What is the vertices ((x+3)^2)/(49)+(y^2)/(36)=1 ?

The vertices ((x+3)^2)/(49)+(y^2)/(36)=1 is (4,0),(-10,0)

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vertices ((x+3)^2)/(49)+(y^2)/(36)=1

vertices

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (- 3, 0), a = 7, b = 6

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

Refine

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

25 x^{2} + 4 y^{2} = 100

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

5 x^{2} + 2 y^{2} = 50

\frac{x ^{2}}{63500} + \frac{y ^{2}}{50900} = 1

3 x^{2} + y^{2} = 84

What is the vertices ((x+3)^2)/(49)+(y^2)/(36)=1 ?
The vertices ((x+3)^2)/(49)+(y^2)/(36)=1 is (4,0),(-10,0)