and the two tangent points.

In this problem, $a = \frac{1}{16}$ and $\theta = \frac{\pi}{5}$ , so the area of the shaded region is $A_{s} = \frac{1}{96(\frac{1}{16})^2 \sin^3 \frac{\pi}{5} \cos^3 \frac{\pi}{5}} \approx \boxed{24.8}$ .