In Exercises 13–26, use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. x^2/100−y^2/64=1

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Welcome back. I am so glad you're here. We are asked to graph the hyperbola defined by the equation X squared, divided by 81 minus Y squared divided by 25 equals one using its vertices and its asom totes. Then determine the equations of the asom totes and the folk. Our answer choices are answer choice. A folk I of negative 106 0 and 0. Ask some totes of Y equals plus or minus 25 81st X. And then there's a graph answer choice B has folk of zero, negative 106 and 0. 106 has some totes of Y equals plus or minus 25th X. And then there is a graph answer to ac has Folk of negative square rut 106 and square rut 106 0. Ask some totes of Y equals plus or minus 5/9 X. And then there is a graph and answer choice D has folk of zero, negative square root 106 and zero square rut as some totes of Y equals plus or minus 9/5 X. And then there is a graph, we'll compare the details of these graphs in a little bit. But right now, let's take a look at this equation. We recognize from previous lessons that this equation of A hyperbola is given as X squared divided by A squared minus Y squared, divided by B squared equals one which is centered about the origin. And so we can recognize that our A squared is equal to 81 because both are underneath the X squared and B squared is equal to 25. Because both of those are underneath the Y squared, we take the square of both sides of both equations to solve for A and B and we get the A is equal to plus or minus nine and B is equal to plus or minus five. Well, how is that helpful? Well, we can use the A and the B to give us our vertices ascent totes. And our, so let's start off with the vertices. We recall from previous lessons that vertices are a units from the center. And in this case, our center is the origin but a units in which direction are they along the X axis or the Y axis? Well, in our equation, X squared comes first. So X is the transverse axis. So these will be XX values for our vertices. So a units from the origin, that's nine units in either direction along the X axis. So our vertices are negative 90 and 90. All right, we've got our vertices. Now, let's work on our as totes, we recall from previous lessons that when X is the transverse axis, then our asymptotes are going to be Y equals plus or minus B divided by A X and our B is five, our A is nine. So our asymptotes are at Y equals plus or minus 5/9 X. Now let's work on our folk. We recall from previous lessons that the folk are sea units from the center, which in this case is the origin. But where's C we haven't talked about C yet. Well, we recall from previous lessons that C squared equals A squared plus B squared for hyperbola. So we know A squared that's 81 we know B squared, that's 25 81 plus 25 is 106. So if C squared equals 106 then we take the square root of both sides to find C and C equals the square rot while plus or minus the square rut of 106. And because the folk are also along the transverse axis, these will be X value. So our folk are at negative square rot, 106 0 and positive square at 106 0. And we have solved for our vertices, our asom totes and our folk eye. Now let's go to a part of the screen where there's a little more room and we can work on graphing this, we'll draw a rough sketch of a coordinate plane. We'll have a vertical Y axis, a horizontal X axis. They come together at the origin in the middle. For our vertices from the origin, we'll have one at negative 90. We'll go to the left nine units and we'll have label this at negative 90. And then for other vertex from the origin, we'll go to the right nine units and label this at 90. Now, for our asom totes, we know that Y equals plus or minus 5/ X is in the format of Y equals M X plus B. Well, our B is zero, there is nothing there for that. And M, our slope is A plus or minus 5/9. So A B of zero, A Y intercept of zero means it's intersecting with our origin and we can start there and draw our two as totes. So for a po a slope of positive five nights, we'll go up five units over to the right nine units. And then we can connect the origin with the dotted line to that point of 95. Mine is a little bit slanted but that's OK. There's one as to and now the one for the negative slope of negative five nights. So from the origin, we can go down five units to the right 59 units and then connect those two points, the origin with our point of nine negative five. And there is our second as to now to draw the lines of our hyperbola, we'll start to the left of the origin. We will have one point coming from a negative infinity for our X values. And it will be above our point negative 90. It's coming from negative infinity for X. But along our as tote of Y equals negative 59 X, then it will intercept our point, our vertex negative 90 and then come down and then come along our as some tote of Y equals five nights X toward negative infinity for the X value. And then it will be mirrored on the other side, the right side of the origin, we can come from positive infinity for our X values above the origin. And then along our ascent tote of Y equals five nines X intercept with our vertex at 90. Turn around come down to negative infinity for Y and positive infinity for X along that asom tote of Y equals negative 5/9 X. And there's our graph now we just need to find out what matches. So let's start looking at answer choice. A answer choice. A has folk of negative 106 0 and 0. OK. Well, they didn't take square ruts. So that's too big. We need the square rut of negative or the negative square root of 106 and the square root of 106. So we can eliminate answer choice. A answer. Choice B very similarly though correctly having well, no, this is totally wrong. So this one has of zero, negative 106 and 0 106. Again, didn't take the square roots, we can eliminate answer choice B. Now looking at answer traces C and D, these did correctly take the square roots of every everything. What's the main difference between C and D? Well, answer choice C has X as the transverse axis and answer choice D has Y as the transverse axis. And we know that the correct answer is with X as the transverse axis. It is F of negative square root 0 and positive square root 106 0 as some totes of Y equals plus or minus +59 X and it has vertices of negative 90 and 90. So answer choice C is our correct answer. Well done. We'll catch you on the next one.

In Exercises 13–26, use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. x^2/100−y^2/64=1

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Key Concepts

Hyperbola Definition

Asymptotes of a Hyperbola

Foci of a Hyperbola