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

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

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

vertices

\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1

(0, 5), (0, - 5)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 4

(0, 0 + 5), (0, 0 - 5)

Refine

(0, 5), (0, - 5)

y^{2} - 6 y - 4 x + 5 = 0

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

16 x^{2} + 9 y^{2} = 144

f (x) = x^{2} + 4 x - 1

(x - h)^{2} = 4 p (y - k)

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