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

Problem

(x^2)/9−(y^2)/16=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=9 and b^2=16 which gives a=3 and b=4

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

c^2=9+16=25

c=5

4. Find the vertices and foci. The vertices are at (±a,0) and the foci are at (±c,0)

Vertices=(±3,0)

Foci=(±5,0)

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

y=±4/3*x

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

e=5/3

Final Answer

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

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