Graph (y^2)/16-(x^2)/144=1

Problem

(y2)/16−(x2)/144=1

Solution

1. Identify the type of conic section. Since the equation is in the form (y2)/(a2)−(x2)/(b2)=1 it represents a vertical hyperbola centered at the origin (0,0)

2. Determine the values of a and b We have a2=16 which gives a=4 and b2=144 which gives b=12

3. Locate the vertices. For a vertical hyperbola, the vertices are at (0,a) and (0,−a)

(0,4)* and *(0,−4)

4. Find the foci using the relation c2=a2+b2

c2=16+144=160

c=√(,160)=4√(,10)≈12.65

The foci are at (0,4√(,10)) and (0,−4√(,10))

5. Determine the equations of the asymptotes. For a vertical hyperbola, the asymptotes are given by y=±a/b*x

y=±4/12*x

y=±1/3*x

6. Sketch the graph by plotting the vertices, drawing the asymptotes, and drawing the two branches of the hyperbola opening upward and downward from the vertices.

Final Answer

(y2)/16−(x2)/144=1⇒Vertical hyperbola with vertices *(0,±4)* and asymptotes *y=±1/3*x

Want more problems? Check here!