What is ((x+3)^2)/(16)+((y-5)^2)/(25)=1 ?

The solution to ((x+3)^2)/(16)+((y-5)^2)/(25)=1 is Ellipse with (h,k)=(-3,5),a=4,b=5

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((x+3)^2)/(16)+((y-5)^2)/(25)=1

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

(h, k) = (- 3, 5), a = 4, b = 5

Solution steps

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{4 ^{2}} + \frac{( y - 5 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 5), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 5), b = 5, a = 4

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

y \geq x^{2}

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

x^{2} + 8 y = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

What is ((x+3)^2)/(16)+((y-5)^2)/(25)=1 ?
The solution to ((x+3)^2)/(16)+((y-5)^2)/(25)=1 is Ellipse with (h,k)=(-3,5),a=4,b=5