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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the area of the smaller region bounded by the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}=$$1 $ and the straight line $\large\frac{x}{a}+\frac{y}{b}$$=1.$

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Toolbox:
  • Suppose we are given two curves represented by $y=f(x)$ and $y=g(x)$,where $f(x)\geq g(x)$ in [a,b] the point of intersection of these two curves are given by $x=a$ and $x=b$,by taking the common values of y from the given equation of two values.
  • The required area $A=\int_a^b[f(x)-g(x)]dx.$
Step 1:
The area of the smaller region bounded by the ellipse $\large\frac{x^2}{a^2}+\large\frac{y^2}{b^2}$$=1,$and the line $\large\frac{x}{a}+\frac{y}{b}$$=1$
Hence the required area is the area enclosed between the straight line and the ellipse.
To find the limits,the semi major axis extends from 0 to a and the x intercept of the straight line is also a.Hence the limits are 0 to a.
The required area is $A=\int_a^b[f(x)-g(x)]dx$
Step 2:
Here a=0,b=a
$f(x)=\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
$g(x)=\large\frac{x}{a}+\frac{y}{b}$$=1$
$A=\int_0^a\frac{b}{a}\sqrt{a^2-x^2}dx-\frac{b}{a}\int_0^a(a-x)dx$
Integrating we get,
$A=\large\frac{b}{a}\left \{\begin{bmatrix}\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\end{bmatrix}-\begin{bmatrix}ax-\frac{x^2}{2}\end{bmatrix}\right\}_0^a$
Step 3:
$A=\large\frac{b}{a}\left \{\begin{bmatrix}\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\end{bmatrix}-\begin{bmatrix}ax-\frac{x^2}{2}\end{bmatrix}\right\}_0^a$
$\;\;\;=\large\frac{b}{a}\bigg[\frac{a^2\pi}{4}-\frac{a^2}{2}\bigg]$
$\;\;\;=\large\frac{ba^2}{2a}\bigg[\frac{\pi}{2}$$-1\bigg]$
$\;\;\;=\large\frac{ab}{2}\bigg[\frac{\pi}{2}-1\bigg]=\frac{ab}{4}$$(\pi-2)sq.units.$
Hence the required area is $\large\frac{ab}{4}$$(\pi-2)sq. units.$
answered Oct 1, 2013 by sreemathi.v
 

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