Precalculus Help???? Graph the hyperbola????

Question

Graph the hyperbola.

9x² − 18x − 16y² + 96y − 279 = 0    

 

 

If the hyperbola is nondegenerate, specify the following: vertices, foci, lengths of transverse and conjugate axes, eccentricity, and equations of the asymptotes. Also specify the center. (If an answer does not exist, enter DNE.)

vertex	(x, y) =	(smaller x-value)
	(x, y) =	(larger x-value)
focus	(x, y) =	(smaller x-value)
	(x, y) =	(larger x-value)
transverse axis length
conjugate axis length
eccentricity
asymptote	y =	(smaller slope)
asymptote	y =	(larger slope)
center	(x, y) =

 

Answer 1

The standard form of hyperbola is .

• Where, "a " is the number in the denominator of the positive term, If the x - term is positive, then the hyperbola is horizontal
• a = semi - transverse axis , b = semi - conjugate axis
• Center: (h, k ) Vertices: (h + a, k ), (h - a, k )
• Foci: (h + c, k ), (h - c, k )
•  Asympototes of hyperbola is
• Eccentricity (e >1) = √ [1 + (b²/a²)].

The hyperbola equation is 9x² - 18x - 16y² + 96y - 279 = 0.
Write the equation : 9x² - 18x - 16y² + 96y - 279 = 0 in standard form of hyperbola : .

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 - coeffient)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 = (- 6/2)²= 9.

Add 9 and - 144 to each side of the equation.

9(x² - 2x + 1) - 16(y² - 6y + 9) - 279 = 0 + 9 - 144

9(x - 1)² - 16(y - 3)² = - 135 + 279

9(x - 1)² - 16(y - 3)² = 144

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

(x - 1)²/(16) - (y - 3)²/(9) = 1

(x - 1)²/(4)² - (y - 3)²/(3)² = 1.

Compare the above equation with standard form 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.