What is ((y-3)^2}{16}+\frac{(x+2)^2)/9 =1 ?

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

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((y-3)^2}{16}+\frac{(x+2)^2)/9 =1

\frac{( y - 3 ) ^{2}}{16} + \frac{( x + 2 ) ^{2}}{9} = 1

(h, k) = (- 2, 3), a = 3, b = 4

Solution steps

\frac{( y - 3 ) ^{2}}{16} + \frac{( x + 2 ) ^{2}}{9} = 1

Rewrite

\frac{( y - 3 ) ^{2}}{16} + \frac{( x + 2 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 3), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 3), b = 4, a = 3

7 (x - 2004)^{2} = 23 - y^{2}

\frac{( x - 6 ) ^{2}}{25} + \frac{( y - 7 ) ^{2}}{9} = 1

x^{2} + 2 y^{2} + 8 x - 4 y - 6 = 0

(\frac{y - 4}{16})^{2} + (\frac{x - 6}{4})^{2} = 1

8 x^{2} + y^{2} - 32 x - 16 y = - 16

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