What is 1=((x-3)^2}{16}+\frac{y^2)/4 ?

The solution to 1=((x-3)^2}{16}+\frac{y^2)/4 is Ellipse with (h,k)=(3,0),a=4,b=2

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1=((x-3)^2}{16}+\frac{y^2)/4

1 = \frac{( x - 3 ) ^{2}}{16} + \frac{y ^{2}}{4}

(h, k) = (3, 0), a = 4, b = 2

Solution steps

1 = \frac{( x - 3 ) ^{2}}{16} + \frac{y ^{2}}{4}

Rewrite

1 - \frac{( x - 3 ) ^{2}}{16} - \frac{y ^{2}}{4} = 0

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 0), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, 0), a = 4, b = 2

(3 x - 3)^{2} + (- 2 y - 4)^{2} = 43 + 9 + 16

36 x^{2} + 81 y^{2} = 49

49 x^{2} + 16 y^{2} - 784 = 0

9 x^{2} + 4 y^{2} - 54 x - 16 y + 61 = 0

\frac{y ^{2}}{16} + \frac{z ^{2}}{9} = 1

What is 1=((x-3)^2}{16}+\frac{y^2)/4 ?
The solution to 1=((x-3)^2}{16}+\frac{y^2)/4 is Ellipse with (h,k)=(3,0),a=4,b=2