Find the Properties (x^2)/100+(y^2)/64=1

Problem

(x^2)/100+(y^2)/64=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 with a plus sign and a^2≠b^2 it is a horizontal ellipse centered at the origin (0,0)

2. Determine the semi-major axis a and semi-minor axis b We have a^2=100 and b^2=64

a=√(,100)=10

b=√(,64)=8

3. Calculate the distance from the center to the foci c using the relationship c^2=a^2−b^2

c^2=100−64=36

c=√(,36)=6

4. Find the vertices and co-vertices. The vertices are at (±a,0) and the co-vertices are at (0,±b)

Vertices=(±10,0)

Co-vertices=(0,±8)

5. Locate the foci. Since the major axis is horizontal, the foci are at (±c,0)

Foci=(±6,0)

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

e=6/10=0.6

Final Answer

Center: *(0,0), Vertices: *(±10,0), Foci: *(±6,0), Eccentricity: *0.6

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