Question

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.

Answer 1

Since the height and width of the arc 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 $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,y\ge 0$ where $a$ is the semi-major axis.

Accordingly, $2a=8$

$a=4,b=2$

Therefore, the equation of the semi-ellipse is $\frac{x^2}{16}+\frac{y^2}{4}=1,y\ge 0$ ...(1)

Let $B$ be a point on the major axis such that $AB=1.5$ m

Draw $BC$ $\perp$ $OA$

$OB=(4-1.5)$ $m$ $=2.5$ $m$

The $x$-coordinate of point $C$ is $-2.5$ $m$.

On substituting the value of $x$ with $-2.5$ in equation (1), we obtain

$\frac{{(-2.5)}^2}{16}+\frac{y^2}{4}=1$

$\Rightarrow \frac{6.25}{16}+\frac{y^2}{4}=1$

$\Rightarrow y^2=4(1-\frac{6.25}{16})$

$\Rightarrow y^2=4\left(\frac{9.75}{16}\right)$

$\Rightarrow y^2=2.4375$

$\Rightarrow y=1.56$ (approx) $(\because y\ge 0)$

$\therefore AC=1.56$ $m$

Thus the height of the arch at a point $1.5$ m from one end is approximately $1.56$ $m$.