What is the center 16x^2+25y^2-64x+150y-11=0 ?

The center 16x^2+25y^2-64x+150y-11=0 is (2,-3)

Popular >

center 16x^2+25y^2-64x+150y-11=0

center

16 x^{2} + 25 y^{2} - 64 x + 150 y - 11 = 0

(2, - 3)

Solution steps

16 x^{2} + 25 y^{2} - 64 x + 150 y - 11 = 0

Rewrite

16 x^{2} + 25 y^{2} - 64 x + 150 y - 11 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 3), a = \frac{5 3}{2}, b = 23

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 3), a = \frac{5 3}{2}, b = 23

And the center is:

(2, - 3)

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

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

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

x^{2} + 10 x + 9 y^{2} + 16 = 0

9 x^{2} + 16 y^{2} - 18 x - 64 y - 71 = 0

What is the center 16x^2+25y^2-64x+150y-11=0 ?
The center 16x^2+25y^2-64x+150y-11=0 is (2,-3)