Find the vertices, foci and eccentricity of the ellipse. 9x2 + y2 = 81

Question

Find the vertices, foci and eccentricity of the ellipse. 9x2 + y2 = 81 I would like to see the steps and the answer to better understand it

Answer 1

The equation may be written as 9x^2 + y^2 = 81; dividing both sides by 81 we have ((x^2)/9)+ (Y^2)/81 =1 which gives us that the ellipse is located at the origin (0,0) due to the fact that The form ((x-h)^2)/a^2 + ((y-k)^2)/b^2 =1 shows that h(x) and k(y) are 0, Then a which is the distance to the vertex over the major axis, will be √81 = 9 and a which is the distance over the minor axis will be b=√9=3; then the vertices will be located at the points (0,9) and (0,-9); the distance to the foci will be c=√(a^2-b^2)=√(81-9)=6√2 so they will be located at the points (0,6√2) and (0,-(6√2)), the eccentricity  is equal to the distance to the foci divided by the distance to the vertices f=eccentricity= (6√2)/9 = 0.9428