What is the vertices ((x+2)^2)/(16)+(y^2)/(49)=1 ?

The vertices ((x+2)^2)/(16)+(y^2)/(49)=1 is (-2,7),(-2,-7)

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vertices ((x+2)^2)/(16)+(y^2)/(49)=1

vertices

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1

(- 2, 7), (- 2, - 7)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (- 2, 0), b = 7, a = 4

(- 2, 0 + 7), (- 2, 0 - 7)

Refine

(- 2, 7), (- 2, - 7)

(\frac{x}{7})^{2} + (\frac{y}{5})^{2} = 1

36 x^{2} + 25 y^{2} - 50 y - 875 = 0

4 (x - 5)^{2} + 3 (y - 5)^{2} = 192

9 x^{2} + 16 y^{2} + 36 x - 160 y + 292 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1

What is the vertices ((x+2)^2)/(16)+(y^2)/(49)=1 ?
The vertices ((x+2)^2)/(16)+(y^2)/(49)=1 is (-2,7),(-2,-7)