In Exercises 27–32, find the standard form of the equation of eac...

Question

In Exercises 27–32, find the standard form of the equation of each hyperbola.

Accepted Answer

Hey, everyone here we are given a graph of a hyperbola. And we are asked to determine the standard form of its equation. Here we have four answer choice options, answer choice A Y squared divided by five minus X squared, divided by seven is equal to one answer. Choice by squared divided by seven minus X squared, divided by five is equal to one cy squared divided by 25 minus X squared, divided by 49 is equal to one. And lastly answer choice dy squared divided by 49 minus X squared divided by 25 is equal to one. So now taking a look at our given graph here, the first thing we want to observe is that the center of the hyperbola is at the origin. So the center of our hyperbola is at the 0.00. So now with this center identified, we can recall the standard form of the equation of a hyper with the center at the origin which are X squared divided by A squared minus Y squared, divided by B squared is equal to one or we also have Y squared divided by A squared minus X squared divided by B squared is equal to one. So now we also can notice here with our formulas that the vertices are a units from the center and the are C units from the center. So now we want to look back back at our graph again and we see that the vertices are the points zero and seven and the 0.0 and negative seven. And so, since the vertices lie on the y axis, then the transverse axis also lies on the Y axis. And so we know the equation of the hyperbole that we are using for our graph is going to be the second formula where we have Y squared divided by A squared minus X squared divided by B squared is equal to one. And so now like also from the vertices, since they are 07 and zero negative seven, we see here that the value of A is just going to be seven. So we have already identified our value for A and now to find our value for B, we need to look back at, at our graph again and we see that the rectangle on the graph passes through the points negative five and zero and the 0.50. And we see that both of these points are five units to the left and five units to the right of the center. And so this tells us that B is going to be five. So now we have identified the correct formula for us to use as well as the values for A and B. So all we need to do is substitute the value for A and the value for B into our formula here. So doing this, we have Y squared divided by seven squared minus X squared divided by five squared is equal to one. Now, we just want to simplify our equation by squaring the values in the denominator. So simplifying me, see our final equation becomes Y squared divided by 49 minus X squared divided by 25 is equal to one. So now with this finalized sim and simplified equation here we see we are left with answer choice D where again our final equation is Y squared divided by 49 minus X squared divided by 25 is equal to one. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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In Exercises 27–32, find the standard form of the equation of each hyperbola.