What is the vertices ((x+4)^2)/(25)+((y-1)^2)/(16)=1 ?

The vertices ((x+4)^2)/(25)+((y-1)^2)/(16)=1 is (1,1),(-9,1)

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vertices ((x+4)^2)/(25)+((y-1)^2)/(16)=1

vertices

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

(1, 1), (- 9, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 4, 1), a = 5, b = 4

(- 4 + 5, 1), (- 4 - 5, 1)

Refine

(1, 1), (- 9, 1)

6 x^{2} + 9 y^{2} - 54 = 0

+ 36 x^{2} + y^{2} = 36

\frac{( x + 5 ) ^{2}}{9} + \frac{y ^{2}}{4} = 1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

16 x^{2} - 96 x + 144 + 25 y^{2} = 400

What is the vertices ((x+4)^2)/(25)+((y-1)^2)/(16)=1 ?
The vertices ((x+4)^2)/(25)+((y-1)^2)/(16)=1 is (1,1),(-9,1)