What is ((x+2)^2)/(16)+((y-4)^2)/(25)=1 ?

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

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((x+2)^2)/(16)+((y-4)^2)/(25)=1

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

(h, k) = (- 2, 4), a = 4, b = 5

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 4), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 4), b = 5, a = 4

\frac{( x - 3 ) ^{2}}{169} + \frac{( y + 4 ) ^{2}}{144} = 1

9 x^{2} + 7 y^{2} = 42

\frac{( x - 2.1 ) ^{2}}{2.25} + \frac{( y + 1.6 ) ^{2}}{6.16} = 1

x^{2} - 2 x + 3 y^{2} - 2 y - 5 = 0

\frac{( x - 5 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{1 0 ^{2}} = 1

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