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In a model, it is shown that an arc of a bridge is semielliptical with major axis horizontal. If the length of the base is $9 \mathrm{~m}$ and the highest part of the bridge is $3 \mathrm{~m}$ from horizontal; the best approximation of the height of the arch, $2 \mathrm{~m}$ from the centre of the base is
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The correct answer is:
$\frac{8}{3} \mathrm{~m}$
Equation of the semielliptical bridge
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ ...(1)
Here, $2 \mathrm{a}=9$
$\begin{array}{l}
\therefore \mathrm{a}=\frac{9}{2}, \quad \mathrm{~b}=3 \\
\therefore \frac{\mathrm{x}^{2}}{\frac{81}{4}}+\frac{\mathrm{y}^{2}}{9}=1
\end{array}$
$\Rightarrow \frac{4 \mathrm{x}^{2}}{81}+\frac{\mathrm{y}^{2}}{9}=1$ ...(2)
Here, $\mathrm{OQ}=2 \mathrm{~m}$, let $\mathrm{PQ}=\mathrm{y}_{1}$
$\therefore \mathrm{P}\left(2, \mathrm{y}_{1}\right)$
Since point P lies on the ellipse (2)
$\therefore \frac{4 \times 4}{81}+\frac{\mathrm{y}_{1}^{2}}{9}=1$
$\Rightarrow \frac{y_{1}^{2}}{9}=1-\frac{16}{81}=\frac{81-16}{81}=\frac{65}{81}$
$\Rightarrow \mathrm{y}_{1}^{2}=\frac{65}{9} \Rightarrow \mathrm{y}_{1}=\frac{\sqrt{65}}{3} \simeq \frac{8}{3} \mathrm{~m}$
Hence, best approximation of the height of the arch $=\frac{8}{3} \mathrm{~m}$
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ ...(1)
Here, $2 \mathrm{a}=9$
$\begin{array}{l}
\therefore \mathrm{a}=\frac{9}{2}, \quad \mathrm{~b}=3 \\
\therefore \frac{\mathrm{x}^{2}}{\frac{81}{4}}+\frac{\mathrm{y}^{2}}{9}=1
\end{array}$
$\Rightarrow \frac{4 \mathrm{x}^{2}}{81}+\frac{\mathrm{y}^{2}}{9}=1$ ...(2)
Here, $\mathrm{OQ}=2 \mathrm{~m}$, let $\mathrm{PQ}=\mathrm{y}_{1}$
$\therefore \mathrm{P}\left(2, \mathrm{y}_{1}\right)$
Since point P lies on the ellipse (2)
$\therefore \frac{4 \times 4}{81}+\frac{\mathrm{y}_{1}^{2}}{9}=1$
$\Rightarrow \frac{y_{1}^{2}}{9}=1-\frac{16}{81}=\frac{81-16}{81}=\frac{65}{81}$
$\Rightarrow \mathrm{y}_{1}^{2}=\frac{65}{9} \Rightarrow \mathrm{y}_{1}=\frac{\sqrt{65}}{3} \simeq \frac{8}{3} \mathrm{~m}$
Hence, best approximation of the height of the arch $=\frac{8}{3} \mathrm{~m}$
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