What is ((x+3)^2)/(49)+((y-5)^2)/(16)=1 ?

The solution to ((x+3)^2)/(49)+((y-5)^2)/(16)=1 is Ellipse with (h,k)=(-3,5),a=7,b=4

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((x+3)^2)/(49)+((y-5)^2)/(16)=1

\frac{( x + 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{16} = 1

(h, k) = (- 3, 5), a = 7, b = 4

Solution steps

\frac{( x + 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{7 ^{2}} + \frac{( y - 5 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 5), a = 7, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 3, 5), a = 7, b = 4

18 x^{2} - 72 x + 4 y^{2} + 4 y + 37 = 0

6 (x + 1)^{2} + 2 (y - 1)^{2} = 72

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

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

\frac{x ^{2}}{3 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

What is ((x+3)^2)/(49)+((y-5)^2)/(16)=1 ?
The solution to ((x+3)^2)/(49)+((y-5)^2)/(16)=1 is Ellipse with (h,k)=(-3,5),a=7,b=4