Find the Properties (x^2)/16-(y^2)/9=1

Problem

(x^2)/16−(y^2)/9=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x^2)/(a^2)−(y^2)/(b^2)=1 it is a horizontal hyperbola centered at the origin (0,0)

2. Determine the values of a and b From the denominators, a^2=16 and b^2=9 which gives a=4 and b=3

3. Calculate the distance to the foci c using the relation c^2=a^2+b^2

c^2=16+9=25

c=5

4. Identify the vertices. For a horizontal hyperbola, the vertices are located at (±a,0)

Vertices=(±4,0)

5. Identify the foci. The foci are located at (±c,0)

Foci=(±5,0)

6. Determine the asymptotes. The equations for the asymptotes of a horizontal hyperbola are y=±b/a*x

y=±3/4*x

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

e=5/4

Final Answer

Center: *(0,0), Vertices: *(±4,0), Foci: *(±5,0), Asymptotes: *y=±3/4*x, Eccentricity: 5/4

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