What is ((x-3)2+y^2)/4 =1 ?

The solution to ((x-3)2+y^2)/4 =1 is Parabola with (h,k)=(5,0),p=-1/2

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((x-3)2+y^2)/4 =1

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

(h, k) = (5, 0), p = - \frac{1}{2}

Solution steps

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

Rewrite

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

in the standard form

4 (- \frac{1}{2}) (x - 5) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (5, 0), p = - \frac{1}{2}

y^{2} + 33 = - 8 x - 23

y^{2} + 6 y + 2 x + 1 = 0

x^{2} - 6 x - y + 33 = 0

y^{2} + 2 x + 6 y + 17 = 0

9 x^{2} - 6 y - 9 - x = 0

What is ((x-3)2+y^2)/4 =1 ?
The solution to ((x-3)2+y^2)/4 =1 is Parabola with (h,k)=(5,0),p=-1/2