An arch is in the form of a semi-cllipse. 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.
An arch is in the form of a semi-cllipse. 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.
since the height and width of the are from the centre is $2\, m$ and $8\, m$ respectively, it is clear that the length of the major axis is $8\, m ,$ while the length of the semi-minor axis is $2 \,m$ The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the $x-$ axis. Hence, the semi-ellipse can be diagrammatically represented as
The equation of the semi-ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 y \geq 0,$ where a is the semimajor axis
Accordingly, $2 a=8 \Rightarrow a=4$ $b=2$
Therefore, the equation of the semi-ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1, y \geq 0$ ....... $(1)$
Let A be a point on the major axis such that $AB =1.5 \,m$
Draw $AC \perp OB$.
$OA =(4-1.5)\, m =2.5 \,m$
The $x-$ coordinate of point $C$ is $2.5$
On substituting the value of $x$ with $2.5$ in equation $(1),$ we obtain
$\frac{(2.5)^{2}}{16}+\frac{y^{2}}{4}=1$
$\Rightarrow \frac{6.25}{16}+\frac{y^{2}}{4}=1$
$\Rightarrow y^{2}=4\left(1-\frac{6.25}{16}\right)$
$\Rightarrow y^{2}=4\left(\frac{9.27}{16}\right)$
$\Rightarrow y^{2}=2.4375$
$\Rightarrow y=1.56 $ (approx.)
$\therefore AC =1.56 \,m$
Thus, the height of the arch at a point $1.5 \,m$ from one end is approximately $1.56 \,m$
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