What is the vertices ((x+3)^2)/(16)+((y-5)^2)/(25)=1 ?

The vertices ((x+3)^2)/(16)+((y-5)^2)/(25)=1 is (-3,10),(-3,0)

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vertices ((x+3)^2)/(16)+((y-5)^2)/(25)=1

vertices

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

(- 3, 10), (- 3, 0)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (- 3, 5), b = 5, a = 4

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

Refine

(- 3, 10), (- 3, 0)

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

y^{2} = 5 x

9 x^{2} - 4 y^{2} = 36

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

y^{2} - 4 y + 4 x = - 16

What is the vertices ((x+3)^2)/(16)+((y-5)^2)/(25)=1 ?
The vertices ((x+3)^2)/(16)+((y-5)^2)/(25)=1 is (-3,10),(-3,0)