Graph the hyperbola with equation

Question

((x+1)^2)/9 - ((y+4)/25)=1

Answer 1

The given hyperbola  equation is ((x+1)^2)/9 - (y+4)^2/25 = 1

Draw the coordinate plane

Draw  the  hyperbola  equation ((x+1)^2)/9 - (y+4)^2/25 = 1

 

Answer 2

The hyperbola equation is .

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 .

Write the equation in standard form of hyperbola : .

.

Compare the above equation with standard form of hyperbola equation.

a = semi - transverse axis = 3 ,

b = semi - conjugate axis = 5

Center: (h, k ) = (- 1, - 4)

Vertices: (h + a, k ) = (2, - 4), and (h - a, k ) = (- 4, - 4).

c² = a²+ b².

c² =  5² + 3²

= 25 + 9 = 34

⇒ c = √34.

Foci: (h + c, k ) = (- 1 + √34, - 4), and (h - c, k ) = (- 1 - √34, - 4).

 Asympototes of hyperbola is .

.

Graph of hyperbola:

• 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.
• 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.