Question

Find the area of the region common to the parabolas 4y² = 9x and 3x² = 16y.

Answer 1

\[3 x^2 = 16 y . . . \left( 1 \right)\text{ is a parabola with vertex at (0, 0) opening upwards and symmetrical about + ve }y -\text{ axis }\]
\[4 y^2 = 9x . . . \left( 2 \right)\text{ is a parabola with vertex at (0, 0) opening sideways and symmetrical about + ve }x -\text{ axis }\]
\[\text{ Solving the equations }\left( 1 \right)\text{ and }\left( 2 \right), \text{ we get the points of intersection of the two parabolas O(0, 0) and A(4, 3 ) }\]
\[\text{ Consider a vertical strip of length }= \left| y_2 - y_1 \right|\text{ and width }= dx\text{ such that }P(x, y_1 )\text{ lies on }\left( 1 \right)\text{ and }Q(x, y_2 )\text{ lies on }\left( 2 \right)\]
\[ \Rightarrow\text{ area of approximating rectangle }= \left| y_2 - y_1 \right| dx\]
\[ \text{ Approximating rectangle moves from }x = 0\text{ to }x = 4\]
\[ \Rightarrow\text{ Area of the shaded region }= \int_0^4 \left| y_2 - y_1 \right| dx = \int_0^4 \left( y_2 - y_1 \right) dx ..................\left[\text{ As, }\left| y_2 - y_1 \right| = y_2 - y_1\text{ for }y_2 - y_1 > 0 \right] \]
\[ \Rightarrow A = \int_0^4 \left( \sqrt{\frac{9}{4}x} - \frac{3}{16} x^2 \right) dx \]
\[ \Rightarrow A = \frac{3}{2} \int_0^4 \sqrt{x} dx - \frac{3}{16} \int_0^4 x^2 dx \]
\[ \Rightarrow A = \frac{3}{2} \left[ \frac{x^\frac{3}{2}}{\frac{3}{2}} \right]_0^4 - \frac{3}{16} \left[ \frac{x^3}{3} \right]_0^4 \]
\[ \Rightarrow A = 4^\frac{3}{2} - \left( \frac{1}{16} \times 4^3 \right)\]
\[ \Rightarrow A = 8 - 4 = 4 \text{ sq . units }\]
\[ \therefore \text{ Area bound by the two curves = 4 sq . units }\]