Solving a Conics Question: Bridge Arch Height at Point 30m from Center

[msimard8]

Hello,

I started this problem. I don't know really how to set it up. I attached my work. I know it is wrong but I do not know where. the correct answer is 42.2.

Here is the question:

A bridge over a river is supported by a hyperbolic arch which is 200 m wide at the base. The maximum height of the arch is 50 m. How high is the arch at a point 30 m from the center.

I drew a diagram (which I know is incorrect because my work assumes the center is (0,0)

Help.

If you have any ideas without looking at my work, anything will be appreciated. Thanks

 

[quantumdude]

Hi, sorry for the late response. Blame it on the summer!

Anyway let's take a look at the standard form of the equation of a hyperbola with a vertical transverse axis.

$$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$$

You've drawn one of the vertices at the origin, which is fine. But then you set $(h,k)=(0,0)$ which is not fine. Those are the coordinates of the center, which certainly does not coincide with either of the vertices. You've also misidentified $a$ and $b$. They are not the distances given in the problem.

Here's what I would do. Start from the diagram that you've drawn (with the vertex at the origin). That means that the center of the hyperbola is on the y-axis, which implies that $h=0$ in the above equation. Then use the 3 points on your diagram to find $a$, $b$, and $k$. You have 3 points and you need to find 3 constants. That should be feasible.

 

[Architeuthis Dux]

Hello,

I can see that you are trying to solve a conics question involving a bridge arch. Firstly, it's great that you have attempted to draw a diagram and set up the problem. However, as you mentioned, your diagram and work assume that the center of the arch is at (0,0), which may not be the case. To solve this problem, we need to use the equation of a hyperbola in standard form:

(x-h)^2/a^2 - (y-k)^2/b^2 = 1

where (h,k) is the center of the hyperbola, a is the distance from the center to the vertex, and b is the distance from the center to the asymptote.

In this case, we know that the center of the hyperbola is at (0,0) and the distance from the center to the vertex is 50 m. We can also see from the given information that the width of the base is 200 m, which means the distance from the center to the asymptote is 100 m. Now, we can plug these values into the equation and solve for the height at a point 30 m from the center.

(x-0)^2/50^2 - (y-0)^2/100^2 = 1

Simplifying, we get:

x^2/2500 - y^2/10000 = 1

Now, we can plug in x = 30 and solve for y:

(30)^2/2500 - y^2/10000 = 1

900/2500 - y^2/10000 = 1

-1600/10000 = -y^2/10000

y^2 = 1600

y = ±40

Since we are looking for the height at a point 30 m from the center, we will take the positive value, which is 40 m. Therefore, the height of the arch at a point 30 m from the center is 40 m.

I hope this helps and clarifies the process for solving this type of problem. Keep up the good work!