What is ((x-3)^2)/9+((y+4))/(25)=1 ?

The solution to ((x-3)^2)/9+((y+4))/(25)=1 is Parabola with (h,k)=(3,21),p=-9/100

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((x-3)^2)/9+((y+4))/(25)=1

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 4 )}{25} = 1

(h, k) = (3, 21), p = - \frac{9}{100}

Solution steps

\frac{( x - 3 ) ^{2}}{9} + \frac{y + 4}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{9} + \frac{y + 4}{25} = 1

in the standard form:

4 (- \frac{9}{100}) (y - 21) = (x - 3)^{2}

Therefore parabola properties are:

(h, k) = (3, 21), p = - \frac{9}{100}

t^{2} = s

x = \frac{1}{2} \cdot y (y - 2)

4 y^{2} - 8 y - x + 1 = 0

8 (x - 0) = (y - 0)^{2}

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

What is ((x-3)^2)/9+((y+4))/(25)=1 ?
The solution to ((x-3)^2)/9+((y+4))/(25)=1 is Parabola with (h,k)=(3,21),p=-9/100