Find the Properties (x^2)/25-(y^2)/16=1

Problem

(x2)/25−(y2)/16=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 denominators, a2=25 and b2=16 Taking the square roots gives a=5 and b=4

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

c2=25+16

c2=41

c=√(,41)

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

Vertices⇒(±5,0)

Foci⇒(±√(,41),0)

5. Determine the equations of the asymptotes. For a horizontal hyperbola, the asymptotes are y=±b/a*x

y=±4/5*x

6. Identify the transverse and conjugate axes. The transverse axis is along the xaxis with length 2*a=10 The conjugate axis is along the yaxis with length 2*b=8

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

e=√(,41)/5

Final Answer

Center: *(0,0), Vertices: *(±5,0), Foci: *(±√(,41),0), Asymptotes: *y=±4/5*x, Eccentricity: √(,41)/5

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