Find the Properties (x^2)/81+(y^2)/225=1

Problem

(x^2)/81+(y^2)/225=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=225 and b^2=81 which gives a=15 and b=9

3. Calculate the focal length c using the relationship c^2=a^2−b^2

c^2=225−81

c^2=144

c=12

4. Find the vertices and co-vertices. The vertices are located at (0,±a) which are (0,15) and (0,−15) The co-vertices are located at (±b,0) which are (9,0) and (−9,0)

5. Find the foci. The foci are located at (0,±c) which are (0,12) and (0,−12)

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

e=12/15

e=0.8

Final Answer

Center: *(0,0), Vertices: *(0,±15), Foci: *(0,±12), Eccentricity: *0.8

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