Question

Find the area of the region bounded by x² = 4ay and its latusrectum.

Answer 1

\[x^2 = 4ay\text{ represents a parabola with vertex O(0, 0) opening upwards and symmetrical about } y - \text{ axis . }\]
\[F(0, a )\text{ is the focus of the parabola and }y = \text{ a its latus rectum }\]
\[\text{ Consider a horizontal strip of length }= \left| x \right| \text{ and width dy in the first quadrant }\]
\[ \therefore\text{ Area of approximating rectangle }= \left| x \right| dy\]
\[\text{ The approximating rectangle moves from }y = 0\text{ to } y = a \]
\[ \Rightarrow \text{ Area OAB }= \int_0^a \left| x \right| dy\]
\[ \Rightarrow \text{ Area OAA'O }= 2 \times\text{ Area OAB }\]
\[ \Rightarrow A = 2 \int_0^a \left| x \right| dy = \int_0^a x dy ................\left[ As, x > 0 \Rightarrow \left| x \right| = x \right]\]
\[ \Rightarrow A = 2 \int_0^a \sqrt{4ay} dy\]
\[ \Rightarrow A = 2 \times 2\sqrt{a} \int_0^a \sqrt{y} dy\]
\[ \Rightarrow A = 4\sqrt{a} \left[ \frac{y^\frac{3}{2}}{\frac{3}{2}} \right]_0^a \]
\[ \Rightarrow A = 4\sqrt{a} \frac{2}{3}\left[ a^{{}^\frac{3}{2}} - 0 \right]\]
\[ \Rightarrow A = \frac{8}{3} a^2\text{ sq . units }\]