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hyperbolas 1

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Graph the hyperbola with equation quantity x plus one squared divided by nine minus the quantity of y plus four squared divided by twenty five = 1

asked Aug 2, 2014 in CALCULUS by Tdog79 Pupil

1 Answer

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The hyperbola equation image

image

In the above equation x term is positive, then the hyperbola is horizontal.

Compare it to  standard form of horizontal hyperbola is image

"a " is the number in the denominator of the positive term

center: (h, k ) Vertices: (h + a, k ), (h - a, k )

Foci: (h + c, k ), (h - c, k )

 Asympototes of hyperbola is image

In this case a = 3, b = 5, (h, k) = (-1,- 4)

Vertices (-1 + 3, -4) ( -1 - 3 , - 4) = ( 2 , - 4) ( -4 , -4)

image

Foci: (-1+6, - 4), ( -1-6, - 4 ) = (5, -4) , (-7, -4)

 Asympototes of hyperbola is image

image

image

image

Asympototes are image

Graph

Draw the coordinate plane.

Plot the center of hyperbola (-1,-4).

To graph the hyperbola go 5 units up and down from center point and 3 units left and right from center point(since a = 3, b = 5.)

Use these points to draw a rectangle .

Draw diagonal lines through the center and the corner of the rectangle. These are asymptotes.

Plot the vertices and foci of hyperbola.

Draw the curves, beginning at each vertex separately, that hug the asymptotes the farther away from the vertices the curve gets.

The graph approaches the asymptotes but never actually touches them.

answered Aug 2, 2014 by david Expert
edited Aug 2, 2014 by bradely
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