What is the vertices ((x-1)^2)/(16)+((y+2)^2)/(25)=1 ?

The vertices ((x-1)^2)/(16)+((y+2)^2)/(25)=1 is (1,3),(1,-7)

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vertices ((x-1)^2)/(16)+((y+2)^2)/(25)=1

vertices

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

(1, 3), (1, - 7)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (1, - 2), b = 5, a = 4

(1, - 2 + 5), (1, - 2 - 5)

Refine

(1, 3), (1, - 7)

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

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1

25 x^{2} + 49 y^{2} = 1225

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1 x

\frac{x ^{2}}{12} + \frac{y ^{2}}{144} = \frac{y}{12}

What is the vertices ((x-1)^2)/(16)+((y+2)^2)/(25)=1 ?
The vertices ((x-1)^2)/(16)+((y+2)^2)/(25)=1 is (1,3),(1,-7)