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

Problem

(x^2)/16+(y^2)/25=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 this is a vertical ellipse centered at the origin (0,0)

2. Determine the values of a and b We have a^2=25 so a=5 We have b^2=16 so b=4

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

c^2=25−16

c^2=9

c=3

4. Identify the vertices and co-vertices. The vertices are located at (0,±a) which are (0,5) and (0,−5) The co-vertices are located at (±b,0) which are (4,0) and (−4,0)

5. Identify the foci. The foci are located at (0,±c) which are (0,3) and (0,−3)

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

e=3/5

Final Answer

Center: *(0,0), Vertices: *(0,±5), Co-vertices: *(±4,0), Foci: *(0,±3), Eccentricity: 3/5

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