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

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 2 ) ^{2}}{36} = 1

\frac{y ^{2}}{36} + \frac{x ^{2}}{100} = 1

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

(x^{2} + 9 x) + (3 y^{2} + 10 y) = 2

\frac{x ^{2}}{100} + \frac{y ^{2}}{50} = 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