Test Your Hyperbola Skills!Here is your chance to test your hyperbola knowledge and skills after our lesson on the conic. This is an application problem and is used in an attempt to bring real world situations to this area of mathematics. Luckily, as is the case for many conics, there is prevalent hyperbola usage around the world everyday. From contemporary artwork and home furnishing, to the LORAN marine navigation system, hyperbolas are just as exciting as any other conic. In this case, we will examine the hyperbolic shape of semifloating automotive axles. Don't be intimidated if you have no idea what a car axle is, you soon will find out. Let's get started!

The SemiFloating Automotive Axle
In general terms, an axle is a shaft on which your car's wheels revolve. Picture it as a long stemmed rose with the rose head as the wheel, attached to the stem, the metaphorical axle. A semifloating axle is a special kind of axle that is used to do many tasks at once such as holding your wheels, driving them, and supporting them. The picture on the right is an example of a semifloating axle. On the interior of the blue rubber dust covers, opposing parabolas lie along the shaft, thus creating a hyperbola. Again, in general terms, the parabolas are knuckles which connect to gearing at the end of the shaft to lighten the tension load before they must support and drive the wheels. In this application problem we will attempt to help some dumbfounded engineers at completing a new product. Let's get started! 
The Question!
The engineers do not seem to be the brightest in the bunch, but have decided that they want to design a new set of axles for the automotive market. After attempting to steal the blueprints of another company's design, all they could manage to get was some data on the conjugate axis and asymptotes, all of which they have ruled unhelpful. Luckily they have our extensive conic knowledge to aid in their design process. They know that the length of the conjugate axis is 6 inches, top to bottom, while the equation for one of the asymptotes is "y = 1/3(x)." They need to find the foci of the hyperbola in order to have a place to connect the tension gears. They also need to know the length of the shaft, from vertex to vertex to build the proper product.. To the right is a diagram of this data so that we can solve this problem and help save this company from bankruptcy. Good luck! 
The Answer!
This type of question, while seemingly difficult, is much easier than it seems once you have grasped a knowledge of basic hyperbolas. To find the foci we must first find our equation and some variables. Since our conjugate runs along the yaxis ( "top to bottom"), we know that b will equal half of its length, thus b = 3 inches. to find a we must use the property that an asymptote is expressed as "y = + b/a (x)." Since we know that the positive asymptote is y = 1/3 (x), we know can figure that the slope has been reduced, since b is 3, not 1. To get the correct answer for a we just need to multiply 3, the reduced a value, by 3, to get our real answer of a = 9 inches. We now have our equation: "(x)^2/81)  (y)^2/9) = 1." We can find the length of the shaft by knowing the length of the transverse axis, vertex to vertex, is equal to 2a, thus the shaft equals 18 inches. To find the foci where the engineers must place the tension gears we must use our equation of "b^2 = c^2 a^2." Now we just plug in values to solve for c: "9 = c^2 81." Adding 81 to the left side and taking the square root of both gives of a c value of c = 9.5 inches. The engineers now know that their shaft must be 18 inches, and the tension gears must be set 9.5 inches left and right of the center of the shaft. Good job! 