What is the center ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1 ?

The center ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1 is (8,-5)

Popular >

center ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1

center

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1

(8, - 5)

Solution steps

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

\frac{( x - 8 ) ^{2}}{7 ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (8, - 5), a = 7, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (8, - 5), a = 7, b = 3

And the center is:

(8, - 5)

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

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

13 x^{2} + 49 y^{2} = 637

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

What is the center ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1 ?
The center ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1 is (8,-5)