What is the vertices ((x-3)^2)/(16)+((y-2)^2)/(20)=1 ?

The vertices ((x-3)^2)/(16)+((y-2)^2)/(20)=1 is (3,2+2sqrt(5)),(3,2-2sqrt(5))

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vertices ((x-3)^2)/(16)+((y-2)^2)/(20)=1

vertices

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{20} = 1

(3, 2 + 25), (3, 2 - 25)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{20} = 1 :

Ellipse with center

(h, k) = (3, 2), b = 25, a = 4

(3, 2 + 25), (3, 2 - 25)

\frac{x ^{2}}{64} + \frac{y ^{2}}{9} = 1

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

\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1

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

5 x^{2} + 9 y^{2} - 20 x + 54 y + 56 = 0

What is the vertices ((x-3)^2)/(16)+((y-2)^2)/(20)=1 ?
The vertices ((x-3)^2)/(16)+((y-2)^2)/(20)=1 is (3,2+2sqrt(5)),(3,2-2sqrt(5))