What is the vertices ((x+1)^2)/(49)+((y-2)^2)/(48)=1 ?

The vertices ((x+1)^2)/(49)+((y-2)^2)/(48)=1 is (6,2),(-8,2)

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vertices ((x+1)^2)/(49)+((y-2)^2)/(48)=1

vertices

\frac{( x + 1 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{48} = 1

(6, 2), (- 8, 2)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{48} = 1 :

Ellipse with center

(h, k) = (- 1, 2), a = 7, b = 43

(- 1 + 7, 2), (- 1 - 7, 2)

Refine

(6, 2), (- 8, 2)

3 x^{2} + 4 y^{2} = 48

2 x^{2} + y^{2} - 98 = 0

3 x^{2} - 6 x + y^{2} + 4 y - 11 = 0

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

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

What is the vertices ((x+1)^2)/(49)+((y-2)^2)/(48)=1 ?
The vertices ((x+1)^2)/(49)+((y-2)^2)/(48)=1 is (6,2),(-8,2)