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

The solution to (x-3)2+(y^2)/4 =1 is Parabola with (h,k)=(7/2 ,0),p=-2

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

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

(h, k) = (\frac{7}{2}, 0), p = - 2

Solution steps

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

Rewrite

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

in the standard form:

4 (- 2) (x - \frac{7}{2}) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{2}, 0), p = - 2

x = 2 x - y^{2}

(y + 2)^{2} = 16 x

x^{2} - 6 x + y + 5 = 0

x^{2} - k (x) - x = 0

xx - y = 5

What is (x-3)2+(y^2)/4 =1 ?
The solution to (x-3)2+(y^2)/4 =1 is Parabola with (h,k)=(7/2 ,0),p=-2