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# The area of the region bounded by the ellipse $\large\frac{x^2}{25}+\frac{y^2}{16}=1$ is \begin{array}{1 1}(A)\;2 \pi sq.units & (B)\;20{\pi}^2 sq.units\\(C)\;16{\pi}^2 sq.units & (D)\;25\pi sq.units \end{array}

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• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big)$$+c Step 1: Given curve is an ellipse \large\frac{x^2}{25}+\frac{y^2}{16}$$=1$.
The equation of the ellipse can be rewritten as $\large\frac{y^2}{16}$$=1-\large\frac{x^2}{25}=\frac{25-x^2}{25} \Rightarrow y^2=\large\frac{16}{25}$$(25-x^2)\Rightarrow y=\large\frac{4}{5}$$\sqrt{25-x^2} Clearly the ellipse is along the x-axis as the major axis and is bounded between x=5 and x=-5. Hence the required area is the shaded portion shown in the fig. Step 2: Area of the shaded portion is A=4\times \int_0^5\large\frac{4}{5}$$\sqrt{25-x^2}dx.$
$\qquad\qquad\qquad\qquad\qquad\quad=\large\frac{16}{5}\int_0^5$$\sqrt{25-x^2}dx. We know \int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big) On integrating we get, A=\large\frac{16}{5}\begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{25-x^2}+\large\frac{25}{2}\normalsize \sin^{-1}\big(\large\frac{x}{a}\big)\end{bmatrix}_0^5 Step 3: On applying limits we get, \;\;\;=\large\frac{16}{5}\begin{bmatrix}\large\frac{5}{2}\normalsize \sqrt{25-25}+\large\frac{25}{2}\normalsize \sin^{-1}\big(\frac{5}{5}\big)-\large\frac{0}{2}\normalsize\sqrt{25-0}+\large\frac{25}{2}\normalsize\sin^{-1}(0)\end{bmatrix} On simplifying we get, \;\;\;=\large\frac{16}{5}[\large\frac{25}{2}$$\sin^{-1}(1)]$
But $\sin^{-1}(1)=\large\frac{\pi}{2}$
Hence $A=\large\frac{16}{5}\times\large\frac{25}{2}\times\frac{\pi}{2}$=$20\pi$ sq.units.
Hence the correct option is A.
answered May 7, 2013