What is ((x-3)^2)/7+(y^2)/(15)=1 ?

The solution to ((x-3)^2)/7+(y^2)/(15)=1 is Ellipse with (h,k)=(3,0),a=sqrt(7),b=sqrt(15)

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((x-3)^2)/7+(y^2)/(15)=1

\frac{( x - 3 ) ^{2}}{7} + \frac{y ^{2}}{15} = 1

(h, k) = (3, 0), a = 7, b = 15

Solution steps

\frac{( x - 3 ) ^{2}}{7} + \frac{y ^{2}}{15} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{7} + \frac{y ^{2}}{15} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{( 7 ) ^{2}} + \frac{( y - 0 ) ^{2}}{( 15 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 0), a = 7, b = 15

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 0), b = 15, a = 7

x^{2} - 6 x + 4 y^{2} + 16 y + 21 = 0

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

16 x^{2} + 36 y^{2} - 96 x + 36 y + 9 = 0

\frac{x ^{2}}{144} + \frac{y ^{2}}{16} = 1

(2 x - 2)^{2} + (3 y - 6)^{2} = 36

What is ((x-3)^2)/7+(y^2)/(15)=1 ?
The solution to ((x-3)^2)/7+(y^2)/(15)=1 is Ellipse with (h,k)=(3,0),a=sqrt(7),b=sqrt(15)