Find the Properties 25x^2+9y^2-100x+54y-44=0

Problem

25*x2+9*y2−100*x+54*y−44=0

Solution

1. Group the x and y terms and move the constant to the right side of the equation.

25*x2−100*x+9*y2+54*y=44

2. Factor out the leading coefficients from the x and y groups.

25*(x2−4*x)+9*(y2+6*y)=44

3. Complete the square for both variables by adding ((−4)/2)2=4 to the x group and (6/2)2=9 to the y group, ensuring the right side is balanced.

25*(x2−4*x+4)+9*(y2+6*y+9)=44+25*(4)+9*(9)

4. Simplify the equation into the standard form of an ellipse.

25*(x−2)2+9*(y+3)2=225

5. Divide both sides by 225 to set the equation equal to 1

((x−2)2)/9+((y+3)2)/25=1

6. Identify the center (h,k) the semi-major axis a and the semi-minor axis b Since 25>9 the ellipse is vertical.

h=2

k=−3

a2=25⇒a=5

b2=9⇒b=3

7. Calculate the distance to the foci c using the relationship c2=a2−b2

c2=25−9=16⇒c=4

8. Determine the vertices, co-vertices, and foci based on the center and the values of a b and c

Center:(2,−3)

Vertices:(2,−3±5)⇒(2,2),(2,−8)

Co-vertices:(2±3,−3)⇒(5,−3),(−1,−3)

Foci:(2,−3±4)⇒(2,1),(2,−7)

9. Calculate the eccentricity e=c/a

e=4/5=0.8

Final Answer

((x−2)2)/9+((y+3)2)/25=1

Center:(2,−3),Vertices:(2,2),(2,−8),Foci:(2,1),(2,−7),Eccentricity:0.8

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