What is ((x-6)^2}{25}+\frac{(y-7)^2)/9 =1 ?

The solution to ((x-6)^2}{25}+\frac{(y-7)^2)/9 =1 is Ellipse with (h,k)=(6,7),a=5,b=3

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((x-6)^2}{25}+\frac{(y-7)^2)/9 =1

\frac{( x - 6 ) ^{2}}{25} + \frac{( y - 7 ) ^{2}}{9} = 1

(h, k) = (6, 7), a = 5, b = 3

Solution steps

\frac{( x - 6 ) ^{2}}{25} + \frac{( y - 7 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 6 ) ^{2}}{25} + \frac{( y - 7 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

\frac{( x - 6 ) ^{2}}{5 ^{2}} + \frac{( y - 7 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (6, 7), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (6, 7), a = 5, b = 3

x^{2} + 2 y^{2} + 8 x - 4 y - 6 = 0

(\frac{y - 4}{16})^{2} + (\frac{x - 6}{4})^{2} = 1

8 x^{2} + y^{2} - 32 x - 16 y = - 16

6 x^{2} + 3 y^{2} + 3 y + 6 x = 15

\frac{x ^{2}}{16} + \frac{y ^{2}}{48} = 1

What is ((x-6)^2}{25}+\frac{(y-7)^2)/9 =1 ?
The solution to ((x-6)^2}{25}+\frac{(y-7)^2)/9 =1 is Ellipse with (h,k)=(6,7),a=5,b=3