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
\[\left( \frac{x^2}{16a} \right)^2 = 16ax\]
\[ \Rightarrow x^4 = \left( 16a \right)^3 x\]
\[ \Rightarrow x^4 - \left( 16a \right)^3 x = 0\]
\[ \Rightarrow x\left[ x^3 - \left( 16a \right)^3 \right] = 0\]
\[ \Rightarrow x = 0\text{ or }x = 16a\]
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
\[= \int_0^{16a} \sqrt{16ax}dx - \int_0^{16a} \frac{x^2}{16a}dx\]
\[ = \left.4\sqrt{a} \times \frac{x^\frac{3}{2}}{\frac{3}{2}}\right|_0^{16a} - \left.\frac{1}{16a} \times \frac{x^3}{3}\right|_0^{16a} \]
\[ = \frac{8\sqrt{a}}{3}\left[ \left( 16a \right)^\frac{3}{2} - 0 \right] - \frac{1}{48a}\left[ \left( 16a \right)^3 - 0 \right]\]
\[ = \frac{8\sqrt{a}}{3} \times 64a\sqrt{a} - \frac{256 a^2}{3}\]
\[ = \frac{512 a^2}{3} - \frac{256 a^2}{3}\]
\[ = \frac{256 a^2}{3}\text{ square units }\]
But,
Area enclosed by the given parabolas = \[\frac{1024}{3}\] square units           ..........(Given)
\[\therefore \frac{256 a^2}{3} = \frac{1024}{3}\]
\[ \Rightarrow a^2 = \frac{1024}{256} = 4\]
\[ \Rightarrow a = 2 ............\left( a > 0 \right)\]

Thus, the value of a is 2.