What is the vertices 27x^2+48y^2-54x-192y-213=0 ?

The vertices 27x^2+48y^2-54x-192y-213=0 is (5,2),(-3,2)

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

vertices

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

(5, 2), (- 3, 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (1, 2), a = 4, b = 3

(1 + 4, 2), (1 - 4, 2)

Refine

(5, 2), (- 3, 2)

\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

100 x^{2} + 96 y^{2} - 8840 = 0

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{25} = 1

What is the vertices 27x^2+48y^2-54x-192y-213=0 ?
The vertices 27x^2+48y^2-54x-192y-213=0 is (5,2),(-3,2)