Question

If the area enclosed by the parabolas y² = 16ax and x² = 16ay, a > 0 is $\frac{1024}{3}$ square units, find the value of a.

Answer 1

The parabola y² = 16ax opens towards the positive x-axis and its focus is (4a, 0).

The parabola x² = 16ay opens towards the positive y-axis and its focus is (0, 4a).

Solving y² = 16ax and x² = 16ay, we get

$$\begin{gathered} {\left(\frac{x^2}{16a}\right)}^2=16ax \\ \Rightarrow x^4={\left(16a\right)}^{3}x \\ \Rightarrow x^4-{\left(16a\right)}^{3}x=0 \\ \Rightarrow x\left[x^3-{\left(16a\right)}^3\right]=0 \\ \Rightarrow x=0\mathrm{or}x=16a \end{gathered}$$

So, the points of intersection of the given parabolas are O(0, 0) and A(16a, 16a).



Area enclosed by the given parabolas

= Area of the shaded region

$$\begin{gathered} ={\int }_0^{16a}\sqrt{16ax}dx-{\int }_0^{16a}\frac{x^2}{16a}dx \\ =4\sqrt{a}\times {\overline{)\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}}}_0^{16a}-\frac{1}{16a}\times {\overline{)\frac{x^3}{{\displaystyle 3}}}}_0^{16a} \\ =\frac{8\sqrt{a}}{3}\left[{\left(16a\right)}^{\frac{3}{2}}-0\right]-\frac{1}{48a}\left[{\left(16a\right)}^3-0\right] \\ =\frac{8\sqrt{a}}{3}\times 64a\sqrt{a}-\frac{256a^2}{3} \\ =\frac{512a^2}{3}-\frac{256a^2}{3} \\ =\frac{256a^2}{3}\mathrm{square}\mathrm{units} \end{gathered}$$

But,

Area enclosed by the given parabolas = $\frac{1024}{3}$ square units (Given)
$$\begin{gathered} \therefore \frac{256a^2}{3}=\frac{1024}{3} \\ \Rightarrow a^2=\frac{1024}{256}=4 \\ \Rightarrow a=2\left(a>0\right) \end{gathered}$$

Thus, the value of a is 2.