What is the vertices ((x-7)^2)/4+((y+3)^2)/(16)=1 ?

The vertices ((x-7)^2)/4+((y+3)^2)/(16)=1 is (7,1),(7,-7)

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vertices ((x-7)^2)/4+((y+3)^2)/(16)=1

vertices

\frac{( x - 7 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 7 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (7, - 3), b = 4, a = 2

(7, - 3 + 4), (7, - 3 - 4)

Refine

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

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{49} = 1

9 x^{2} + 4 y^{2} - 36 x + 40 y + 100 = 0

9 x^{2} + 16 y^{2} + 96 y = 0

49 x^{2} + 25 y^{2} - 1225 = 0

What is the vertices ((x-7)^2)/4+((y+3)^2)/(16)=1 ?
The vertices ((x-7)^2)/4+((y+3)^2)/(16)=1 is (7,1),(7,-7)