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

The solution to ((x-3^2))/4+((y-1)^2)/3 =1 is Parabola with (h,k)=(13,1),p=-3/16

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

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

(h, k) = (13, 1), p = - \frac{3}{16}

Solution steps

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

Rewrite

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

in the standard form:

4 (- \frac{3}{16}) (x - 13) = (y - 1)^{2}

Therefore parabola properties are:

(h, k) = (13, 1), p = - \frac{3}{16}

y > - x^{2} + 4 x - 6

y^{2} - 4 x + 4 y + 8 = 0

x^{2} - 10 x - 4 y + 17 = 0

y^{2} - 20 x = 69

15 x^{2} + 7 y = 13

What is ((x-3^2))/4+((y-1)^2)/3 =1 ?
The solution to ((x-3^2))/4+((y-1)^2)/3 =1 is Parabola with (h,k)=(13,1),p=-3/16