What is the vertices 3x^2+12y^2-6x-48y+33=0 ?

The vertices 3x^2+12y^2-6x-48y+33=0 is (1+sqrt(6),2),(1-sqrt(6),2)

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vertices 3x^2+12y^2-6x-48y+33=0

vertices

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0

(1 + 6, 2), (1 - 6, 2)

Solution steps

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

Calculate ellipse properties

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0 :

Ellipse with center

(h, k) = (1, 2), a = 6, b = \frac{3}{2}

(1 + 6, 2), (1 - 6, 2)

36 x^{2} + 25 y^{2} = 900

x^{2} + 2 y^{2} + 2 x - 20 y + 43 = 0

x^{2} + y^{2} + 8 x = 32

4 x^{2} + 9 y^{2} - 16 x + 36 y + 16 = 0

25 x^{2} + 81 y^{2} = 2025

What is the vertices 3x^2+12y^2-6x-48y+33=0 ?
The vertices 3x^2+12y^2-6x-48y+33=0 is (1+sqrt(6),2),(1-sqrt(6),2)