Locus and area

I've found a solution slightly different than mr @Chris Lewis. Here it is:

Since the derivative of $y = x^2 + \frac14$ with respect to $x$ is equal to $2x$ , the tangents to the curve in a point $(t, t^2+\frac14)$ will have equation:

$\large \displaystyle y - \left (t^2 - \frac14 \right) = (2t)\cdot (x-t)$

$\large \displaystyle y = 2tx - t^2 + \frac14$

The line which is perpendicular to the aforementioned tangent line and passes to the origin must have inclination equal to the negative inverse inclilation of the line and constant term equal to 0. So:

$\large \displaystyle y = - \frac{1}{2t} x$

The point P $(x_P, y_P)$ of intersection among those lines is found by:

$\large \displaystyle 2tx - t^2 + \frac14 = - \frac{1}{2t} x$

$\large \displaystyle x_P = \frac{4t^3-t}{2(4t^2+1)}$

$\large \displaystyle y_P = \frac{-4t^2+1}{4(4t^2+1)}$

Each point of C will then be given by $2$ times vector $\overrightarrow{OP}$ :

$\large \displaystyle x = \frac{4t^3-t}{4t^2+1}$

$\large \displaystyle y = \frac{-4t^2+1}{2(4t^2+1)}$

This curve passes through $(0,0)$ when $t=-0.5$ , goes to $(0,0.5)$ when $t = 0$ and then goes back to $(0,0)$ when $t= 0.5$ , enclosing an area. The integral we're looking for is then:

$\large \displaystyle A = \left | \int y dx \right |$

$\large \displaystyle A = \left | \int_{-0.5}^{0.5} y \left ( \frac{dx}{dt} \right ) dt \right |$

$\large \displaystyle A = \left | \frac12 \int_{-0.5}^{0.5} \frac{-64t^6 -48t^4 + 20t^2 - 1}{(4t^2+1)^3} dt \right |$

The reason for the absolute value is that the integral itself can be positive or negative depending on the orientation chosen, but an area must be positive. Expanding in partial fractions:

$\large \displaystyle A = \left | \frac12 \left ( \int_{-0.5}^{0.5} \left [ \frac{8}{(4t^2+1)^2} - \frac{8}{(4t^2+1)^3} - 1 \right ] dt \right ) \right |$

$\large \displaystyle A = \left | \frac12 \left ( \int_{-0.5}^{0.5} \left [ \frac{8}{(4t^2+1)^2} - \frac{8}{(4t^2+1)^3} \right ] dt -1 \right ) \right |$

And then making $t = \frac{\tan(u)}{2}$ , one gets:

$\large \displaystyle A = \left | 2 \left ( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} ( \cos^2(u) - \cos^4(u) ) du \right ) - \frac12 \right |$

Since $\cos(4u) = 8 \cos^4(u) - 8 \cos^2(u) + 1$ , then $\cos^2(u) - \cos^4(u) = \frac18 (-\cos(4u)+ 1)$ . Then:

$\large \displaystyle A = \left | 2 \left ( \frac18 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (- \cos(4u) + 1) du \right ) - \frac12 \right |$

$\large \displaystyle A = \left | \frac14 \left ( - \frac{\sin(4u) }{4} + u \right )_{-\frac{\pi}{4}}^{\frac{\pi}{4}} - \frac12 \right |$

$\color{#20A900} \boxed{ \large \displaystyle A = \frac12 - \frac18 \pi }$

So:

$\color{#3D99F6} \large \displaystyle a = \frac12, b = \frac18 \rightarrow \boxed{ \large \displaystyle \frac{a}{b} = 4}$

A bit of a sketch solution but the missing derivations are easy to check.

The image of the origin reflected in the line $ax+by+c=0$ has coordinates $\left( \frac{-2ac}{a^2+b^2}, \frac{-2bc}{a^2+b^2} \right)$ .

The tangent to the curve $L$ at the point $P(t,t^2+\frac14)$ has equation $2tx-y+\frac14-t^2=0$ .

Putting these two together, we find the curve $C$ is defined parametrically by

$\begin{aligned} x&=t \cdot \frac{4t^2-1}{4t^2+1} \\ y&=-\frac12 \cdot \frac{4t^2-1}{4t^2+1} \end{aligned}$

Noting the similarity with the tangent-half angle substitution formulas , we put $2t=\tan \theta$ to get

$\begin{aligned} x&=-\frac12 \cos 2\theta \tan \theta \\ y&=\frac12 \cos 2\theta \end{aligned}$

The next useful observation is that

$\begin{aligned} \frac{x}{\sin \theta}&=-\frac12\cos 2\theta \sec \theta \\ \frac{y}{\cos \theta} &=\frac12 \cos 2\theta \sec \theta \end{aligned}$

so the curve in question can be obtained from the polar curve $C':\quad r=\frac{\cos 2\phi}{2\cos \phi}$ by applying a rotation through $90^{\circ}$ .

This transformation does not affect the enclosed area of $C'$ (which lies in the interval $-\frac{\pi}{4} \le \phi \le \frac{\pi}{4}$ ), so the required area (using the standard formula for areas in polar coordinates) is given by

$\begin{aligned} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac12 r^2 d\phi &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac12 \left(\frac{\cos 2\phi}{2\cos \phi}\right)^2 d\phi \\ &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac18 \frac{\left(2\cos^2 \phi-1 \right)}{\cos^2 \phi} d\phi \\ &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac18 \frac{4\cos^4 \phi-4\cos^2 \phi + 1 }{\cos^2 \phi} d\phi \\ &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac18 \left(4\cos^2 \phi-4 + \sec^2 \phi \right) d\phi \\ &=\frac12-\frac{\pi}{8} \end{aligned}$

and the final answer is $\boxed4$ .