Video: Finding the Equation of a Hyperbola Using the Asymptote’s Equations in a Real-World Context

Suppose that we model an object’s trajectory in the solar system by a hyperbolic path in the coordinate plane, with its origin at the sun and its units in astronomical units (Au). The π‘₯-axis is a line of symmetry of this hyperbola. The object enters in the direction of 𝑦 = 3π‘₯ βˆ’ 9, leaves in the direction of 𝑦 = βˆ’3π‘₯ + 9, and passes within 1 Au of the sun at its closest point. Using the equations of the asymptotes, find the equation of the object’s path.

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Video Transcript

Suppose that we model an object’s trajectory in the solar system by a hyperbolic path in the coordinate plane, with its origin at the sun and its units in astronomical units, Au. The π‘₯-axis is a line of symmetry of this hyperbola. The object enters in the direction of 𝑦 is three π‘₯ minus nine, leaves in the direction of 𝑦 is negative three π‘₯ plus nine, and passes within one Au of the sun at its closest point. Using the equations of the asymptotes, find the equation of the object’s path.

We’re told that the object’s path is a hyperbola, that it enters in the direction of 𝑦 is three π‘₯ minus nine and leaves in the direction of 𝑦 is negative three π‘₯ plus nine. So we can assume that these two equations are the equations of the asymptotes of the hyperbola. In both of these equations, when 𝑦 is equal to zero, we find that π‘₯ is equal to three, so that the lines intersect at the point with coordinates three, zero. Our first line has a slope of three and a 𝑦-intercept at negative nine. And our second asymptote has a slope of negative three and a 𝑦-intercept of plus nine.

We know that our object passes within one Au of the sun at its closest point. This means that one of our vertices must have coordinates one, zero, so that our hyperbola looks something like this. And we’re told that the π‘₯-axis is one of our lines of symmetry. If we take this is as our major or transverse axis, then our hyperbola is horizontal or east west–opening. And we’re asked to use the equations of the asymptotes to find the equation of the object’s path. Now we’ve established that the center of the hyperbola is at the point three, zero. So this is a hyperbola in standard position that’s been shifted three units in the positive π‘₯-direction, where a unit here is an astronomical unit.

Now we know that a hyperbola with center β„Ž, π‘˜ and with this orientation has the equation π‘₯ minus β„Ž squared over π‘Ž squared minus 𝑦 minus π‘˜ squared over 𝑏 squared is equal to one. And we concede that in our case, β„Ž is equal to three and π‘˜ is equal to zero. So that our equation will have the form π‘₯ minus three squared over π‘Ž squared minus 𝑦 squared over 𝑏 squared is equal to one. And all we need to do now is to find the values of π‘Ž and 𝑏. We can do this by using our knowledge of hyperbolas.

Now we know that for any east west–opening hyperbola, the slope of the asymptotes is positive and negative 𝑏 over π‘Ž, where π‘Ž and 𝑏 are both positive. In our case, 𝑏 over π‘Ž is equal to three, so that 𝑏 is equal to three π‘Ž. Just to make some room, I’ll move things round a little bit. We have 𝑏 equals three π‘Ž and π‘₯ minus three squared over π‘Ž squared minus 𝑦 squared over 𝑏 squared is equal to one. So now if we substitute 𝑏 is equal to three π‘Ž into our equation, we have π‘₯ minus three squared over π‘Ž squared minus 𝑦 squared over nine π‘Ž squared is equal to one.

Now we know that the hyperbola passes through the point one, zero because this is our vertex. So we know that when π‘₯ is equal to one, 𝑦 is equal to zero. And if we substitute these into our equation, we have negative two squared over π‘Ž squared is equal to one, so that π‘Ž squared is equal to four, and π‘Ž is equal to two since π‘Ž is positive. And remember that 𝑏 is equal to three times π‘Ž, so 𝑏 is equal to six. And the equation of the object’s path is π‘₯ minus three squared over four minus 𝑦 squared over 36 is equal to one.

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