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

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

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

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

(h, k) = (- 1, - 3), a = 4, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, - 3), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, - 3), a = 4, b = 3

x^{2} + (8 - 6) y^{2} = 1

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{9} = 1

\frac{100}{1 + x ^{2} + 2 y ^{2}} = 1

81 x^{2} + 49 y^{2} + 486 x - 490 y - 2015 = 0

0 = 1 + 2 x^{2} + 2 y^{2}

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