# 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 park. The hedge will follow the asymptotes π¦ = 2/3 π₯ and π¦ = β2/3 π₯, and its closest distance to the center fountain is 12 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 park. The hedge will follow the asymptotes π¦ equals two-thirds π₯ and π¦ equals negative two-thirds π₯. And its closest distance to the center fountain is 12 yards. Find the equation of the hyperbola.

Much like circles and ellipses, hyperbolas have their own equations that you need to be familiar with. We can also find further information from this formula like the foci and transverse and conjugate axes. However, we have no need for these bits of information in this question.

Letβs start applying what we know to these equations. Firstly, weβre given the asymptotes are π¦ equals two-thirds π₯ and π¦ equals negative two-thirds π₯. Comparing these to our equation for the asymptotes of a generic hyperbola, we see that both π and β must be zero since our asymptotes have no constant.

Now that we have the values of π and β, we can use the general form of the vertices to help us find the value of π. β is zero and π is zero, so our vertices become zero plus or minus π and zero. Remember, the vertices are a bit like the turning points of each part of a hyperbola. Weβre told that the closest distance to the fountain is 12 yards, and we also know that the vertices must lie on the π₯-axis because the asymptotes cross at the origin.

Therefore, π must equal 12. All we need to find now is the value of π. Letβs go back to our equation for the asymptotes. We know that π is 12, and π over π must simplify to two-thirds to correspond to the equations weβve been given. Therefore, π must be eight since eight over 12 simplifies to two-thirds. Substituting this into our general formula for the equation of a hyperbola gives us π₯ squared over 12 squared minus π¦ squared over eight squared equals one.

The equation of the hyperbola is π₯ squared over 144 minus π¦ squared over 64 equals one.