What is ((x+7)^2}{16}+\frac{(y+5)^2)/4 =1 ?

The solution to ((x+7)^2}{16}+\frac{(y+5)^2)/4 =1 is Ellipse with (h,k)=(-7,-5),a=4,b=2

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((x+7)^2}{16}+\frac{(y+5)^2)/4 =1

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{4} = 1

(h, k) = (- 7, - 5), a = 4, b = 2

Solution steps

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{4} = 1

Rewrite

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 7 ) ) ^{2}}{4 ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 7, - 5), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 7, - 5), a = 4, b = 2

x^{2} + 3 \cdot y^{2} = 84

y^{2} = 9 - 4 x^{2}

25 x^{2} + 16 y^{2} - 150 x + 64 y = 111

\frac{x ^{2}}{25} + \frac{y ^{2}}{1} = 1

\frac{x ^{2}}{2 7 ^{2}} + \frac{y ^{2}}{2 5 ^{2}} = 1

What is ((x+7)^2}{16}+\frac{(y+5)^2)/4 =1 ?
The solution to ((x+7)^2}{16}+\frac{(y+5)^2)/4 =1 is Ellipse with (h,k)=(-7,-5),a=4,b=2