Find the Properties (x^2)/49+(y^2)/9=1

Problem

(x2)/49+(y2)/9=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x2)/(a2)+(y2)/(b2)=1 with a2>b2 this is a horizontal ellipse centered at the origin (0,0)

2. Determine the values of a and b We have a2=49 and b2=9 which gives a=7 and b=3

3. Calculate the distance to the foci c using the relationship c2=a2−b2

c2=49−9

c2=40

c=√(,40)=2√(,10)

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

Vertices⇒(±7,0)

Co-vertices⇒(0,±3)

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

Foci⇒(±2√(,10),0)

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

Major Axis Length=2*(7)=14

Minor Axis Length=2*(3)=6

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

e=(2√(,10))/7

Final Answer

Center: *(0,0), Vertices: *(±7,0), Foci: *(±2√(,10),0), Eccentricity: (2√(,10))/7

Want more problems? Check here!