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

Problem

(x^2)/25+(y^2)/16=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≠b it is a horizontal ellipse centered at the origin (0,0)

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

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

c^2=25−16

c^2=9

c=3

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

5. Locate the foci. The foci are at (±c,0) which are (±3,0)

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

e=3/5

e=0.6

Final Answer

Center: *(0,0), Vertices: *(±5,0), Co-vertices: *(0,±4), Foci: *(±3,0), Eccentricity: *0.6

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