Question Video: Graphing a Hyperbola given the Equation in Standard Form | Nagwa Question Video: Graphing a Hyperbola given the Equation in Standard Form | Nagwa

# Question Video: Graphing a Hyperbola given the Equation in Standard Form Mathematics

The graph shows a sketch of the hyperbola given by the equation ((π¦ β 2)Β²/25) β ((π₯ + 3)Β²/2) = 1. Give the coordinates of the center πΆ. Give the coordinates of the vertices πβ and πβ. Give the coordinates of the foci or foci πΉβ and πΉβ. Give the equations of the asymptotes π΄β and π΄β.

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### Video Transcript

The graph shows a sketch of the hyperbola given by the equation π¦ minus two squared over 25 minus π₯ plus three squared over two is equal to one. Give the coordinates of the center πΆ. Give the coordinates of the vertices π one and π two. Give the coordinates of the foci or foci πΉ one and πΉ two. And give the equations of the asymptotes π΄ one and π΄ two.

Weβre given both the graph and the equation of a hyperbola, where the equation is π¦ minus two squared over 25 minus π₯ plus three squared over two is equal to one. Weβre asked to find the center πΆ, the vertices π one and π two, the foci πΉ one and πΉ two, and the asymptotes π΄ one and π΄ two. We can see from the graph that this hyperbola has its transfers or major axis, thatβs the axis containing the foci and the vertices, parallel to the π¦-axis. The equation also tells us that this is the case, since itβs the π¦-term thatβs positive. Our hyperbola is, therefore, north-south opening, that is, vertical.

Itβs worth pointing out that if the π₯- and π¦-terms were swapped, in the equation, our hyperbola would be east-west opening with the transfers or major axis parallel to the π₯-axis. The two branches face east and west, and this is called a horizontal hyperbola. In our case, however, the hyperbola is north-south opening. Our equation therefore takes the form π¦ minus π squared over π squared minus π₯ minus β squared over π squared is equal to one. Comparing the equation weβve been given with this, we can see that π is equal to two, β is equal to negative three, π squared is equal to 25, and π squared is equal to two.

Weβre asked to give the coordinates of the center πΆ. And for a hyperbola to find in this way, the center is the point with coordinates β, π. So we can see that with our values of β is equal to negative three and π is equal to two, our center is the point negative three, two. Weβre next asked to give the coordinates of the vertices π one and π two. And for a hyperbola of this type, the vertices have coordinates π one is β, π plus π and π two is β, π minus π. Noting that since π and π are both positive, if π squared is 25, then π is equal to five. And with π squared equal to two, that means π is the positive square root of two. The π₯-coordinate of our vertex π one is negative three. And the π¦-coordinate, which is π plus π, is two plus five so that π one has coordinates negative three, seven.

Similarly, for π two, our π₯-coordinate β is equal to negative three, and the π¦-coordinate is π minus π which is two minus five. Two minus five is negative three. So our vertex π two is negative three, negative three. And so we have the coordinates of our vertices π one and π two. Now, weβre asked to give the coordinates of the foci πΉ one and πΉ two. Now for a hyperbola of this type, the foci πΉ one and πΉ two have π₯-coordinate β and π¦-coordinate π plus or minus π, where π squared is π squared plus π squared, and so that π, which is positive, is the square root of π squared plus π squared. In our case then, π is the positive square root of 25 plus two, which is the square root of 27. And in its simplest form, this is three times the square root of three.

So now we have everything we need to work out our foci πΉ one and πΉ two. We have β is negative three, π is two, and π is three times the square root of three so that our first focal point πΉ one has coordinates π₯ is equal to negative three and π¦ is two plus three root three. And our second focal point πΉ two has coordinates π₯ is negative three and π¦ is two minus three root three.

Lastly, weβre asked to find the equations of the asymptotes π΄ one and π΄ two. And for a hyperbola of this form, the asymptotes are given by π¦ is equal to positive or negative π over π times π₯ minus β plus π. And with our values of π, π, β, and π, our first asymptote π΄ one has the equation π¦ is equal to five over the square root of two times π₯ minus negative three plus two. That is, π¦ is five over the square root of two times π₯ plus three plus two. And our second asymptote π΄ two has the equation π¦ is equal to negative five over root two times π₯ plus three plus two. We can rewrite these as π¦ minus two is equal to five over the square root of two times π₯ plus three and π¦ minus two is negative five over root two times π₯ plus three.

For the hyperbola shown then with the equation π¦ minus two squared over 25 minus π₯ plus three squared over two is equal to one, the coordinates of the center are π₯ is negative three, π¦ is two. The vertices have coordinates π one is negative three, seven and π two is negative three, negative three. The foci or foci have coordinates πΉ one is equal to negative three, two plus three root three. And πΉ two is equal to negative three, two minus three root three. And the asymptotes π΄ one and π΄ two are the lines π¦ minus two is equal to five over the square root of two times π₯ plus three and π¦ minus two is equal to negative five over the square root of two times π₯ plus three.

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