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. y=±√x^2−2

Accepted Answer

Hey, everyone here we are asked to graph the hyperbola, which is defined by the equation Y is equal to plus or minus the square root of X squared minus six using its vertices and asymptotes. Then we are asked to determine the of the asymptotes and the FCI. Here we have four answer choice options. Each of which display a graph representing our hyperbola given by the equation and each solution states the foci and the asymptotes. So to begin solving this problem, we first need to take our given equation which as we recall is Y is equal to plus or minus the square root of X squared minus six. And so we want to take this equation and manipulate it so that it resembles some equation of a hyperbola. So we can do this by first taking these square of both sides to get rid of the square root. So we have Y squared is equal to X squared minus six. Now we want X squared and Y squared to be on the same side of the equation. So we can subtract X squared from both sides and we get Y squared minus X squared is equal to negative six. And now we want the right side of the equation to be equal to one. So we just need to divide both sides by negative six. So we have Y squared, negative six minus X squared divided by negative six is equal to one. And now just simplifying our equation, we see we have X squared divided by six minus Y squared, divided by six is equal to one. And now looking at this equation that we have, we want to make one more manipulation by making the denominators of both the X squared and Y squared to be squared terms in the denominator. So we know that six as the quantity of the square root of six squared. So we can rewrite our equation as X squared divided by the quantity of the square root of six squared minus Y squared divided by the quantity of the square root of six squared. And this is still equal to one. And so now this equation that we have, we see that our hyperbola is going to be centered at the origin. And since it is centered at the origin, we can recall the standard form of the equation of a hyperbola which is centered at the origin, which is X squared divided by A squared minus Y squared, divided by B squared is equal to one. Or we also have Y squared divided by A squared minus X squared divided by B squared is equal to one And here we have that the vertices are a units from the center and the foci are C units from the center from these formulas here. So the foci can be found using C squared is equal to A squared plus B squared. And so now in the, we found, we see that the X squared term is preceded by a plus sign. And so we know that the transverse axis lies along the X axis. And so we see our equation matches our first identified formula. And so now we want to identify our values for A and B. And we see here in our equation that A squared is equal to B squared, since both denominators are the same and both of these denominators are the quantity of the square root of six squared. And so now just taking the square root of Here, we see that a is equal to B which is equal to the square root of six. And now we want to identify our vertices for our parabola. And we can do this by recalling that the vertices are going to be the points negative A and zero and the point positive A and zero. So now just substituting in our value identified for A, which is the square root of six, we have the vertices at the points negative square root of six and zero and the point the square root of six and zero. So now we can draw a rectangle to find the asymptotes by using the negative square root of six and the positive square root of six for the axis and for the Y axis, or in other words, we're going to graph X is equal to the negative square root of six and the positive square root of six and Y is equal to the negative square root of six and the positive square root of six. And graphing these lines, we will get a square. And so we see here we have our square and we also can note that our vertices have a decimal approximation of negative 2.449 and zero and positive 2.449 and zero. So these points are shown on the graph as well as this square that we made using our X and Y values. And now using this square, we can draw in our asymptotes by drawing in diagonals of which pass through the corners of the square. So drawing in these asymptotes, we see we now have two asymptotes that are two different lines again that pass through the corners of the square. So next, we want to find the equations of our two as and toads. And we can do this using the formulas where Y is equal to the quantity of B divided by A times X and Y is equal to the negative quantity of B divided by A times X. So now with these two formulas, we can just substitute in our values for A and B. So starting with our first equation, we have Y is equal to the square root of six divided by the square root of six times X. And simplifying we just get Y is equal to X. And now for our second equation, we have Y is equal to the negative quantity of the square root of six divided by the square root of six times X. And simplifying, we see that our second as is for when Y is equal to negative X. So now we have our two asymptotes, one is Y is equal to X and one is Y is equal to negative X. And so now we want to identify our full and if we recall the are located at the point negative C and zero and the point positive C and zero. And as we recall from earlier, we can find the, using the formula C squared is equal to A squared plus B squared. And again, recalling from earlier, we had a squared is equal to B squared which is equal to the quantity of the square root of six squared. And so now just substituting in this value for A squared and B squared, we have C squared is equal to the quantity of the square root of six, six squared plus the quantity of the square root of six squared, which simplifies +26 plus six, which sums to 12. So we see C squared is equal to 12. But taking the square root of both sides, we see C is equal to plus or minus two times the square root of three. And so, with this value of C, we can determine our points of the full side which are now at negative two times the square root of three and zero. And the other focus is at two times the square root of three and zero. And so now that we have identified our and our asymptotes, we want to finish up our graph here. So finishing our graph, we want to draw the two branches of the hyperbola. And we do this by starting at each vertex and then approaching but not crossing the asymptotes. So here we can see we have our solid curved lines that start at our vertexes and approach the asymptotes that we drew in from earlier. And so now we see we have our final graph here but just removing that square in the middle, we see our finalized graph including our two asymptotes, which we identified as Y is equal to X and Y is equal to negative X. And we also have our vertexes identified at negative 2.449 and zero and positive 2.449 and zero. So now with all of the information that we found here, we see that we are left with answer choice B where our full are at negative two times the square root of three and zero and the 00.2 times the square root of three and zero and the asom totes are at plus or minus X. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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

Key Concepts

Hyperbola

Asymptotes

Foci

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