Find the Properties (x^2)/16-(y^2)/20=1

Problem

(x2)/16−(y2)/20=1

Solution

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

2. Determine the values of a and b From the equation, a2=16 and b2=20 Taking the square roots gives a=4 and b=√(,20)=2√(,5)

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

c2=16+20=36

c=6

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

Vertices:(±4,0)

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

Foci:(±6,0)

6. Determine the equations of the asymptotes. For a horizontal hyperbola centered at the origin, the asymptotes are y=±b/a*x

y=±(2√(,5))/4*x

y=±√(,5)/2*x

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

e=6/4=1.5

Final Answer

Center: *(0,0), Vertices: *(±4,0), Foci: *(±6,0), Asymptotes: *y=±√(,5)/2*x, Eccentricity: *1.5

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