What is ((x+2))/(16)+((y-3)^2)/(49)=1 ?

The solution to ((x+2))/(16)+((y-3)^2)/(49)=1 is Parabola with (h,k)=(14,3),p=-49/64

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((x+2))/(16)+((y-3)^2)/(49)=1

\frac{( x + 2 )}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

(h, k) = (14, 3), p = - \frac{49}{64}

Solution steps

\frac{x + 2}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

Rewrite

\frac{x + 2}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

in the standard form:

4 (- \frac{49}{64}) (x - 14) = (y - 3)^{2}

Therefore parabola properties are:

(h, k) = (14, 3), p = - \frac{49}{64}

y = - 0.005 (x)^{2} + 190

y^{2} - x - 16 = 0

(y - 3)^{2} = 16 (x - 6)

2 x^{2} + 3 y + 19 = 0

(x - 2)^{2} = - 33 (y - 4)

What is ((x+2))/(16)+((y-3)^2)/(49)=1 ?
The solution to ((x+2))/(16)+((y-3)^2)/(49)=1 is Parabola with (h,k)=(14,3),p=-49/64