Boxed parabolas

Take two points on a parabola, and draw the tangents to the parabola at these points. Then the area of the parabolic segment is equal to 2/3 the area of the triangle formed by the two points and the intersection of the tangents.

Hence, the area is simply $2/3 \cdot 12 = 8$ .

Instead of finding a parabola to fit a box, I'm going to start with a parabola and put a box around it.

The parabola will be of the form $x^2=4ay$ , so that any point on the parabola can be expressed in the parametric form $P=(2ap,ap^2)$ where $p$ is a parameter. From this, we know a bunch of important equations, which can be proven fairly easily:

• The equation of the chord between the points $P(2ap,ap^2)$ and $Q(2aq,aq^2)$ is $y=\frac{1}{2}(p+q)x-apq$
• The equation of the tangent to the parabola at a point P is $y=px+ap^2$
• The coordinates of the intersection of the tangents at points P and Q is $T(a(p+q),apq)$
• If the chord PQ is a focal chord (i.e. passes through the point $(0,a)$ ), then $pq=-1$
• Combining the two statements above, if the chord PQ is a focal chord then the intersection of the tangents of P and Q is $(a(p+q),-a)$

Using this we can find values of $a$ , $p$ and $q$ so that the tangents are of length 3 and 4 and meet at right angles. So we need to know the distance from each point on the parabola to the intersection of the tangents. $\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ $\begin{aligned}PT&=\sqrt{(2ap-a(p+q))^2+(ap^2+a)^2}&QT&=\sqrt{(2aq-a(p+q))^2+(aq^2+a)^2}\\ &=\sqrt{a^2(p-q)^2+a^2(p^2+1)^2}&&=\sqrt{a^2(q-p)^2+a^2(q^2+1)^2}\\ &=a\sqrt{(p-q)^2+(p^2+1)^2}&&=a\sqrt{(q-p)^2+(q^2+1)^2}\\ &&\text{But }pq=-1,\text{so }q=-\frac{1}{p}\\ PT&=a\sqrt{\left(p+\frac{1}{p}\right)^2+(p^2+1)^2}&QT&=a\sqrt{\left(-\frac{1}{p}-p\right)^2+\left(\frac{1}{p^2}+1\right)^2}\\ &=a\sqrt{\frac{1}{p^2}(p^2+1)^2+(p^2+1)^2}&&=a\sqrt{p^2\left(\frac{1}{p^2}+1\right)^2+\left(\frac{1}{p^2}+1\right)^2}\\ &=a\sqrt{(p^2+1)^2\left(\frac{1}{p^2}+1\right)}&&=a\sqrt{(p^2+1)\left(\frac{1}{p^2}+1\right)^2}\end{aligned}$ Now we simply let PT=3 and QT=4 and solve simultaneously for $a$ and $p$ $\begin{aligned}a\sqrt{(p^2+1)^2\left(\frac{1}{p^2}+1\right)}=3&&(1)\\a\sqrt{(p^2+1)\left(\frac{1}{p^2}+1\right)^2}=4&&(2)\\(1)÷(2)\end{aligned}$ $\begin{aligned}\sqrt{\frac{(p^2+1)^2\left(\frac{1}{p^2}+1\right)}{(p^2+1)\left(\frac{1}{p^2}+1\right)^2}}&=\frac{3}{4}\\ \frac{(p^2+1)}{\left(\frac{1}{p^2}+1\right)}&=\frac{9}{16}\\ \frac{p^2(p^2+1)}{(1+p^2)}&=\frac{9}{16}&\text{substitute }p\text{ into }(1)\\ p^2&=\frac{9}{16}&a\sqrt{\left(\left(\frac{3}{4}\right)^2+1\right)^2\left(\left(\frac{4}{3}\right)^2+1\right)}&=3\\ p&=\frac{3}{4}\nearrow&a&=\frac{144}{125}\end{aligned}$ Now we can find the equation of both the parabola and the chord PQ, which we will need in a moment. $\begin{aligned}x^2=4ay&&y=\frac{1}{2}(p+q)x-apq\\ y=\frac{x^2}{4a}&&y=\frac{1}{2}\left(p-\frac{1}{p}\right)x+a\\ y=\frac{125}{576}x^2&&y=-\frac{7}{24}x+\frac{144}{125}\end{aligned}$ We can also get the x coordinates of the points P and Q, which we will use as the terminals of an integral to find the area. $\begin{aligned}x_1&=2ap&x_2&=2aq\\ &=\frac{216}{125}&&=-\frac{2a}{p}\\ &&&=-\frac{384}{125}\end{aligned}$ Now we can finally put together an integral. The area we want is the curved shape shown here: The integral will simply be the equation of the chord minus the equation of the parabola, between the two points we worked out above: $\displaystyle\int_{-\frac{384}{125}}^\frac{216}{125}\left(-\frac{7}{24}x+\frac{144}{125}-\frac{125}{576}x^2\right)dx=4$ This is easy to evaluate using the power rule, albeit with some annoying fractions, but you should end up with an answer of 4. This is only half the area we need though, as we get the shape above from cutting the rectangle along the diagonal, so the final answer will be 8 .

Let's consider the rectangle in the $xy$ plane with $(0,0)$ , $(3,0)$ , $(3,4)$ , $(0,4)$ as vertices. We can find the parabola parametric equation using the Bèizer curve equation

$\displaystyle c(t)=\sum_{j=0}^{2} \binom{2}{j}t^j(1-t)^{2-j}\vec{p}_j$

where $\vec{p}_0=(3,0)$ , $\vec{p}_1=(0,0)$ , $\vec{p}_2=(0,4)$ . Solving the equation we have

$c(t)=(3t^2-6t+3,4t^2)$ , $t\in\mathbb{R}$ , $t\in[0,1]$ .

Now, the area of the interested region can be considered as the difference between the area of the rectangle and the area of the two triangoloid in the picture. So we can write

$\displaystyle Area= 12-2\int_{1}^{0} 4t^2(6t-6)dt=8$ .

The integral has been derived substituting $f(x)=4t^2$ and $x=3t^2-6t+3$ into $\int_{a}^{b} f(x)dx$ . I inverted the integral index because the curve is traveled from $(3,0)$ to $(0,4)$ , otherwise, using $\int_{0}^{1}$ the triangoloid area would have been negative.

I've got a cheating solution, making use of the format of the answer required being an integer. Look below and you will get it immediately. Basically you can see the integer that best fit this approximation is 8. Even if it was 7 or 9 you got 3 tries ;) .