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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 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

$\begin {array} {1 1} (A)\;1.56 \: m & \quad (B)\;3.12 \: m \\ (C)\;0.78 \: m & \quad (D)\;2.43 \: m \end {array}$

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  • Equation of an ellipse whose major axis is along the x - axis is given by $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
Step 1 :
It is given that the height of the arc from the centre is 2m.
The width of the arc from the centre is 8 m
The length of the major axis is 8 m and length of the semi - minor axis is 2m.
Let the origin of the coordinates plane be the centre of the ellipse.
The major axis is along the x - axis.
The semi - ellipse is as shown in the fig.
Step 2 :
Since the major axis is along the x - axis, the equation of the semi ellipse is $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$. Where a is the semi major axis.
$ \therefore 2a = 8$
$ \Rightarrow a = 4$ and b = 2
$ \therefore $ Equation of the semi ellipse is
$ \large\frac{x^2}{16}$$+\large\frac{y^2}{4}$$=1$
$ y \geq 0$
Step 3 :
Let us take a point B on x - axis.
Such that AB = 1.5 m
$AC \perp 0B$
$ \therefore $ 0A = 4-1.5
= 2.5 m
$ \therefore $ The x - coordinates of point C is 2.5 .
Substituting the value of x with 2.5 in the equation of semi ellipse we get,
$ \large\frac{(2.5)^2}{16}$$+ \large\frac{y^2}{4}$$=1$
(i.e) $\large\frac{y^2}{4}$$=1- \large\frac{6.25}{16}$
$ \therefore y^2 = 4 \bigg( \large\frac{16-6.25}{16} \bigg)$
$ = \large\frac{9.75}{4}$
$ \therefore y^2 = 2.4375$
$ \Rightarrow y = 1.56$ ( approx )
$ \therefore AC = 1.56 m$
Thus the length of the arc at a point 1.5 m from one end is approximately 1.56 m
answered Jul 23, 2014 by thanvigandhi_1
 

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