What is (x^2)/(16)+(y-4)^2=1 ?

The solution to (x^2)/(16)+(y-4)^2=1 is Ellipse with (h,k)=(0,4),a=4,b=1

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(x^2)/(16)+(y-4)^2=1

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

(h, k) = (0, 4), a = 4, b = 1

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - 4 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 4), a = 4, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 4), a = 4, b = 1

\frac{x ^{2}}{169} + \frac{y ^{2}}{144} = 1

x^{2} + 2 y^{2} + 2 x - 20 y + 43 = 0

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

x^{2} + \frac{1}{3} y^{2} = x + y

36 x^{2} + 9 y^{2} + 48 x - 36 y + 43 = 0

What is (x^2)/(16)+(y-4)^2=1 ?
The solution to (x^2)/(16)+(y-4)^2=1 is Ellipse with (h,k)=(0,4),a=4,b=1