center 2x^2+y^2-2x+16y+1=0

Lösung

center

2 x^{2} + y^{2} - 2 x + 16 y + 1 = 0

(\frac{1}{2}, - 8)

Schritte zur Lösung

2 x^{2} + y^{2} - 2 x + 16 y + 1 = 0

Rewrite

2 x^{2} + y^{2} - 2 x + 16 y + 1 = 0

in the form of the standard ellipse equation

\frac{( x - \frac{1}{2} ) ^{2}}{( \frac{127}{2} ) ^{2}} + \frac{( y - ( - 8 ) ) ^{2}}{( \frac{127}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (\frac{1}{2}, - 8), a = \frac{127}{2}, b = \frac{127}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (\frac{1}{2}, - 8), b = \frac{127}{2}, a = \frac{127}{2}

And the center is:

(\frac{1}{2}, - 8)

ce n t er 25 x^{2} + 16 y^{2} - 150 x + 128 y + 81 = 0

ce n t er 16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

ce n t er x^{2} + y^{2} - 10 y = 0

ce n t er \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

ce n t er 6 x^{2} + 9 y^{2} - 18 y - 45 = 0