Find the Properties (x^2)/36+(y^2)/100=1

Problem

(x^2)/36+(y^2)/100=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x^2)/(b^2)+(y^2)/(a^2)=1 with a^2>b^2 it is a vertical ellipse centered at the origin (0,0)

2. Determine the values of a and b We have a^2=100 so a=10 We have b^2=36 so b=6

3. Calculate the distance to the foci c using the formula c^2=a^2−b^2

c^2=100−36

c^2=64

c=8

4. Find the vertices and co-vertices. The vertices are located at (0,±a) which are (0,10) and (0,−10) The co-vertices are located at (±b,0) which are (6,0) and (−6,0)

5. Locate the foci. The foci are located at (0,±c) which are (0,8) and (0,−8)

6. Determine the lengths of the axes. The major axis length is 2*a=20 The minor axis length is 2*b=12

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

e=8/10

e=0.8

Final Answer

Center: *(0,0), Vertices: *(0,±10), Foci: *(0,±8), Eccentricity: *0.8

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