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

Problem

9*x^2−16*y^2=144

Solution

1. Divide both sides by 144 to write the equation in the standard form of a hyperbola (x^2)/(a^2)−(y^2)/(b^2)=1

(9*x^2)/144−(16*y^2)/144=144/144

(x^2)/16−(y^2)/9=1

2. Identify the values of a^2 and b^2 to determine the dimensions of the hyperbola.

a^2=16⇒a=4

b^2=9⇒b=3

3. Determine the center and the orientation. Since the x^2 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 c^2=a^2+b^2

c^2=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*x^2−16*y^2=144:Center *(0,0),Vertices *(±4,0),Foci *(±5,0),Asymptotes *y=±3/4*x,e=5/4

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