Find the Properties (x^2)/81+(y^2)/9=1

Problem

(x^2)/81+(y^2)/9=1

Solution

1. Identify the type of conic section by observing the equation is in the form (x^2)/(a^2)+(y^2)/(b^2)=1 with a^2>b^2 which represents a horizontal ellipse centered at the origin (0,0)

2. Determine the semi-major and semi-minor axes by taking the square roots of the denominators.

a^2=81⇒a=9

b^2=9⇒b=3

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

c^2=81−9=72

c=√(,72)=6√(,2)

4. Identify the vertices along the major axis (x-axis) using (±a,0)

Vertices=(±9,0)

5. Identify the co-vertices along the minor axis (y-axis) using (0,±b)

Co-vertices=(0,±3)

6. Identify the foci along the major axis using (±c,0)

Foci=(±6√(,2),0)

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

e=(6√(,2))/9=(2√(,2))/3

Final Answer

Center: *(0,0), Vertices: *(±9,0), Foci: *(±6√(,2),0), Eccentricity: (2√(,2))/3

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