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. Foci: (0, −3), (0, 3) ; vertices: (0, −1), (0, 1)

Accepted Answer

Hello everyone in this video, we're going to be looking at this practice problem where we want to find the standard form equation of hyperbole to with the conditions that it has focused at zero negative five and zero positive five and vertex is at zero negative two and zero positive two. So in order to find the equation for this, let's first determine if this hyperbole A is transverse across the X axis or the y axis. And in this case because all of the X values are staying the same. This hyperbole is transverse across the y axis. And because it's transverse across the y axis, we can use the standard form equation for the y axis, which dictates y squared is first so y squared divided by a squared minus X squared divided by b squared Is equal to one. And the book I are given in terms of H and K plus or minus C. And the vertex is are given in terms of H and K plus or minus A. And the center is at H K. So we're going to use this to help us find our equation. So if we look at the folk, I You can see that one folk, I is that negative five. And the other folk, I or the other focus is at positive five. So if we find the midpoint between the two folk, I you can see that it lies at zero. So if we were looking at this on a graph, we have a focus down here at zero negative five. And focus up here at zero positive five. So the center of this hyper villa is the midpoint between these two. And if we go five units towards the middle We find that the center is here at zero. So ours center for this hyperbole. It's actually at the origin or 00 which means that H is equal to zero and K is equal to zero. But from here we still need to find A. And B. In order to solve our standard form equation so we can solve A. Using the vertex or the vertex is that we're given? So C. R. U zero and two. We can compare that to H. K plus A. And I know that H and K are both zero, so H is equal to zero and now I have K plus A. Is equal to two. But I know that K is actually zero. So I have A. Is equal to two and I can't solve for B. But recall that C squared is equal to a squared plus B squared and hyper bliss. So I can soul for see using my folk. I so if I look at one focus Say zero and positive five, I can compare that to H and K plus C. Where I already knew that H. was or is zero. But now I have K plus C. Is equal to five and again K is zero. So I'm off with C is equal to five. So now I can plug in both the A value and the C value into this equation to solve for B. So if I solve for B using this equation it becomes b squared is equal to c squared minus a squared. I'll take the square root of both sides. So I'm left with B is equal to the square root of c squared minus a squared. And I can plug in the values of C and A. If I do that, I have B is equal to the square root of C squared which is five squared minus A squared which is two squared. So B is equal to the square root of 25 -4, Which means B is equal to the square root of 21. And now I have a value for both A and B. So I can plug that into my standard equation. So have Y squared divided by a squared which is two and minus X squared divided by B squared, which is the square root of That's equal to one. I can simplify the denominators. So I'll have y squared divided by two squared is two times two which is four minus x squared divided by the square root of 21 times the square root of 21 leaves me with just equal to one. So this becomes my final answer for the standard form equation for this approval. Um And you can see that 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. Foci: (0, −3), (0, 3) ; vertices: (0, −1), (0, 1)