What is the axis ((x+5)^2)/(25)+((y-5)^2)/(16)=1 ?

The axis ((x+5)^2)/(25)+((y-5)^2)/(16)=1 is semi-major axis a=5, semi-minor axis b=4

Popular >

axis ((x+5)^2)/(25)+((y-5)^2)/(16)=1

axis

\frac{( x + 5 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{16} = 1

semi - major axis a = 5, semi - minor axis b = 4

Solution steps

\frac{( x + 5 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 5 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

9 x^{2} + 36 y^{2} + 36 x + 432 y = - 1008

6 x^{2} + 8 y^{2} + 12 x - 32 y = 178

3 x^{2} + 12 x + 4 y^{2} - 40 y + 100 = 0

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 6 ) ^{2}}{100} = 1

2 x^{2} + 7 y^{2} = 14

What is the axis ((x+5)^2)/(25)+((y-5)^2)/(16)=1 ?
The axis ((x+5)^2)/(25)+((y-5)^2)/(16)=1 is semi-major axis a=5, semi-minor axis b=4