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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{a - 16} = 1

5 y^{2} - 6 x - 3 x^{2} - 10 y - 13 = 0

[f (x)]^{2} = 5 x^{2} - x + 3

2 y^{2} + \frac{10}{2} x^{2} - 8 + 1 = 3 y^{2} - 4

3 x^{2} - 4 y^{2} = 36

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