Find the Properties 2x^2+8y^2=16

Problem

2*x^2+8*y^2=16

Solution

1. Divide both sides of the equation by 16 to put the equation into the standard form of an ellipse.

(2*x^2)/16+(8*y^2)/16=16/16

2. Simplify the fractions to identify the denominators.

(x^2)/8+(y^2)/2=1

3. Identify the center (h,k) from the standard form ((x−h)^2)/(a^2)+((y−k)^2)/(b^2)=1

Center=(0,0)

4. Determine the values of a^2 and b^2 Since 8>2 the major axis is horizontal.

a^2=8⇒a=√(,8)=2√(,2)

b^2=2⇒b=√(,2)

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

c^2=8−2=6⇒c=√(,6)

6. Find the vertices by adding and subtracting a from the xcoordinate of the center.

Vertices=(±2√(,2),0)

7. Find the co-vertices by adding and subtracting b from the ycoordinate of the center.

Co-vertices=(0,±√(,2))

8. Find the foci by adding and subtracting c from the xcoordinate of the center.

Foci=(±√(,6),0)

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

e=√(,6)/(2√(,2))=√(,3)/2

Final Answer

Center: *(0,0), Vertices: *(±2√(,2),0), Foci: *(±√(,6),0), Eccentricity: √(,3)/2

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