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. 4y^2−x^2=1

Accepted Answer

Welcome back. I am so glad you're here. We are asked to graph the hyperbola defined by the equation Y squared minus nine X squared equals one using its vertices and its asom totes. Then determine the asom totes and the folk for our answer choices. Answer choice A has folk at negative 13th, zero and 13th zero as some totes of Y equals plus or minus 1/18 X. And then there is a graph answer choice B has folk of zero negative 13th and zero 13th as some totes of Y equals plus or minus X. And then there is a graph answer choice C has folk of negative square root, 13th, zero and positive square root, 13th zero. Ask some totes of Y equals plus or minus one third X. And then there is a graph and answer choice D has folk of zero, negative square at 13th and zero square. It 13th ascent totes of Y equals plus or minus three X. And then there is a graph, we will go over all the details of those graphs a little bit later. But first, let's take a look at our equation. We recall from previous lessons that when our Y squared comes first and the X squared is negative, we are going to use the equation for a hyperbola, which is Y squared divided by A squared minus X squared, divided by B squared equals one. And ours is already equal to one. That's great. But there's nothing underneath our Y squared and our X squared and that X squared has a coefficient. So what do we do? Well, we can rewrite these a little bit. We've got Y squared that's divided by one. And for this nine X squared, we know that that's the same as X squared divided by 1/9 because X squared divided by 1/9 is the same as X squared multiplied by the reciprocal of 1/9 which is nine and X squared multiplied by nine is nine X squared. So we can say Y squared divided by one minus X squared, divided by 1/9 equals one. And now we can identify our A squared and our B squared A squared equals one. They're both underneath the Y squared and B squared equals 1/9. They're both underneath the X squared. Now, we can solve for A and B taking the square root of both sides of both equations. A equals plus or minus one and B equals plus or minus one third. So why do we need to know those? Well, those help us find our vertices asom totes And folk, let's start with the vertices, we recall from previous lessons that the vertices are a units from the center. And where is the center? Well, the center in this case is the origin knowing that from our equation of the hyperbole that we're using in the format that it's in. So our vertices are a units from the origin. But along which axis, well, when Y is first in the equation, the Y is our transverse axis. So these are Y values, they are found along the Y axis. So they're a units from the center. That means they're going to be one unit above and one unit below. So our vertices are going to be zero, negative one and zero, positive one. We found our vertices. Now let's work on our asom totes. We recall from previous lessons that the asymptotes when we have Y as the transverse axis are going to be Y equals plus or minus A divided by B X. We know our A and our B. So we've got a is one that's divided by B which is one third. Well, we know that one divided by one third is the same as one multiplied by the reciprocal of one third, which is three and one multiplied by three is three. So for our asom totes, we have Y equals plus or minus three X. Now, for our fka, we recall from previous lessons that the folky are C units from the center the center in this case being the origin, but we haven't figured out c yet. Well, c we recall from previous lessons, we know C squared is equal to A squared plus B squared. We know what A squared and B squared are. A squared is one B squared is 1/9. In order to add, when we've got a fraction, we need a common denominator. So this is 9/9 plus 1/9 which is equal to 19th. So we know that if C squared equals 19th to sol for C, we take the square root of both sides. And so we will have C equal to plus or minus the square root of 19th. And we can simplify that that's the same as plus or minus the squared of divided by the square root of nine. And we know that the square root of nine is three. So we've got plus or minus the square root of thirds for our C value. And so we know that our folk, I are also along the transverse axis, the Y axis in this case. So we have folk I of zero, negative square at 10 thirds and zero positive square root 10 thirds. And we've got our vertices, our asymptotes and our folk. Now we can graph this thing. Let's move to a different part of our screen where we'll have a little bit more room. And so now we can 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. We know our transverse axis is our Y axis. We have vertices at zero negative one and zero positive one. So from the origin, we'll go down one unit for our first vertex at zero negative one. And then we'll go from the origin up one unit for our other vertex at zero, positive one. And now we need to graph our asom totes. We have Y equals plus or minus three X. That's in the format Y equals M X plus B and our B is zero. So this goes through the origin and our m our slope is plus or minus three. So from the origin, we'll go up 3/1, up 3/1. So they're relatively steep asom totes. Then we draw a dotted line to show our asom totes and then we'll have another one with a negative slope and that one's just going down 3/1 down 3/1. And we can connect those dots. All right, we've got our as toads drawn. Now, we just need to draw the lines of our hyperbola. So from negative infinity for our Y values, this hyperbole is going to coast along this asom tote of X or Y equals three X and then it's going to intersect with our vertex at zero, negative one, then turn around and start heading down again toward negative infinity along that as some tote of Y equals negative three X. And we'll have the same thing going on in that is two above, excuse me, above our origin. So coming from positive infinity down along this Y equals negative three X asom tote, hitting our vertex at 01, turning around, coming back up along our asy tote of Y equals positive three X and heading toward positive infinity. And there we have our graph. So this is great. We just need to find the one that matches now and we can go to the different part of our screen. Let's take a look at answer choice. A answer choice. A had folk of negative 13th, zero and positive 13th zero. They didn't take that square rut. So those are incorrect and looking at the graph, they have vertices of negative 20 and 20. They're along the X axis for a transverse axis. They're the wrong points altogether. A is wrong. Now, looking at answer trace B that one has the Y axis is the transverse axis. However, it has vertices at 02 and zero, negative two. So that one is incorrect. And it also has the folk I listed as zero, negative 13th, zero, 13th instead of the square root of 10. So we can eliminate answer choice. B let's move these down to a different part of the screen and we'll check out answer choices C and D. All right answer choice C has folk of a negative square at 13th. So they took that square out of 10. But these are along the x axis that has x axis is the transverse axis. And that is incorrect looking at answer choice D that correctly has the Y axis is the transverse axis. It has the correct folk, the correct asom tote the correct vertices. Answer choice D is our correct answer. Well done. We'll catch you on the next one.

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In Exercises 13–26, use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. 4y^2−x^2=1

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