Find the Properties 2x^2+7y^2=14

Problem

2*x2+7*y2=14

Solution

1. Divide both sides of the equation by 14 to put the equation into the standard form of an ellipse, (x2)/(a2)+(y2)/(b2)=1

(2*x2)/14+(7*y2)/14=14/14

(x2)/7+(y2)/2=1

2. Identify the center (h,k) by looking at the numerators x2 and y2

Center=(0,0)

3. Determine the semi-major axis a and semi-minor axis b by taking the square roots of the denominators. Since 7>2 the major axis is horizontal.

a2=7⇒a=√(,7)

b2=2⇒b=√(,2)

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

c2=7−2

c=√(,5)

5. Locate the vertices and co-vertices by adding and subtracting a and b from the center coordinates.

Vertices=(±√(,7),0)

Co-vertices=(0,±√(,2))

6. Locate the foci by adding and subtracting c from the center along the major axis.

Foci=(±√(,5),0)

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

e=√(,5)/√(,7)=√(,35)/7

Final Answer

Center: *(0,0), Vertices: *(±√(,7),0), Foci: *(±√(,5),0), Eccentricity: √(,35)/7

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