What is the vertices ((x-4)^2)/4+((y-3)^2)/9 =1 ?

The vertices ((x-4)^2)/4+((y-3)^2)/9 =1 is (4,6),(4,0)

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vertices ((x-4)^2)/4+((y-3)^2)/9 =1

vertices

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

(4, 6), (4, 0)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (4, 3), b = 3, a = 2

(4, 3 + 3), (4, 3 - 3)

Refine

(4, 6), (4, 0)

9 x^{2} + 4 y^{2} - 18 x + 16 y - 11 = 0

3 x^{2} + 7 y^{2} = 63

\frac{x ^{2}}{4} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{2} + \frac{y ^{2}}{4} = 1

x^{2} + 2 y^{2} = 3

What is the vertices ((x-4)^2)/4+((y-3)^2)/9 =1 ?
The vertices ((x-4)^2)/4+((y-3)^2)/9 =1 is (4,6),(4,0)