Question

The area of the region bounded by the ellipse ${\displaystyle \frac{x^2}{25}+\frac{y^2}{16}}$ = 1 is ______.

विकल्प

• 20π sq.unit

• 20π² sq.units

• 16π² sq.units

• 25π sq.units

Answer 1

The area of the region bounded by the ellipse ${\displaystyle \frac{x^2}{25}+\frac{y^2}{16}}$ = 1 is 20π sq.unit.

Explanation:

Given equation of ellipse is ${\displaystyle \frac{x^2}{25}+\frac{y^2}{16}}$ = 1

⇒ ${\displaystyle \frac{y^2}{16}=1-\frac{x^2}{25}}$

⇒ y² = ${\displaystyle \frac{16}{25}(25-x^2)}$

∴ y = ${\displaystyle \frac{4}{5}\sqrt{25-x^2}}$

∴ Since the ellipse is symmetrical about the axes.

∴ Required area = ${\displaystyle 4\times {\int }_0^5\frac{4}{5}\sqrt{25-x^2} \text{d}x}$

= ${\displaystyle 4\times \frac{4}{5}{\int }_0^5\sqrt{{\left(5\right)}^2-x^2} \text{d}x}$

= ${\displaystyle \frac{6}{5}{[\frac{x}{2}\sqrt{{\left(5\right)}^2-x^2}+\frac{25}{2}{\sin }^{-1} \frac{x}{5}]}_0^5}$

= ${\displaystyle \frac{16}{5}[0+\frac{25}{2}\cdot {\sin }^{-1}\left(\frac{5}{5}\right)-0-0]}$

= ${\displaystyle \frac{16}{5}[\frac{25}{2}\cdot {\sin }^{-1}\left(1\right)]}$

= ${\displaystyle \frac{16}{5}[\frac{25}{2}\cdot \frac{\pi }{2}]}$

= 20π sq.unit