What is ((x-2)^2)/(49)+((y+4)^2)/(16)=1 ?

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

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((x-2)^2)/(49)+((y+4)^2)/(16)=1

\frac{( x - 2 ) ^{2}}{49} + \frac{( y + 4 ) ^{2}}{16} = 1

(h, k) = (2, - 4), a = 7, b = 4

Solution steps

\frac{( x - 2 ) ^{2}}{49} + \frac{( y + 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{49} + \frac{( y + 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 4), a = 7, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 4), a = 7, b = 4

25 x^{2} + 4 y^{2} - 225 = 0

16 x^{2} + 36 y^{2} - 224 x + 208 = 0

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1

\frac{x ^{2}}{1 2 ^{2}} + \frac{y ^{2}}{5 ^{2}} = 1

36 x^{2} + 16 y^{2} - 576 = 0

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