Find the Properties (x^2)/25+(y^2)/4=1

Problem

(x^2)/25+(y^2)/4=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^2>b^2 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=4 which gives a=5 and b=2

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

c^2=25−4

c^2=21

c=√(,21)

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

Vertices⇒(±5,0)

Co-vertices⇒(0,±2)

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

Foci⇒(±√(,21),0)

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

e=√(,21)/5

Final Answer

Center: *(0,0), Vertices: *(±5,0), Foci: *(±√(,21),0), Eccentricity: √(,21)/5

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