5 Marks Questions

Question 15 Marks

Find the area of the region bounded by the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$.

Answer

$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$
$ = 4\int_0^2 {\frac{2}{3}\sqrt {9 - {y^2}} dx} $
$ = 4\int_0^2 {\frac{2}{3}\sqrt {{3^2} - {y^2}} dx} $
$=4×\frac{2}{3} \left[ {\frac{y}{2}\sqrt{9-y^2}+\frac{9}{2}sin^{-1}\frac{y}{3}} \right]_0^3$
$=4×\frac{2}{3} \left[ {\frac{3}{2}(0)+\frac{9}{2}sin^{-1}(1)- }\left[ {0-0} \right] \right]$
$=4×\frac{2}{3} \left[ {\frac{9}{2}(\frac{π}{2})} \right]$
$ = 6\pi $ sq. units

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Question 25 Marks

Find the area of the region bounded by the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$

Answer

The given equation of the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$, can be represented.
It can be observed that the ellipse is symmetrical about x-axis and y-axis.∴ Area bounded by ellipse = 4 × Area of OAB
Area of $OAB = \int_0^4 {ydx} $
$ = \int_0^4 {3\sqrt {1 - \frac{{{x^2}}}{{16}}} dx} $
$ = \frac{3}{4}\int_0^4 {\sqrt {16 - {x^2}} dx} $
$ = \frac{3}{4}\left[ {\frac{x}{2}\sqrt {16 - {x^2}} + \frac{{16}}{2}{{\sin }^{ - 1}}\frac{x}{4}} \right]_0^4$
$= \frac{3}{4}\left[ {2\sqrt {16 - 16} + 8{{\sin }^{ - 1}}(1) - 0 - 8{{\sin }^{ - 1}}(0)} \right]$
$ = \frac{3}{4}\left[ {\frac{{8\pi }}{2}} \right]$
$ = \frac{3}{4}\left[ {4\pi } \right]$ $ = 3\pi $
Therefore, are a bounded by the ellipse = 4 × 3$\pi$ = 12$\pi$units.

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Maths 5 Marks Questions

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