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

Problem

2*x^2+7*y^2=14

Solution

1. Divide both sides of the equation by 14 to put the equation into the standard form of an ellipse, (x^2)/(a^2)+(y^2)/(b^2)=1

(2*x^2)/14+(7*y^2)/14=14/14

(x^2)/7+(y^2)/2=1

2. Identify the center (h,k) by looking at the numerators x^2 and y^2

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.

a^2=7⇒a=√(,7)

b^2=2⇒b=√(,2)

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

c^2=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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