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

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

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

\frac{x - 3}{2 ^{2}} - \frac{y ^{2}}{2 ^{2}} = 1

(h, k) = (7, 0), p = \frac{1}{4}

Solution steps

\frac{x - 3}{2 ^{2}} - \frac{y ^{2}}{2 ^{2}} = 1

Rewrite

\frac{x - 3}{2 ^{2}} - \frac{y ^{2}}{2 ^{2}} = 1

in the standard form:

4 \cdot \frac{1}{4} (x - 7) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (7, 0), p = \frac{1}{4}

(y - 3)^{2} = 7 (x - 8)

x = \frac{( y ^{(2)} + 5 )}{( 2 )}

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

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

3 y^{2} = 1 - u

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