What is the vertices ((x-3)^2)/4+((y-4)^2)/(25)=1 ?

The vertices ((x-3)^2)/4+((y-4)^2)/(25)=1 is (3,9),(3,-1)

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vertices ((x-3)^2)/4+((y-4)^2)/(25)=1

vertices

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{25} = 1

(3, 9), (3, - 1)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (3, 4), b = 5, a = 2

(3, 4 + 5), (3, 4 - 5)

Refine

(3, 9), (3, - 1)

16 x^{2} - 64 x + 9 y^{2} - 54 y = - 1

\frac{x ^{2}}{15} + \frac{y ^{2}}{16} = 1

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

36 x^{2} + 16 y^{2} - 64 y - 512 = 0

7 x^{2} + 16 y^{2} = 112

What is the vertices ((x-3)^2)/4+((y-4)^2)/(25)=1 ?
The vertices ((x-3)^2)/4+((y-4)^2)/(25)=1 is (3,9),(3,-1)