What is the vertices 9x^2+25y^2-36x-50y+60=0 ?

The vertices 9x^2+25y^2-36x-50y+60=0 is (7/3 ,1),(5/3 ,1)

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vertices 9x^2+25y^2-36x-50y+60=0

vertices

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0

(\frac{7}{3}, 1), (\frac{5}{3}, 1)

Solution steps

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

Calculate ellipse properties

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0 :

Ellipse with center

(h, k) = (2, 1), a = \frac{1}{3}, b = \frac{1}{5}

(2 + \frac{1}{3}, 1), (2 - \frac{1}{3}, 1)

Refine

(\frac{7}{3}, 1), (\frac{5}{3}, 1)

\frac{( x + 1 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{48} = 1

3 x^{2} + 4 y^{2} = 48

2 x^{2} + y^{2} - 98 = 0

3 x^{2} - 6 x + y^{2} + 4 y - 11 = 0

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

What is the vertices 9x^2+25y^2-36x-50y+60=0 ?
The vertices 9x^2+25y^2-36x-50y+60=0 is (7/3 ,1),(5/3 ,1)