In Exercises 13–26, use vertices and asymptotes to graph each hyp...

Question

In Exercises 13–26, use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. x^2/9−y^2/25=1

Accepted Answer

Welcome back. I am so glad you're here. We are asked to graph the hyperbola defined by the equation X squared divided by 36 minus Y squared divided by 64 equals one using its vertices and asymptotes. Then determine the equations of the asymptotes and determine the F for answer choice. A are folk I R negative 10, 0 and 10 0. Ask some totes of Y equals plus or minus four thirds X. And then there is a graph answer choice B has folk I of zero negative 10 and 0 10 as some totes of Y equals plus or minus 3/4 X. And then there is a graph answer choice C has folk I of negative 40 0 and 40 0. Ask some totes at Y equals plus or minus 16 9th X. And then there's a graph and answer choice D has F of zero negative 40 and 0 40. Ask some totes of Y equals plus or minus 9/16 X. And then there is a graph, we will go into more detail comparing the graphs in a little bit. But now let's work on building our own graph. So looking at this equation we recognize from previous lessons that this is the equation of a hyperbola with the X squared first. So we have X squared divided by A squared minus Y squared divided by B squared equals one. And the transverse axis is the X axis because that's listed first for the hyperbola. And this is centered about the origin because H and K are not listed, they're equal to zero here. With that in mind, we can identify that A squared is equal to 36. Both of those are underneath the X squared B squared is equal to 64. Both of those are underneath the Y squared. We can determine what A and B are by taking the square root of both sides of both equations. A is equal to plus or minus six and B is equal to plus or minus eight. And how does that help us? Well, we recall that the vertices are a units from the center and in this case, the center is the origin and A units A is equal to six. So there's six units from the center, but in which direction? Well, we said that X is the transverse axis. So these are along the X axis and they're X values. So we have one vertex at negative 60 and another at 60. All right, we've got the vertices. Let's find the asom totes. We recall from previous lessons that when the ASOM or when the X axis is the transverse axis. Then the asymptotes are found with the formula Y equals plus or minus B divided by A X. We know that B is eight A is six, those are both divisible by two. So that simplifies to four thirds. So the asymptotes here are Y equals plus or minus four thirds X. Now, let's find the folk guy. We recall from previous lessons that the folk are C units from the center, which in this case is the origin, but we haven't found C yet. How do we find C? Well, we know that C squared is equal to A squared plus B squared for a hyperbola. From previous lessons, we know that A squared is 36 B squared is 64 36 plus 64 is 100. So C squared is equal to 100 to solve for C take the square root of both sides and we have C equals plus or minus 10. Now, what values are those, they're C units from the center in which direction again along the transverse axis, which is the X axis. So those C values are XX values. So one focus is at negative 10 0 and the other is at 10 0. All right, we've got the vertices, the asom totes and the folk eye. Now let's work on graphing. Move to a different part of our screen where there's a little more space and we can draw a rough sketch of a coordinate plane will have a vertical Y axis, a horizontal X axis they come together at the origin in the middle. Let's start by graphing the vertices. So the first one negative 60 from the origin will go to the left six units. And then we can label this negative 60. Now, the 0.60 from the origin go to the right six units and label this 60. Now let's graph our asom totes Y equals plus or minus four thirds. X is in the format of Y equals M X plus B where M is our slope and B is our Y intercept. Well, our slope here we have a positive four thirds and then another line with a negative four thirds, our B is equal to zero, which means that the Y intercept is the origin. So graphing this from the origin for the positive slope, we will go up four units to the right three units. So that's approximately here. We can draw a dashed line for asom tote connecting the origin with our point of 34. And there's our first as and tote of Y equals four thirds X. And now for our second one with the negative slope from the origin, we'll go down four units to the right three units approximately here, draw a dash line connecting the origin with our point of three negative four. And we can draw label this our as tote of Y equals negative four thirds X. And now it's just a matter of drawing in our hyperbola. So what will happen is we'll have lines coming from infinity connecting with our vertices. And then heading back to infinity to the left of the origin. We have the hyperbole coming from positive infinity for the Y values and negative infinity for the X values it hugs up along our asom tote of Y equals negative four thirds X till it gets close to the vertex of negative 60 hits that and then comes back along toward negative infinity for the Y and X values along the asom tote of Y equals four thirds X. And then, I mean, mine doesn't really look very good, but it'll be mirrored on the right side of the origin as well. So it's coming from positive infinity for both the X and Y values along the as tote, Y equals four thirds X to the right of it. Then it's coming down along that and then it will head toward the vertex of 60, hit that and then head back down toward negative infinity for the Y values positive infinity for the X values. And along our asom tote of Y equals negative four thirds X. So there is a really rough sketch of our graph. Let's see if we can find the matching answer choice going to start with answer choices C and D because that's where our answers are. Well, for C, they have folk I of negative 40 0 and 40 0. And for d they have folk I of zero, negative 40 and 0 40. That's not even close to our folk I of negative 10 0 and 0. So we can eliminate these answer choices. The graphs are to a much different scale. The numbers are much bigger. Now, let's look at answer choices. B and A move this up to a different part of the screen. All right. So we've got Folky of negative 10, 0 and 10 0 for answer choice A that's looking good. Answer. Choice B has folk of zero, negative 10 and 0 10. So they have used the Y axis as the transverse axis for answer choice B but we have X as the transverse axis. So answer trace B can be eliminated. Now just comparing everything for answer choice A the folk I are correct, ask some totes of Y equals plus or minus four thirds X those are correct. And the vertices looking at the graph are at negative 60 and 60. That all looks correct. So answer choice A is the 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^2/9−y^2/25=1

Key Concepts

Hyperbola Definition

Asymptotes of a Hyperbola

Foci of a Hyperbola

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