What is ((x-3)^2)/(16)+((y-7)^2)/(64)=1 ?

The solution to ((x-3)^2)/(16)+((y-7)^2)/(64)=1 is Ellipse with (h,k)=(3,7),a=4,b=8

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((x-3)^2)/(16)+((y-7)^2)/(64)=1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 7 ) ^{2}}{64} = 1

(h, k) = (3, 7), a = 4, b = 8

Solution steps

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 7 ) ^{2}}{64} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 7 ) ^{2}}{64} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{4 ^{2}} + \frac{( y - 7 ) ^{2}}{8 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 7), a = 4, b = 8

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 7), b = 8, a = 4

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25 y^{2} + 36 (x - 8)^{2} = 900

x^{2} + 3 y^{2} - 4 x + 12 y + 13 = 0

4 x^{2} + 3 y^{2} - 32 x - 18 y + 43 = 0

What is ((x-3)^2)/(16)+((y-7)^2)/(64)=1 ?
The solution to ((x-3)^2)/(16)+((y-7)^2)/(64)=1 is Ellipse with (h,k)=(3,7),a=4,b=8