What is the vertices x^2+4y^2-6x+8y+9=0 ?

The vertices x^2+4y^2-6x+8y+9=0 is (5,-1),(1,-1)

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vertices x^2+4y^2-6x+8y+9=0

vertices

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

(5, - 1), (1, - 1)

Solution steps

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

Calculate ellipse properties

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0 :

Ellipse with center

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

(3 + 2, - 1), (3 - 2, - 1)

Refine

(5, - 1), (1, - 1)

9 x^{2} + 25 y^{2} - 225 = 0

4 x^{2} - 48 x + 9 y^{2} + 72 y = - 144

\frac{( x - 2 ) ^{2}}{16} + \frac{y ^{2}}{9} = 1

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 4 ) ^{2}}{25} = 1

\frac{( x ) ^{2}}{44} + \frac{y ^{2}}{125} = 1

What is the vertices x^2+4y^2-6x+8y+9=0 ?
The vertices x^2+4y^2-6x+8y+9=0 is (5,-1),(1,-1)