Use vertices and asymptotes to graph each hyperbola. Locate th...

Question

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y²/16−x²/36=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 divided by 25 minus X squared divided by 64 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 of negative square root 89 and square root 89 0. Ask some totes of Y equals plus or minus 8/5 X. And then there's a graph answer choice B has folk of zero negative square root 0 square rot 89. Add some totes of Y equals plus or minus five eights X. And then there's a graph answer choice C has folk of negative 89 0 and 89 0 as some totes of Y equals plus or minus 25th X. And then there's a graph and answer choice D has folk of zero negative 0 89 as some totes of Y equals plus or minus 25th, 25 64th X. And then there's a graph we will go in more detail. Comparing those graphs in a little bit. But now let's start on building our own graph. So we look at our equation and we see that that is in the format from previous lessons, we know Y squared divided by A squared minus X squared, divided by B squared equals one which is centered about the origin. And we know looking at ours that A squared is equal to 25 because they're both underneath the Y squared and B squared is equal to 64 because they're both underneath the X squared. Now, we can solve for A and B taking the square roots of both sides of both equations. A is equal to plus or minus five. B is equal to plus or minus eight and solving for those. What does that do? Well, that allows us to find our vertices asom totes and folk. So for our vertices first, we recall from previous lessons that the vertices are a units from the center and in this case, the center is the origin. So in which direction are these going, are these going from the origin along the X axis or the Y axis? Well, when the Y squared comes first, why is the transverse axis? So these are going to be Y values they'll be along the Y axis. So a units from the origin, that'll be five units. So we will have for our vertices zero, negative five and zero, positive five. All right, we've got our vertices. Now let's work on our as totes for our asom totes, we recall from previous lessons that when the Y axis is the transverse axis, our asom totes are going to be Y equals plus or minus A divided by B X. We know that our A is five and our B is eight. So we are going to have asom totes at Y equals plus or minus 5/8 X. Now let's work on our folk. We recall from previous lessons that the folky are going to be C units from the center, which in this case is the origin, but we haven't solved for C yet. Where does that come from? Well, we recall from previous lessons that for hyperbola C squared is equal to A squared plus B squared, we know what A squared is. That's 25. We know what B squared is. That's 64 25 plus 64 is 89. So if C squared equals 89 taking the square root of both sides, C equals plus or minus the square root of 89. So our folk, I again are along the transverse axis, which is the Y axis. So these are Y values. So we've got folk of zero, negative square rot 89 and zero positive square rut 89. No gonna rewrite that below for when I need to grab it and move it to a different part of the screen in a minute. OK. So we've got these figured out the vertices, the ascent totes and the folk eye. Now let's move to a different part of the screen where there's a little more room and we can graph this right. We'll 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 for our vertices. First, these are going to be along the Y axis. So from the origin, we will go up five units for our vertex of 05. And from the origin, we will go down five units for our vertex of zero, negative five. Now, for the asymptotes, we have asom totes of Y equals plus or minus 5/8 X, this is in the format of Y equals M X plus B. Our M that's immediately in front of the X, we have a plus or minus 5/8. That's our slope. We have a positive 5/8 and a negative five eights and our B is zero. So our Y intercept is zero, it's at the origin. So from the origin, we can draw our positive slope first of positive 5/8. So we will go up five units from the origin and to the right eight units. And then we can draw a dotted line connecting the origin with that point. And there is one of our asom totes. Now, from the origin, we can draw our negative 5/8 slope. So we will go down five units into the right eight units and then we can connect those points with a dotted line. And there is our as tote four Y equals and negative five eights X. Now we can draw our lines for our hyperbola. So above the origin, so for our positive Y values, this line is going to come from a positive infinity for the Y values above our asom tote of Y equals negative five eights X. It's going to hug up against that asom tote until it comes close to the vertex, it will hit our vertex at 05 and then come back along heading toward positive infinity along our as tote of Y equals positive 58 X. And then the same thing mirrored will be happening below the origin from negative infinity. For Y this will be below our Asymptote of Y equals five eights X coming from negative infinity along that Asom tote and then hitting our vertex at zero, negative five. And then coming back along our asom tote toward negative infinity for Y along that Y equals negative 58 X as tote. And there is our graph roughly. So now it's just a matter of finding our matching graph. So let's compare all of these. Let's head back up. Look at answer choices A and B. So answer choice A starting just with the folk I, they have negative square rot 89 0 and positive square root 0, that's close. But they have the X axis as the transverse axis. So this one's incorrect because it is drawn with the X axis as the transverse axis. Looking at answer trace B we have folk of zero negative square root 89 0 positive square root 89. That's correct. As some totes of Y equals plus or minus 5/8 X. That's correct. It's drawn with the correct vertices of 05 and zero, negative five transverse axis being the X or excuse me, the Y axis. Answer choice B looks great. Let's look at C and D real quickly to see if we've missed anything. So for answer choice C the folk, I are negative 89 0 and 89 0. They didn't take the square root and we can see that this is drawn with the X axis being the transverse axis. It has incorrect vertices of negative 25 0 and 0. Answer choice D is wrong answer choice D correctly along the Y axis. However, again, they've got that 0 25 0, negative 25 for the vertices. The folk, I are zero negative 89 0 89. So they didn't take the square root of things that needed to be taken. So our answer choice B is correct. Well done. We'll catch you on the next one.

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y²/16−x²/36=1

Key Concepts

Standard Form of a Hyperbola

Foci of a Hyperbola

Equations of the Asymptotes

Watch next

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y2/16−x2/36=1

Key Concepts

Standard Form of a Hyperbola

Foci of a Hyperbola

Equations of the Asymptotes

Watch next

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y²/16−x²/36=1