What is ((x+4)^2)/(16)+((y-5)^2)/(25)=1 ?

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

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((x+4)^2)/(16)+((y-5)^2)/(25)=1

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

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

Solution steps

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

Rewrite

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{y ^{2}}{9} = 1 - \frac{x ^{2}}{4}

8 = \frac{1}{2} x^{2} + \frac{1}{4} y^{2}

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

x^{2} = 1 - 4 y^{2}

5 x^{2} + 9 y^{2} - 18 y - 36 = 0

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