What is the vertices ((x-0)^2)/(25)+((y-3)^2)/(49)=1 ?

The vertices ((x-0)^2)/(25)+((y-3)^2)/(49)=1 is (0,10),(0,-4)

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vertices ((x-0)^2)/(25)+((y-3)^2)/(49)=1

vertices

\frac{( x - 0 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1

(0, 10), (0, - 4)

Solution steps

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

Calculate ellipse properties

\frac{( x - 0 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, 3), b = 7, a = 5

(0, 3 + 7), (0, 3 - 7)

Refine

(0, 10), (0, - 4)

81 x^{2} + 64 y^{2} = 5184

16 x^{2} + 25 y^{2} - 64 x + 150 y = - 279

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{5} = 1

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

25 x^{2} + 9 y^{2} + 200 x - 90 y + 400 = 0

What is the vertices ((x-0)^2)/(25)+((y-3)^2)/(49)=1 ?
The vertices ((x-0)^2)/(25)+((y-3)^2)/(49)=1 is (0,10),(0,-4)