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. 9x^2−4y^2=36

Accepted Answer

Hello, today we're gonna be graphing the Parabola with the given equation then finding the equations of the asymptotes and the locations of the side. So the equation given to us is four X squared minus 25 Y squared is equal to 100. Now, in order for us to get this into a form of the equation of Parabola, we need to set this equation equal to one in order to do that, we can divide every term in the equation by 100. And doing so allows us to simplify the equation of the hyperbola to X squared over 25 minus Y squared over four is equal to one. Now, one thing to note is that this is the equation of a hyperbola where the center is located at the origin. So the center is going to be at zero comma zero. Furthermore, the X term is the leading term before the Y term. So that means that the major axis of this hyperbola is going to be the X axis. And what this means is that this is the equation of a hyperbola where the hyperbola opens up going left to right. So we need to find the vertices for the hyperbolas, we represent the distances of the vertices from the center using an A variable. And we allow the width or the height in this case of the as in total box to be represented by the B variable. Now, the general form of this type of hyperbola is X squared over A squared minus Y squared over B squared is equal to one. So let's go ahead and use the general form to get the A and B values. Well, in this case here, A squared is going to equal to 25. So we have a squared equal to 25 we have B squared equal to four. So we have B squared equal to four. Now, if we solve for A and B in both of the equations, we can do this by taking the square root of both sides of both of the equations. And what we get in return is A is equal to plus or minus five and B is equal to plus or minus two. So what this means is that we have a hyperbola who is centered at the origin, the vertices of the hyperbola are going to be located at five comma zero and negative five comma zero. And the height of the ASOM total box is gonna be defined by zero, comma two and zero, comma negative two. So we can go ahead and connect the dots using our M total box and we can also use this box to make a rough sketch of the diagonal ascent totes. And now we can go ahead and draw the hyperbola with respect to the asymptotes. Now keep in mind that this hyperbola ob opens up from left to right. So the shape of the hyperbola is roughly going to look like this. So now that we have in the shape of the hyperbola, what we need to do now is we need to find the locations of the and we need to find the equations of the diagonal asy totes. Let's go ahead and start with the. Now, for the foci, we represent the distance of the foci from the center with AC variable. And we have an equation to help us find this distance. For hyperbolas, we have C squared is equal to A squared plus B squared. In this case, we already know what our A squared and B squared values are going to be A squared is equal to 25 B squared is equal to four. So if we plug this into the equation, we get C squared is equal to 25 plus four, 25 plus four is going to simplify to give us 29. So we have C squared is equal to 29. And if we take the square roots of both sides, we get C is equal to the square root of 29. Now what this means is that the are located a distance of square root 29 to the left and right of the center of the hyperbola. So what we can go ahead and do is take the value of the square root of add and subtract this value to the X value of the center of the hyperbola. And what this is going to give us is square root of 29 comma zero and negative square root 29 comma zero. This is going to be the locations of the. Now there's one more thing we need to find and that's going to be the equations of the diagonal asymptotes. Now, since this is a hyperbola with the major axis being the X axis and it's centered at the origin, the general form of the equations of the dia A diagonal asymptotes are going to be Y is equal to plus or minus B over A times X. We have our B and A values B is equal to two and A is equal to five. And if we plug this into the positive and negative version of the equation of the diagonal asymptotes, we get Y is equal to positive 2/5 X and Y is equal to negative 2/5 X. These are going to be the equations of the diagonal asymptotes. So now we have a rough diagram of our hyperbola, we have the locations of the site and we have the equations for the Diago asymptotes. Now, what we need to do is we just need to select the option that accurately represents this information. And that option in this case is going to be option a option A has the ci at plus or minus square root 29 comma zero. And the ascent totes at plus or minus 2/5 times X and the vertices of the ascent totes obviously see the vertices of the parabola are located at negative five comma zero and five comma zero. So with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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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. 9x^2−4y^2=36

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