Find the Properties 1/2x^2+1/8y^2=1/4

Problem

1/2*x^2+1/8*y^2=1/4

Solution

1. Standardize the equation by multiplying both sides by 4 to eliminate the fraction on the right side.

2*x^2+1/2*y^2=1

2. Rewrite in standard form for an ellipse, which is (x^2)/(a^2)+(y^2)/(b^2)=1 by expressing the coefficients as denominators.

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

3. Identify the center (h,k) from the terms (x−h)^2 and (y−k)^2

Center=(0,0)

4. Determine the semi-axes by taking the square roots of the denominators. Since 2>1/2 the major axis is vertical.

a^2=2⇒a=√(,2)

b^2=1/2⇒b=√(,2)/2

5. Calculate the focal distance c using the relationship c^2=a^2−b^2

c^2=2−1/2=3/2

c=√(,3/2)=√(,6)/2

6. Identify the vertices by moving a units from the center along the major axis (y-axis).

Vertices=(0,±√(,2))

7. Identify the foci by moving c units from the center along the major axis.

Foci=(0,±√(,6)/2)

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

e=(√(,6)/2)/√(,2)=√(,3)/2

Final Answer

1/2*x^2+1/8*y^2=1/4⇒Ellipse with Center *(0,0), Vertices *(0,±√(,2)), Foci *(0,±√(,6)/2), Eccentricity √(,3)/2

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