What is the vertices (x-2)^2+4(y-3)^2=4 ?

The vertices (x-2)^2+4(y-3)^2=4 is (4,3),(0,3)

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vertices (x-2)^2+4(y-3)^2=4

vertices

(x - 2)^{2} + 4 (y - 3)^{2} = 4

(4, 3), (0, 3)

Solution steps

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

Calculate ellipse properties

(x - 2)^{2} + 4 (y - 3)^{2} = 4 :

Ellipse with center

(h, k) = (2, 3), a = 2, b = 1

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

Refine

(4, 3), (0, 3)

x^{2} + 4 y^{2} - 6 x + 20 y - 2 = 0

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{25} = 1

2 \frac{( x - 6 ) ^{2}}{3} + 7 \frac{( y + 4 ) ^{2}}{4} = 1

What is the vertices (x-2)^2+4(y-3)^2=4 ?
The vertices (x-2)^2+4(y-3)^2=4 is (4,3),(0,3)