What is the vertices ((x-3)^2)/(49)+((y+5)^2)/(25)=1 ?

The vertices ((x-3)^2)/(49)+((y+5)^2)/(25)=1 is (10,-5),(-4,-5)

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vertices ((x-3)^2)/(49)+((y+5)^2)/(25)=1

vertices

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

(10, - 5), (- 4, - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (3, - 5), a = 7, b = 5

(3 + 7, - 5), (3 - 7, - 5)

Refine

(10, - 5), (- 4, - 5)

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

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

4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

What is the vertices ((x-3)^2)/(49)+((y+5)^2)/(25)=1 ?
The vertices ((x-3)^2)/(49)+((y+5)^2)/(25)=1 is (10,-5),(-4,-5)