Question

A railroad tunnel is shaped like a semiellipse



The height of the tunnel at the center is 58 ft and the vertical clearance must be 29 ft at a point 21 ft from the center. Find an equation for the ellipse.

Answer 1

y²/58² + x²/b² = 1 
(0,58), (-21,29), and (21,29) are points on the ellipse. 

29²/58² + 21²/b² = 1 
¼ + 21²/b² = 1 
21²/b² = ¾ 
4·21²/3 = b² 
b² = 588 


y²/3364 + x²/588 = 1

Answer 2

Answer:

Step-by-step explanation:

The equation of ellipse is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$       (1)

Now, from the given information, The ellipse passes through (0, 58), (0, -58), (21, 29), thus equation (1) becomes:

$\frac{0}{a^2}+\frac{(58)^2}{b^2}=1$

⇒$b^2=(58)^2$

⇒$b^2=3364$

Also, $\frac{(21)^2}{a^2}+\frac{(29)^2}{(58)^2}=1$

⇒$\frac{(21)^2}{a^2}=1-\frac{841}{3364}$

⇒$\frac{(21)^2}{a^2}=\frac{3}{4}$

⇒$a^2=\frac{4(441)}{3}$

⇒$a^2=588$

Now, substituting the values of $a^2 and b^2$ in the equation (1), we have

$\frac{x^2}{588}+\frac{y^2}{3364}=1$

which is the required equation for ellipse.