What is ((x-5)^2)/(16)+((y+15)^2)/(25)=1 ?

The solution to ((x-5)^2)/(16)+((y+15)^2)/(25)=1 is Ellipse with (h,k)=(5,-15),a=4,b=5

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((x-5)^2)/(16)+((y+15)^2)/(25)=1

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 15 ) ^{2}}{25} = 1

(h, k) = (5, - 15), a = 4, b = 5

Solution steps

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 15 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 15 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{4 ^{2}} + \frac{( y - ( - 15 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, - 15), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, - 15), b = 5, a = 4

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

\frac{x ^{2}}{3. 5 ^{2}} + \frac{y ^{2}}{2. 5 ^{2}} = 1

12 x^{2} + y^{2} = 24

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

36 x^{2} + 16 y^{2} = 576

What is ((x-5)^2)/(16)+((y+15)^2)/(25)=1 ?
The solution to ((x-5)^2)/(16)+((y+15)^2)/(25)=1 is Ellipse with (h,k)=(5,-15),a=4,b=5