What is the vertices ((x-5)^2)/9+((y-4)^2)/(36)=1 ?

The vertices ((x-5)^2)/9+((y-4)^2)/(36)=1 is (5,10),(5,-2)

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vertices ((x-5)^2)/9+((y-4)^2)/(36)=1

vertices

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1

(5, 10), (5, - 2)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (5, 4), b = 6, a = 3

(5, 4 + 6), (5, 4 - 6)

Refine

(5, 10), (5, - 2)

25 x^{2} + 9 y^{2} - 225 = 0

64 x^{2} + y^{2} = 64

\frac{x ^{2}}{36} + \frac{y ^{2}}{100} = 1

6 x^{2} + 6 y^{2} + 24 x + 36 y = 0

64 x^{2} + 9 y^{2} = 576

What is the vertices ((x-5)^2)/9+((y-4)^2)/(36)=1 ?
The vertices ((x-5)^2)/9+((y-4)^2)/(36)=1 is (5,10),(5,-2)