What is the vertices 3x^2+4y^2-12x=36 ?

The vertices 3x^2+4y^2-12x=36 is (6,0),(-2,0)

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vertices 3x^2+4y^2-12x=36

vertices

3 x^{2} + 4 y^{2} - 12 x = 36

(6, 0), (- 2, 0)

Solution steps

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

Calculate ellipse properties

3 x^{2} + 4 y^{2} - 12 x = 36 :

Ellipse with center

(h, k) = (2, 0), a = 4, b = 23

(2 + 4, 0), (2 - 4, 0)

Refine

(6, 0), (- 2, 0)

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

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

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{4} = 1

x^{2} + 4 y^{2} - 6 x + 8 y - 3 = 0

What is the vertices 3x^2+4y^2-12x=36 ?
The vertices 3x^2+4y^2-12x=36 is (6,0),(-2,0)