Video: Finding the Equation of a Hyperbola

A hedge is to be constructed in the shape of a hyperbola near a fountain at the center of a yard. The hedge will follow the asymptotes 𝑦 = (3/4)π‘₯ and 𝑦 = (βˆ’3/4)π‘₯, and its closest distance to the center fountain is 20 yards. Find the equation of the hyperbola.

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Video Transcript

A hedge is to be constructed in the shape of a hyperbola near a fountain at the center of a yard. The hedge will follow the asymptotes 𝑦 equals three-fourths π‘₯ and 𝑦 equals negative three-fourths π‘₯, and its closest distance to the center fountain is 20 yards. Find the equation of the hyperbola.

Here we have our hedge, and it’s following the asymptotes 𝑦 equals three-fourths π‘₯ and 𝑦 equals negative three-fourths π‘₯. Those are the dashed lines. Those are the lines that they approach but never actually touch.

And it says the closest distance to the center of the yard would be 20 yards, so there are a few things we should write down about hyperbolas, first using equation of a hyperbola that’s horizontal just like what we have. The asymptotes are in the form 𝑦 equals plus or minus 𝑏 over π‘Ž, which we know. Plus or minus three-fourths, we have the positive and the negative, so 𝑏- so 𝑏 could be three and π‘Ž could be four.

Now we put a question mark because three-fourths could be simplified; those numbers could actually be a little bit bigger, because of something called the traverse, which is this entire length, and it’s equal to two π‘Ž, so we know another place where π‘Ž shows up.

So if this entire length is two π‘Ž and that half is 20, so that half is 20, the whole thing is 40. So if we set two π‘Ž equal to 40 and then divide by two, we must have that π‘Ž is 20. So if π‘Ž is 20, π‘Ž would be 20, and we don’t know 𝑏, but we know that it should be equal to three-fourths, so we can set it equal to three-fourths.

So we can find the cross-product and solve for 𝑏. 𝑏 times four is four 𝑏, 20 times three is 60, and now we divide both sides by four, so 𝑏 is 15. So putting 𝑏 over π‘Ž, fifteen twentieths does indeed reduce to three-fourths, because you can divide both of those numbers by five and you get three-fourths, so π‘Ž is 20, 𝑏 is 15, so first π‘Ž squared, which is 20 squared, that’s 400, and then 𝑏 squared, which is 15 squared, is 225. And we set it equal to one, so this would be the equation of the hyperbola.

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