Find the Properties 5x^2+8y^2=40

Problem

5*x2+8*y2=40

Solution

1. Standardize the equation by dividing both sides by 40 to set the right side equal to 1

(5*x2)/40+(8*y2)/40=40/40

2. Simplify the fractions to identify the standard form of an ellipse (x2)/(a2)+(y2)/(b2)=1

(x2)/8+(y2)/5=1

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

Center=(0,0)

4. Determine the semi-axes a and b by taking the square roots of the denominators.

a2=8⇒a=√(,8)=2√(,2)

b2=5⇒b=√(,5)

5. Identify the major axis by comparing a2 and b2 Since 8>5 the major axis is horizontal (along the xaxis).

Major axis length=2*a=4√(,2)

Minor axis length=2*b=2√(,5)

6. Calculate the focal distance c using the relationship c2=a2−b2

c2=8−5=3

c=√(,3)

7. Locate the vertices by adding and subtracting a from the xcoordinate of the center.

Vertices=(±2√(,2),0)

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

Foci=(±√(,3),0)

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

e=√(,3)/√(,8)=√(,3)/(2√(,2))=√(,6)/4

Final Answer

Center: *(0,0), Vertices: *(±2√(,2),0), Foci: *(±√(,3),0), Eccentricity: √(,6)/4

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