Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sket

step 1 The trick to this problem is completing the square. And the first thing I'm going to do is just

Find the center, vertices, foci, and the equations of the...

Key Concepts

Graphing the Hyperbola

Graphing a hyperbola involves plotting its center, vertices, and foci, and then drawing the asymptotes, which act as guidelines for the shape of the hyperbola. The asymptotes help in sketching the approximate curves of the hyperbola, ensuring that the branches converge to these lines at extreme distances from the center.

Asymptotes

Asymptotes are lines that the branches of a hyperbola approach but never intersect. They provide a framework for sketching the hyperbola and define the directions in which the hyperbola opens. Their equations can be derived from the standard form of the hyperbola by considering the behavior of the function as the independent variable approaches infinity.

Completing the Square

Completing the square is a method used to rewrite quadratic expressions in a form that highlights their geometric properties. This technique isolates the squared terms by adding and subtracting appropriate constants, which is essential in converting general conic section equations into their standard forms.

Standard Form of a Hyperbola

The standard form of a hyperbola presents the equation in a way that readily reveals geometric characteristics such as the center, axes, vertices, and asymptotes. Transforming the equation into standard form typically involves completing the square and factoring to rewrite the equation in one of the standard formats for hyperbolas, depending on the orientation (horizontal or vertical).

Center, Vertices, and Foci

The center of a hyperbola is the point about which the hyperbola is symmetric, and it is determined from the completed-square form of the equation. The vertices are points on the hyperbola that lie along its transverse axis and are located a fixed distance from the center. The foci, which also lie along the transverse axis, are located further from the center than the vertices and are central to the definition of the hyperbola based on the difference of distances.

Transcript

00:01 The trick to this problem is completing the square.

00:03 And the first thing i'm going to do is just use the commutative property and move things around.

00:10 Now, normally i would put a space here, but i actually do one more thing as well anyway.

00:16 Same thing with a minus y squared, minus 6y with a blank.

00:25 And i'm just going to subtract 18 to the right side, and 0 minus 18 to 918.

00:32 And what i actually have to do before i complete the square is i need to factor out the leading coefficient.

00:40 That way, my, i need to have a minus, i need to have the leading coefficient be one on each of these.

00:48 So just to make sure that your math is correct, because as you, i guess, factor out that you, like, double check that your math is correct, like 9x squared as i distribute that in.

01:02 So the process is always half of that middle number squared.

01:07 Well, negative 2 squared is 4.

01:09 So i'm going to add 4 in here.

01:11 But the quantity i'm actually adding is 9 times 4, so that's 36.

01:16 And the same thing over here, it's the same process, half of that middle number squared.

01:22 So 3 squared is 9.

01:24 But what's really happening is i'm adding a negative 9 to that side.

01:29 So i need to subtract 9 over.

01:31 So the whole point of doing this, completing the square process, is now you have a perfect square trinomial, x minus 2, x minus 2, but we just write it once.

01:43 Same thing with this piece.

01:46 Y plus 3, y plus 3 is the factored form.

01:50 And then on the right side we have 36 minus 18 is 18, minus 9 is 9...

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