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. 9y^2−25x^2=225

Accepted Answer

Welcome back. I am so glad you're here. We are asked to graph the hyperbola defined by the equation. We're given 64 Y squared minus nine X squared equals 576 using its vertices and asymptotes. Then determine the equations of the asymptotes and the folk for our answer choices, answer choice A has of zero negative square at 73 0 square at 73. And as some totes of Y equals plus or minus 5/ X. And then there is a graph answer choice B has folk of negative square rot 73 0 and square it 0 as some totes of Y equals plus or minus 8/5 X. And then there is a graph answer choice C gives us of zero, negative square at 73 0 square at 73 as some totes of Y equals plus or minus 3/8 X. And then there is a graph and answer choice D has folk of negative square root 73 0 and square root 73 as some totes of Y equals plus or minus eight thirds X. And then there is a graph we will look at the differences of these graphs later on. But now let's start off taking a look at our equation. So our equation for the hyperbola is not yet in standard form, we want something to be equal to one in standard form. And then we recall from previous lessons that we've got options on the other side of the equation. But let's start off getting this equal to one. So we will divide everything by 576. So on the left, we have 64 Y squared divided by 576. Well, both 64 576 are divisible by 64. That's going to come out to Y squared divided by 64. No, it's not divided by nine, excuse me minus. And then the 2nd 19 X squared divided by 576. Those are both divisible by nine. And that will give us an X squared divided by 64 and that all is now equal to one. So when we've got the Y squared on the left, our equation for the hyperbola in standard form is Y squared divided by A squared minus X squared, divided by B squared equals one. So from this, we know that A squared is equal to what's beneath the Y squared, that's going to be nine and that B squared is equal to what's beneath the X squared, which in this case is 64 we can solve for a And for B taking the square root of both sides, so A squared equals nine will be A equals plus or minus three. And B squared equals 64 will be B equals plus or minus eight. OK. So we've got our A and our B now, what do those tell us? Well, we know that when we have a hyperbola, the vertices are a units from the center. So we are going to be along the Y ax axis for our transverse axis because the Y squared comes first in our equation. So the vertices will be our Y values. So we'll have zero for our X values. So we'll have one at zero, negative three and then the other will be at zero positive three. All right. What else do we know? Well, we know that the folk guy are going to be C units from the center. Well, where's C, we only have an A and A B while we also recall from previous lessons that for a hyperbola, we are going to have C squared equal to A squared plus B squared. Well, we know what A squared and B squared are. A squared is nine plus B squared. That's 64. So nine plus 64 is a positive 73. Well, if C squared equals 73 then C equals a plus or minus square ret of 73. So our folk guy, as we said, they're going to be C units from the center, the folk are along the same axis, the transverse axis which is the y axis in this case. So those are going to be our Y values. So we will have zero, negative square 73 and zero positive square root 73 for our folk. Now, we just need to determine the asom totes. Well, we recall from previous lessons that the asom totes for the equation where we have Y squared A are going to be where Y equals A divided by B X and Y equals a negative A divided by B X. We know our A and our B. So our asom totes are going to be Y equals A is three divided by B is eight X and A Y equals a negative 38 X. OK. Now let's graph. I'm going to draw a rough sketch of a coordinate plane. We have a vertical Y axis. A horizontal X axis. They come together at the origin in the middle. Our transverse axis is the Y axis. Our vertices are at zero negative three. So from the origin, we will go down three units and have zero negative three as a point for our vert C vertex. And then from the origin for our 0.0.3, we will go up three units and now drawing our asymptotes. So we have one asom totes where Y equals 3/ X. Well, that tells us that because there's no B that it intersects with the origin and we are going to go up three units to the right eight units for our slope, which is right in front of the X. So up three to the right eight, up three to the right eight. And then we also have Y equals negative three eights X. So that's going to be the same slope but negative, it's heading down so we can draw our asom totes and then our lines for the hyperbola will go through our vertices and then hug up against these as some totes heading toward negative infinity and positive infinity. So there's our graph. All right. So now let's compare our answer choices. Answer choice A has folk at zero, negative square root 73 0 square rot 73. That's pretty good. But as some totes at Y equals plus or minus 5/ X and we had plus or minus 3/8 X. And if you look at the graph, they have vertices at +05 and zero, negative five instead of +03 and zero, negative three. So we can eliminate answer choice. A let's examine answer choice B here. The folk I are negative square red 73 0 and square red 73 0. So those are listed as the X values. The ASOM totes are Y equals plus or minus 8/5 X. And you can see the transverse axis in that graph is along the X axis instead of the Y axis. So we can eliminate. Answer choice B now let's check out C and D going to grab some of this information and move it to a different part of the screen. OK. Answer choice C has the folk at zero square or excuse me, zero, negative square, red 73 0 square. It 73 as some tode at Y equals plus or minus 3/8 X. That's what we've got and looking at the graph, they have the correct correct vertices at zero, negative three and 03. And that graph looks pretty good. Let's just examine answer choice. D really quick to make sure we're not missing anything. Answer choice D has the folk I at negative square root 0 and square root 73 0. So this one's along the X axis instead of the Y axis. So we can eliminate answer choice D and so answer choice C is correct. 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. 9y^2−25x^2=225

Key Concepts

Hyperbola Definition

Asymptotes of a Hyperbola

Foci of a Hyperbola

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