What is ((x-3)^2)/(49)+((y+5)^2)/(25)=1 ?

The solution to ((x-3)^2)/(49)+((y+5)^2)/(25)=1 is Ellipse with (h,k)=(3,-5),a=7,b=5

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((x-3)^2)/(49)+((y+5)^2)/(25)=1

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

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

Solution steps

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

2 x^{2} - 12 x + y^{2} + 6 y + 26 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{49} = 1

2 x^{2} + y^{2} + 12 x - 6 y + 26 = 0

4 x^{2} + 9 y^{2} = 25

9 x^{2} + 49 y^{2} = 441

What is ((x-3)^2)/(49)+((y+5)^2)/(25)=1 ?
The solution to ((x-3)^2)/(49)+((y+5)^2)/(25)=1 is Ellipse with (h,k)=(3,-5),a=7,b=5