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. Endpoints of transverse axis: (0, −6), (0, 6); asymptote: y=2x

Accepted Answer

Hello everyone in this video. We're going to be looking at this practice problem where we want to try to get the standard form equation for hyperbole to with the conditions that it has end points of the transverse axis at zero negative eight and zero positive eight and y equals four X. So in order to get started, I'm gonna first draw it to see what this could look like. So I'm going to do my Y axis and my X axis and I'll go ahead and Just label positive eight on my y axis and negative eight on my Y axis. Two more ticks. So do positive four and -4. And then I will put eight ticks on the y axis. So I'll do 1234, positive negative one negative two negative three, negative four negative. And I'm going to start plotting the points that they gave me. So I have endpoints at zero and negative eight. So if I go to zero negative eight, I'll put a dot there and if I go front to zero to positively I'll put another dot there. And the next thing is that we have an ascent at Y is equal to four X. Which means that if why are effects is equal to one, Why is equal to 4? And if you recall the ascent, Oh, is this sort of dotted line that the curve will follow? So this is what our hyperbole are looking like. I'm not exactly sure, I'm not drawing it to exact but this is giving us a visualization and this is going to help us come up with the equation. So because it has the ascent okay going this way we know that there's a reflection on the negative side. So there's actually also an ascent at Y equals negative four X. And I think my previously drawn ascent is a little bit off because if X is equal to zero, Y is equal to zero. So let me try to draw it better And then the new ascent 0 -4 X will follow a very similar pattern but a sort of mirror reflection. So now I have the two ascent toads and my two endpoints. And from this graph you can see that we need the standard form for hyperbole, that is transverse on the y axis. And the standard form for that is y squared divided by a squared minus X squared divided by b squared is equal to one. So in this case, notice how we want to try to find the values of A and B. In order to find the standard form equation for our hyperbole. But a lot of that relies on us finding the center because A is the value from the center to the sort of vertex of the hyperbole. So were given the value of this upper Vertex negative or positive eight and negative eight. So the center would fall in the midpoint of those two. So if I do the midpoint formula I have negative eight plus positive eight divided by two equal to 0/2 or zero. Now I know that My center is actually at the origin 00. And from this graph you can see that This portion here that is a seems to be eight Units because we're going from the center to the vertex and that's eight units down. And notice again how is also up here as we go from center to this top of vertex, it's also eight units. So now I have the value of A. But I still need the value of B. And we can actually solve this using the formula C squared is equal to a squared plus B squared. However we don't have the value of C but we can get it from the ascent. Okay because recall that Y is equal to a divided by B times X. And were given that Y is equal to four X. So if we plug in four X into this equation, I have four. X. is equal to a divided by B and X. If I divide by X. On both sides I get These exes cancel out. So 4 4 to a divided by B. And multiplied by B. On both sides, I'll have four. B is equal to a divided by four. B is equal to a divided by four. And we know the value of a. is eight. So I can plug that in. So this becomes B is equal to eight divided by four or B is equal to two. So now I have both the value of A and the value of BE. So I can plug those into the standard form. So I have y squared divided by A, which is eight squared minus X squared divided by B squared which is two Is equal to one. So now I have my final standard form equation for my hyperbole. And you can see that this alliance with answer choice B. 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. Endpoints of transverse axis: (0, −6), (0, 6); asymptote: y=2x

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