Find the Properties 9x^2-16y^2=144

Problem

9*x2−16*y2=144

Solution

1. Divide both sides by 144 to write the equation in the standard form of a hyperbola (x2)/(a2)−(y2)/(b2)=1

(9*x2)/144−(16*y2)/144=144/144

(x2)/16−(y2)/9=1

2. Identify the values of a2 and b2 to determine the dimensions of the hyperbola.

a2=16⇒a=4

b2=9⇒b=3

3. Determine the center and the orientation. Since the x2 term is positive and there are no shifts, the center is at (0,0) and the transverse axis is horizontal.

Center=(0,0)

4. Calculate the distance to the foci c using the relation c2=a2+b2

c2=16+9=25

c=5

5. Find the vertices by moving a units from the center along the transverse axis.

Vertices=(±4,0)

6. Find the foci by moving c units from the center along the transverse axis.

Foci=(±5,0)

7. Determine the equations of the asymptotes using the formula y=±b/a*x

Asymptotes:y=±3/4*x

8. Calculate the eccentricity e using the formula e=c/a

e=5/4

Final Answer

Properties of *9*x2−16*y2=144:Center *(0,0),Vertices *(±4,0),Foci *(±5,0),Asymptotes *y=±3/4*x,e=5/4

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