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Focus Question: How can factoring help me graph hyperbolas? On a separate sheet of paper, write a letter to a student who knows how to factor and how to graph a hyperbola using a calculator, but wants to know how the ideas relate to one another. Include at least five key ideas that you have learned. Make your letter interesting and fun to read.

asked Nov 30, 2014 in PRECALCULUS by anonymous

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I consider an example of the hyperbola equation is 9x2 - 18x - 16y2 + 96y - 279 = 0.

9x2 - 18x - 16y2 + 96y - 279 = 0

9(x2 - 2x) - 16(y2 - 6y) - 279 = 0

To change the expression into a perfect square  add (half the x coefficient)² and (half the y - Coefficient)² to each side of the expression.

Here x coefficient = - 2, so, (half the x coefficient)² = (- 2/2)2= 1.

Here y coefficient = - 6, so, (half the x coefficient)² = (- 96/2)2= 9.

x coefficient is multiplied with 9.So add 9*1 = 9 to each side of the equation.

y coefficient is multiplied with 16.So add -16*9 = -144 to each side of the equation.

9(x2 - 2x) + 9 - 16(y2 - 6y) - 144 - 279 = 0 + 9 - 144

9(x2 - 2x + 1) - 16(y2 - 6y + 9) - 279 = - 135

9(x2 - 2x + 1²) - 16(y2 - 2*3y + 3²) - 279 = - 135

9(x - 1)2 - 16(y - 3)2 = - 135 + 279

9(x - 1)2 - 16(y - 3)2 = 144

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Compare the above equation with standard form ( image ) of hyperbola equation.

a = semi - transverse axis = 4

b = semi - conjugate axis = 3

Center: ( h, k ) = (1, 3)

Vertices: ( h + a , k  ) and (h - a, k) = (5, 3) and (- 3, 3).

c2 = a2+ b2.

c2 = (4)2 + (3)2

= 16 + 9 = 25

⇒ c = √25 = 5.

Foci : (h + c, k) and (h - c, k ) : (6, 3) and (- 4, 3).

Eccentricity (e) = √ [1 + (b2/a2)] = √ [1 + (32/42)] = 5/4 = 1.025 > 1.

Asympototes of hyperbola are : y = k ± [(b/a)(x - h)].

y = 3 ± [(3/4)(x - 1)].

Graph of hyperbola:

Draw the coordinate plane.

Plot the center of hyperbola (1, 3).

To graph the hyperbola go 3 units up and down from center point and 4 units left and right from center point.

Use these points to draw a rectangle .

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

The graph approaches the asymptotes but never actually touches them.

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

Plot the vertices and foci of hyperbola.

 

answered Dec 2, 2014 by Shalom Scholar
edited Dec 2, 2014 by Shalom

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