What is ((x-2)^2)/9+((y-4)^2)/7 =1 ?

The solution to ((x-2)^2)/9+((y-4)^2)/7 =1 is Ellipse with (h,k)=(2,4),a=3,b=sqrt(7)

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((x-2)^2)/9+((y-4)^2)/7 =1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{7} = 1

(h, k) = (2, 4), a = 3, b = 7

Solution steps

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{7} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{7} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{3 ^{2}} + \frac{( y - 4 ) ^{2}}{( 7 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, 4), a = 3, b = 7

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, 4), a = 3, b = 7

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

9 x^{2} + 49 y^{2} - 98 y - 392 = 0

x^{2} + 49 y^{2} = 49

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

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

What is ((x-2)^2)/9+((y-4)^2)/7 =1 ?
The solution to ((x-2)^2)/9+((y-4)^2)/7 =1 is Ellipse with (h,k)=(2,4),a=3,b=sqrt(7)