Graph this

Question

Graph the following equation: (x-2)^2 / 81 - (y+1)^2 / 16 = 1

Answer 1

The standard form of hyperbola :

The hyperbola equation is (x - 2)²/81 - (y + 1)² /16 = 1.

Write the equation : (x - 2)²/81 - (y + 1)² /16 = 1 in standard form of hyperbola : .

(x - 2)²/(9)² - (y - (- 1))²/(4)² = 1.

Compare the above equation with standard form of hyperbola equation.

a = semi - transverse axis = 9,

b = semi - conjugate axis = 4,

Center: (h, k ) = (2, - 1),

Vertices: (h + a , k  ) and (h - a, k) = (11, - 1) and (- 7, - 1).

c² = a²+ b².

c² = (9)² + (4)²

= 81 + 16

= 97

⇒ c = √(97).

Foci: (h + c, k) and (h - c, k ) : (2 + √(97), - 1) and (2 - √(97), - 1).

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

y = - 1 ± [(4/9)(x - 2)].

Graph of hyperbola:

• Draw the coordinate plane.
• Plot the center of hyperbola (2, - 1).
• To graph the hyperbola go 4 units up and down from center point and 9 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.