What is ((x-3)^2)/5+((y-1)^2)/(15)=1 ?

The solution to ((x-3)^2)/5+((y-1)^2)/(15)=1 is Ellipse with (h,k)=(3,1),a=sqrt(5),b=sqrt(15)

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((x-3)^2)/5+((y-1)^2)/(15)=1

\frac{( x - 3 ) ^{2}}{5} + \frac{( y - 1 ) ^{2}}{15} = 1

(h, k) = (3, 1), a = 5, b = 15

Solution steps

\frac{( x - 3 ) ^{2}}{5} + \frac{( y - 1 ) ^{2}}{15} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{5} + \frac{( y - 1 ) ^{2}}{15} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{( 5 ) ^{2}} + \frac{( y - 1 ) ^{2}}{( 15 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 1), a = 5, b = 15

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 1), b = 15, a = 5

9 x^{2} + 4 y^{2} + 36 x - 24 y + 36 = 0

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

\frac{x ^{2}}{169} + \frac{y ^{2}}{25} = 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{36} = 1

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

What is ((x-3)^2)/5+((y-1)^2)/(15)=1 ?
The solution to ((x-3)^2)/5+((y-1)^2)/(15)=1 is Ellipse with (h,k)=(3,1),a=sqrt(5),b=sqrt(15)