Find the Properties 5x^2+3y^2=15

Problem

5*x^2+3*y^2=15

Solution

1. Standardize the equation by dividing both sides by 15 to set the right side equal to 1

(5*x^2)/15+(3*y^2)/15=15/15

2. Simplify the fractions to identify the standard form of an ellipse (x^2)/(b^2)+(y^2)/(a^2)=1

(x^2)/3+(y^2)/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 by taking the square roots of the denominators, where a^2 is the larger value.

a^2=5⇒a=√(,5)

b^2=3⇒b=√(,3)

5. Identify the orientation of the major axis. Since a^2 is under the y^2 term, the ellipse is vertical.

Major Axis: Vertical

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

c^2=5−3=2

c=√(,2)

7. Find the vertices by moving a units from the center along the major axis (ydirection).

Vertices=(0,±√(,5))

8. Find the foci by moving c units from the center along the major axis (ydirection).

Foci=(0,±√(,2))

9. Find the co-vertices by moving b units from the center along the minor axis (xdirection).

Co-vertices=(±√(,3),0)

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

e=√(,2)/√(,5)=√(,10)/5

Final Answer

Center: *(0,0), Vertices: *(0,±√(,5)), Foci: *(0,±√(,2)), Eccentricity: √(,10)/5

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