What is the center 16y^2+25x^2-80x+36y+80=0 ?

The center 16y^2+25x^2-80x+36y+80=0 is (8/5 ,-9/8)

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center 16y^2+25x^2-80x+36y+80=0

center

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

(\frac{8}{5}, - \frac{9}{8})

Solution steps

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

Rewrite

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

in the form of the standard ellipse equation

\frac{( x - \frac{8}{5} ) ^{2}}{( \frac{17}{10} ) ^{2}} + \frac{( y - ( - \frac{9}{8} ) ) ^{2}}{( \frac{17}{8} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (\frac{8}{5}, - \frac{9}{8}), a = \frac{17}{10}, b = \frac{17}{8}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (\frac{8}{5}, - \frac{9}{8}), b = \frac{17}{8}, a = \frac{17}{10}

And the center is:

(\frac{8}{5}, - \frac{9}{8})

9 x^{2} + y^{2} = 1

9 x^{2} + y^{2} = 9

(\frac{X + 3}{25})^{2} + (\frac{Y - 7}{36})^{2} = 1

\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{16} = 1

What is the center 16y^2+25x^2-80x+36y+80=0 ?
The center 16y^2+25x^2-80x+36y+80=0 is (8/5 ,-9/8)