What is ((x-2)^2}{16}+\frac{(y-4)^2)/7 =1 ?

The solution to ((x-2)^2}{16}+\frac{(y-4)^2)/7 =1 is Ellipse with (h,k)=(2,4),a=4,b=sqrt(7)

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((x-2)^2}{16}+\frac{(y-4)^2)/7 =1

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{7} = 1

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

Solution steps

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{7} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{7} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{16} = 1

4 x^{2} + 9 y^{2} = 40

2 x^{2} + 2 y^{2} + 4 y = 0

7 x^{2} + 16 y^{2} - 28 x - 128 y + 172 = 0

x^{2} + 5 y^{2} + 6 x - 40 y - 167 = 0

What is ((x-2)^2}{16}+\frac{(y-4)^2)/7 =1 ?
The solution to ((x-2)^2}{16}+\frac{(y-4)^2)/7 =1 is Ellipse with (h,k)=(2,4),a=4,b=sqrt(7)