# The arch of a bridge is in the shape of a 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 .

Toolbox:
• Standard forms of equation of the ellipse with major axes 2a, minor axis 2b $(a >b)$ eccentricity e and $b^2=a^2(1-e^2)$ or $e^2=1-\large\frac{b^2}{a^2}$
• $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 • http://clay6.com/mpaimg/21_2_Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum \; (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/21_2_Toolbox1.png
• Foci$(0,\pm ae),$ center $(0,0)$,vertices $(0,\pm a)$
• End points of Latus Rectum $(\pm \large\frac{b^2}{a}$$,ae) and (\pm \large\frac{b^2}{a}$$,-ae)$
• Directrices $y=\pm \large\frac{a}{e}$.
• The major axis is $x=0$ (y- axis) and the minor axes is $y=0$ (x- axis)
Step 1:
Let $A'A$ be the span of the area.$C$ the centre of the ellipse(the origin) and $B$ the highest point.
$CA=CA'=20=a$ the length of the semi major axis.
$CB=16=b$ is the length of the semi minor axis.
Step 2:
The equation of the ellipse is $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}=$$1 Substitute the values of 'a' and 'b' we get, \large\frac{x^2}{400}+\frac{y^2}{256}$$=1$
Step 3:
Let $PM$ be the position $9$ feet to the right of $C$.
$P$ has coordinates $(9,y)$ substituting we get
$\large\frac{81}{400}+\frac{y^2}{256}$$=1 \large\frac{y^2}{256}=1-\large\frac{81}{400}=\large\frac{319}{400} y^2=\large\frac{256\times 319}{400} The height PM=\large\frac{\sqrt{256\times 319}}{400} \Rightarrow \large\frac{16}{20}$$\sqrt{319}$
$\Rightarrow \large\frac{4}{5}$$\sqrt{319}$ft
edited Jun 18, 2013