What is ((x+4)^2)/9+((y-4)^2)/(16)=1 ?

The solution to ((x+4)^2)/9+((y-4)^2)/(16)=1 is Ellipse with (h,k)=(-4,4),a=3,b=4

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((x+4)^2)/9+((y-4)^2)/(16)=1

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

(\frac{x - 1}{36})^{2} + (\frac{y + 2}{25})^{2} = 1

9 x^{2} + 36 y^{2} + 36 x - 216 y + 36 = 0

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

25 x^{2} + 16 y^{2} - 400 = 0

4 x^{2} + 9 y^{2} - 24 x - 36 y + 36 = 0

What is ((x+4)^2)/9+((y-4)^2)/(16)=1 ?
The solution to ((x+4)^2)/9+((y-4)^2)/(16)=1 is Ellipse with (h,k)=(-4,4),a=3,b=4