What is the center 8x^2+2y^2+16x=1 ?

The center 8x^2+2y^2+16x=1 is (-1,0)

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center 8x^2+2y^2+16x=1

center

8 x^{2} + 2 y^{2} + 16 x = 1

(- 1, 0)

Solution steps

8 x^{2} + 2 y^{2} + 16 x = 1

Rewrite

8 x^{2} + 2 y^{2} + 16 x = 1

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{( \frac{3}{2 2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{3}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = \frac{3}{2 2}, b = \frac{3}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), b = \frac{3}{2}, a = \frac{3}{2 2}

And the center is:

(- 1, 0)

25 x^{2} + 16 y^{2} + 50 x - 64 y - 311 = 0

4 x^{2} + 9 y^{2} - 16 x - 18 y - 11 = 0

\frac{( x - 7 ) ^{2}}{25} + \frac{( y - 6 ) ^{2}}{39} = 1

\frac{( x - 1 ) ^{2}}{100} + \frac{( y - 2 ) ^{2}}{36} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1