An equation of a hyperbola is given. (a) Find the center, vertices, foci, and asymptotes of the

An equation of a hyperbola is given. (a) Find the center,...

Key Concepts

Asymptotes

Asymptotes are the lines that the branches of a hyperbola approach but never intersect. They act as a guide for sketching the hyperbola and are derived from the slopes given by ±b/a (for a horizontal hyperbola) or ±a/b (for a vertical hyperbola), with the asymptotes passing through the center of the hyperbola.

Foci

The foci are two fixed points located along the transverse axis of the hyperbola. The distance from the center to each focus, denoted by 'c', is found using the relationship c² = a² + b². These points are essential in defining the hyperbola as the difference in distances from any point on the hyperbola to the two foci is constant.

Vertices

Vertices are the points on a hyperbola where the branches are closest to the center, lying along the transverse axis. They are determined by the parameter 'a' in the standard form, which represents the distance from the center to each vertex.

Hyperbola

A hyperbola is a type of conic section defined as the set of points for which the absolute difference of the distances to two fixed points (foci) is constant. It consists of two separate branches and its analysis involves identifying key features such as the center, vertices, foci, and asymptotes.

Completing the Square

Completing the square is an algebraic technique used to transform a quadratic expression into a perfect square plus or minus a constant. This method is essential in rewriting the general equation of a conic section into its standard form, which then allows for straightforward identification of the hyperbola's key features.

Standard Form of Hyperbola

The standard form of a hyperbola's equation clearly displays its geometric properties. For a hyperbola centered at (h, k), the standard forms are either (x-h)²/a² - (y-k)²/b² = 1 for a horizontal transverse axis or (y-k)²/a² - (x-h)²/b² = 1 for a vertical transverse axis. This form makes it easier to identify the center, vertices, and other critical aspects of the hyperbola.

Transcript

00:02 36 x squared plus 72x minus 4 y square plus 32y square plus 32y is plus 1 116 is equal to 0 this can be written here we take the common here 36 and this is here x square and we take here 36 so this becomes here 2 multiplied x and this is a formula of a plus 3 whole square so 2 multiplied x multiply 1 that is we add 1 square and subtract 1 square so this becomes a x plus 2 whole square and we take here common minus 4 and we take here minus 4 so the equation becomes here that is equation is equal to here minus 4 y square take here minus minus here this is minus 4 take here this is y squared minus 2 multiplied 2 y and this is here so this is here 2 multiplied this is here this is 2 multiplied y into 4 plus 4 square minus 4 a square plus 4 square plus 4 square and plus 1 1 6 is equal to 0 so the equation becomes here that is 36 and this is here x plus this is x plus 1 whole square minus 1 and here y minus 4 take common here so this is here here so this becomes here 4 so this y minus 4 whole square and minus 16 and plus 116 is equal to 0 and so the equation becomes here that is 36 multiplied x plus 1 whole square and this is minus 4 multiply y minus 4 whole a square and this is here minus 36 and here minus 64 and plus 116 this is this is here minus 16 this becomes here goes to right side and this becomes minus 144 and so the equations if we divide minus 144 both sides so the equation becomes a standard equation of hyperbola becomes y minus 4 whole square divided by 36 is minus and x plus 1 whole square divided by here this is y minus 4 that is 36 and this is here 4 this is equal to 1 so here a is equal to 6 and b is equal to here 2 and c equals to here root under 40 and this here root under 2 to root 10 so here center c is equal to here h k and h is equal to here 4 and minus 1 and 4 so this is here minus 1 and 4 and and vertices here, v1 is equal to here, that is vertex is equal to h, kp here, h and k plus minus a here, vertices.

03:29 So, v1 is equal to here, minus 1, and 4 minus 6 here.

03:35 So this is a 4 minus 6 equals to here, minus 2.

03:40 And v2 is equal to here, minus 1, and 4 plus 6.

03:46 4 plus 6 equals to here 10.

03:48 So this is here plus minus so v1 and v2 vertex here and fokey equals to here foky is equal to here that is h and k plus minus c so fokey f1 is equal to here minus 1 4 minus 2 root 10 and f2 is equal to here minus 1 and 4 plus 2 root 10 so so, asymptotes here, as aemtotor's, y minus, this is here, y minus 4 plus minus, this is a by b, so a is here 6 and b equals to 2.

04:30 So this becomes 3 and x is here that is x plus 1.

04:35 So this is x plus 1.

04:37 If we take one times negative, then positive another.

04:40 So if we take negative sign, then y is equal to here minus 3x plus 1 and y equals to if you take positive, then y equals to here 3x plus here 7.

04:54 So this is the asymptotes here and we have to draw the graph.

04:59 So this is x -axis, this is x -tas, and this is y, and this is here, y -dice.

05:06 So vertex is here minus 1 and minus 2 and with vertex is equal to 10.

05:11 So we take the point here, this is origin, this is 2, this is 4, 6, 6, and we take the point here...

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