What is (x-1)+4(y-2)^2=10 ?

The solution to (x-1)+4(y-2)^2=10 is Parabola with (h,k)=(11,2),p=-1/16

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(x-1)+4(y-2)^2=10

(x - 1) + 4 (y - 2)^{2} = 10

(h, k) = (11, 2), p = - \frac{1}{16}

Solution steps

x - 1 + 4 (y - 2)^{2} = 10

Rewrite

x - 1 + 4 (y - 2)^{2} = 10

in the standard form:

4 (- \frac{1}{16}) (x - 11) = (y - 2)^{2}

Therefore parabola properties are:

(h, k) = (11, 2), p = - \frac{1}{16}

x + (y - 6)^{2} = 0

(y + 2)^{2} = - 2 (x - 3)

(y + 3)^{2} = 6 (x + 12)

5 (y + 1) 3 + x^{2} = 1

(x - h)^{2} = - 4 p (y - k)

What is (x-1)+4(y-2)^2=10 ?
The solution to (x-1)+4(y-2)^2=10 is Parabola with (h,k)=(11,2),p=-1/16