Question

Find the area of the region bounded by x² + 16y = 0 and its latusrectum.

Answer 1

\[x^2 + 16 y = 0 \Rightarrow x^2 = - 16 y\]
\[\text{ Comparing it with equation of parabola }x^2 = 4ay \Rightarrow a = - 4\]
\[\text{ Thus, }x^2 + 16 y = 0 \text{ represents a parabola, opening downwards, with vertex at O(0, 0) and - ve }y -\text{ axis being its axis of symmetry }\]
\[\text{ Focus of the parabola is F(0, - 4)}\]
\[y = - 4 \text{ is the latus rectum of the parabola }\]
\[\text{ The latus rectum cuts the parabola at B }(8, - 4)\text{ and B'}( - 8, - 4)\]
\[x = 8\text{ cuts the }x -\text{ axis at A(8, 0) }\]
\[\text{ Area of the curve bound by latus rectum = Shaded area BOB'B }= 2 \left(\text{ Area OBF }\right) . . . \left( 1 \right)\]
\[\text{ Consider a vertical strip of length }= \left| y \right|\text{ and width }= dx \text{ in shaded area OAB such that point P}(x, y )\text{ lies on the parabola }\]
\[\text{ The area of the approximating rectangle }= \left| y \right| dx\]
\[\text{ But the approximating rectangle moves from }x = 0\text{ to }x = 8\]
\[ \therefore \text{ Area of the shaded region OAB }= \int_0^8 \left| y \right| dx \]
\[ \Rightarrow A = \int_0^8 \left| - \frac{x^2}{16} \right| dx ....................\left[ \because x^2 = - 16 y \Rightarrow y = - \frac{x^2}{16} \right] \]
\[ \Rightarrow\text{ Area of the shaded region OAB }= \int_0^8 \frac{x^2}{16} dx = \frac{1}{16} \times \frac{1}{3} \left[ x^3 \right]_0^8 = \frac{8 \times 8 \times 8}{16 \times 3} = \frac{32}{3} \text{ sq . units }. . . \left( 2 \right)\]
\[\text{ So, area of rectangle OABF = }\overline{OA} \times \overline{AB} = 8 \times 4 = 32\text{ sq . units . }. . \left( 3 \right)\]
\[\text{ From }\left( 2 \right)\text{ and }\left( 3 \right)\]
\[\text{ Area of OBF = Area of rectangle OABF - Area of the shaded region OAB }= 32 - \frac{32}{3} = \frac{64}{3}\text{ sq . units }\]
\[ \Rightarrow\text{ Shaded area BOB'B }= 2\left(\text{ Area OBF }\right) = 2 \times \frac{64}{3} = \frac{128}{3} \text{ sq . units }\]
\[\text{ Thus, area of the curve }x^2 + 16 y = 0\text{ bound by its latus rectum }= \frac{128}{3}\text{ sq . units }\]