In Exercises 5–12, find the standard form of the equation of each...

Question

In Exercises 5–12, find the standard form of the equation of each hyperbola satisfying the given conditions. Center: (4, −2); Focus: (7, −2); vertex: (6, −2)

Accepted Answer

Hello everyone in this video. We're going to be looking at this practice problem. We want to find the standard form equation for hyperbole to with the conditions that it has the center at three negative one. Focus at nine negative one and a vertex at seven negative one. And for this we want to remember what the standard form for the hyperbole. It looks like, remember it's x minus h squared divided by a squared minus y minus k squared divided by b squared. And that is equal to one. And in this case remember that the center is H. K and A and B are the lengths of the semi major and semi minor axis axes. So in terms of the information that we were given, we can plug in the center Where we have a church is equal to three and K is equal to negative one. So if I plug those values in, I have x minus H which is three squared divided by a squared minus y minus K, which is negative one squared divided by b squared is equal to one. If I simplify that second fraction, I'll still have x minus three squared divided by a squared minus. Why minus negative one becomes plus one squared divided by B squared is equal to one and notice I still need to find A and I still need to find B. In order to find A and B. Let's take a look at the other information that we're given. So we're also given the focus and the vertex. So for the standard form call that the folk I are given in terms of H plus or minus C and K. And the vertex are H plus or minus A. Okay, I should say vortices not for a text. So using that we're going to determine the values of C and a since we're given a focus and a vertex, so our focus is nine negative one and that is the equivalent of H plus C. Okay. So we've already established from our center that K is equal to negative one. And now we have a church plus C is equal to nine. And from our center we know that H is equal to three so I can plug in three into H. So I have three plus C equals nine. Once I subtract the three I get C is equal to six and I can do the same sort of solving using my vertex where I have seven negative one and that is the equivalent of H plus A and K. So I still have K. is equal to negative one and now I have a church Plus A is equal to seven. And notice how I have these two equations that both have the same sort of format and same as before, H is still equal to three. So when I plug in three for this H All subtract three from both sides. Leaving me with a is equal to four. So now I have a value for C and a value for a but I still need my B value and we can determine that using the equation, C squared is equal to a squared plus B squared. So in this case I have all the values except for B. So if I solve for B, this becomes b squared is equal to c squared minus a squared, I take the square root of both sides, B is equal to the square root of c squared minus a squared and I can go ahead and plug in those values. So this becomes B is equal to the square root of c squared which is six squared minus a squared, which is four squared, B is equal to the square root of 36 minus 16, Or B is equal to the square root of 20. Once I do 36 -16. So now I have a more simplified answer And 20 is the same thing as four times 5. So if I separate The square roots I have B is equal to the square root of four times the square root of five. Or my final simplification will be B is equal to two because the square root of four is two times the square root of five. And now I can plug in the value of B. And the value of a back into my standard equation. So I have x minus three squared divided by a squared which we solved two, B four minus y plus one squared divided by B squared, which we found to be two square root of five. So this becomes our final answer for the standard form of the ellipse described, and you can see that if we look at the answer choices, this aligns with answer choice A. So hopefully this video helped and see you next time.

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In Exercises 5–12, find the standard form of the equation of each hyperbola satisfying the given conditions. Center: (4, −2); Focus: (7, −2); vertex: (6, −2)

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