What is the vertices x^2+4y^2-6x-16y+21=0 ?

The vertices x^2+4y^2-6x-16y+21=0 is (5,2),(1,2)

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vertices x^2+4y^2-6x-16y+21=0

vertices

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

(5, 2), (1, 2)

Solution steps

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

Calculate ellipse properties

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0 :

Ellipse with center

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

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

Refine

(5, 2), (1, 2)

100 x^{2} + 36 y^{2} = 225

x^{2} + y^{2} + 2 x = 3 y

\frac{x ^{2}}{49} + \frac{y ^{2}}{169} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

What is the vertices x^2+4y^2-6x-16y+21=0 ?
The vertices x^2+4y^2-6x-16y+21=0 is (5,2),(1,2)