Question

The area bounded by the parabola y² = 4ax and x² = 4ay is ___________ .

Options

• \[\frac{8 a^3}{3}\]

• \[\frac{16 a^2}{3}\]

• \[\frac{32 a^2}{3}\]

• \[\frac{64 a^2}{3}\]

Answer 1

\[\frac{16 a^2}{3}\]

 

To find the point of intersection of the parabolas substitute \[y = \frac{x^2}{4a}\] in \[y^2 = 4ax\]  we get
\[\frac{x^4}{16 a^2} = 4ax\]
\[ \Rightarrow x^4 - 64 a^3 x = 0\]
\[ \Rightarrow x\left( x^3 - 64 a^3 \right) = 0\]
\[ \Rightarrow x = 0\text{ or }x = 4a\]
\[ \Rightarrow y = 0 \text{ or }y = 4a\]
Therefore, the required area ABCD,
\[A = \int_0^{4a} \left( y_1 - y_2 \right) d x ...........\left(\text{Where, }y_1 = 2\sqrt{ax}\text{ and }y_2 = \frac{x^2}{4a} \right)\]
\[ = \int_0^{4a} \left( 2\sqrt{ax} - \frac{x^2}{4a} \right) d x\]
\[ = \left[ \frac{4\sqrt{a}}{3} x^\frac{3}{2} - \frac{x^3}{12a} \right]_0^{4a} \]
\[ = \left[ \frac{4\sqrt{a}}{3} \left( 4a \right)^\frac{3}{2} - \frac{\left( 4a \right)^3}{12a} \right] - \left[ \frac{4\sqrt{a}}{3} \left( 0 \right)^\frac{3}{2} - \frac{\left( 0 \right)^3}{12a} \right]\]
\[ = \left[ \frac{4\sqrt{a}}{3}8 a^\frac{3}{2} - \frac{64 a^3}{12a} \right] - 0\]
\[ = \frac{32 a^2}{3} - \frac{16 a^2}{3}\]
\[ = \frac{16 a^2}{3}\text{ square units }\]