What is the vertices x^2+4y^2-6x-16y=36 ?

The vertices x^2+4y^2-6x-16y=36 is (3+sqrt(61),2),(3-sqrt(61),2)

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vertices x^2+4y^2-6x-16y=36

vertices

x^{2} + 4 y^{2} - 6 x - 16 y = 36

(3 + 61, 2), (3 - 61, 2)

Solution steps

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

Calculate ellipse properties

x^{2} + 4 y^{2} - 6 x - 16 y = 36 :

Ellipse with center

(h, k) = (3, 2), a = 61, b = \frac{61}{2}

(3 + 61, 2), (3 - 61, 2)

(\frac{x + 5}{4})^{2} + (\frac{y - 4}{16})^{2} = 1

x^{2} + 25 y^{2} = 25

x^{2} + 9 y^{2} + 4 x = 5

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

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

What is the vertices x^2+4y^2-6x-16y=36 ?
The vertices x^2+4y^2-6x-16y=36 is (3+sqrt(61),2),(3-sqrt(61),2)