Question

The arch of a bridge is in the shape of semi-ellipse having a horizontal span of $40ft$ and $16ft$, high at the centre. How high is the arch, $9ft$ from the right or left of the centre?

Answer 1

Take the mid-point of the base as the centre $C(0,0)$.

Since the base is $40ft$, the vertices $A$ and $A^{\prime }$ are $(20,0)$ and $(-20,0)$.
Therefore $2a=40$

$\Rightarrow$ $a=20,b=16$
Equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\therefore \frac{x^2}{400}+\frac{y^2}{256}=1$
Let $y_1$ be the height of the arch from $9ft$ right of the centre.

$\therefore$ $(9,4)$ is a point on the semi-ellipse.
The equation of ellipse is $\frac{x^2+y^2}{a^2+b^2}=1$
$\Rightarrow \frac{9^2}{400}+\frac{y_1^2}{256}=1$
$\Rightarrow \frac{9^2}{400}+\frac{y_1^2}{256}=1$
$\Rightarrow \frac{y_1^2}{256}=1-\frac{81}{400}=\frac{319}{400}$
$\Rightarrow y_1^2=\frac{256\times 319}{400}$
$\Rightarrow y_1=\frac{16}{20}\sqrt{319}=\frac{4}{5}\sqrt{319}$
$=\frac{4}{5}\sqrt{319}ft$
$\therefore$ the height of the arch $9$ ft from the right of the centre is $=\frac{4}{5}\sqrt{319}ft$.