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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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An arch is in the form of a semi ellipse. It is 8m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

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The problem can be represented diagrammatically as follows:
We can see that the length of the major axis is 8m, and the minor axis is 4m.
$2a = 8 \rightarrow a = 4$ and $b = 2 \rightarrow b = 2$.
Therefore the equation of the semi-ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2} $$ = 1, y \geq 0 \rightarrow \large\frac{x^2}{16}$$+\large\frac{y^2}{4} $$ = 1$
We need to find $AC$, given $AB = 1.5m$.
$x$ coordinate of point $C$ is $2.5\;$i.e., $(OB - AB = 4 - 1.5 = 2.5)$
$\Rightarrow y\;$ coordinate (AC) $= \large\frac{2.5^2}{16}$$+\large\frac{y^2}{4}$ $ = 1 \rightarrow y^2 = \large\frac{16-6.25}{4}$$ = \large\frac{9.75}{4} $$ \rightarrow y \approx 1.56$
answered Apr 8, 2014 by balaji.thirumalai
 

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