center 16x^2+25y^2+32x-384=0

Lösung

center

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

(- 1, 0)

Schritte zur Lösung

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

Rewrite

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{5 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = 5, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), a = 5, b = 4

And the center is:

(- 1, 0)

ce n t er 4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0

ce n t er x^{2} + 36 y^{2} - 16 x - 72 y + 64 = 0

ce n t er 9 X^{2} + 4 Y^{2} = 36

ce n t er 2 x^{2} + y^{2} - 2 x + 16 y + 1 = 0

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