What is ((x-5)^2)/(25)+((y+7)^2)/(16)=1 ?

The solution to ((x-5)^2)/(25)+((y+7)^2)/(16)=1 is Ellipse with (h,k)=(5,-7),a=5,b=4

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((x-5)^2)/(25)+((y+7)^2)/(16)=1

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

(h, k) = (5, - 7), a = 5, b = 4

Solution steps

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 7 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, - 7), a = 5, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, - 7), a = 5, b = 4

\frac{( y + 2 ) ^{2}}{12} + \frac{x ^{2}}{2} - 1 = 0

x^{2} - x + 4 y^{2} + 24 y + 30 = 4

4 x^{2} + y^{2} = 2 y

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

5 x^{2} + 9 y^{2} - 8 x - 4 = 0

What is ((x-5)^2)/(25)+((y+7)^2)/(16)=1 ?
The solution to ((x-5)^2)/(25)+((y+7)^2)/(16)=1 is Ellipse with (h,k)=(5,-7),a=5,b=4