Solution
Step 1
The general equation of the hyperbola is given below
[tex]\frac{(x-h)^2}{a^2}-\text{ }\frac{(y-k)^2}{b^2}\text{ = 1}[/tex]
Where (h,k) is the center and a and b are the length of semi-major and semi-minor
Step 2
K = 0 and h = 6
[tex]\begin{gathered} (h\text{ + 6\rparen}^2\text{ = a}^2\text{ + b}^2 \\ (h\text{ - 6\rparen}^2\text{ = a}^2\text{ + b}^2 \\ We\text{ know that }\frac{b}{a}\text{ = 1} \\ b\text{ = a} \end{gathered}[/tex]
Step 3
[tex]\begin{gathered} a^2\text{ + b}^2\text{ = 6}^2 \\ 2a^2\text{ = 36} \\ a^2\text{ = }\frac{36}{2} \\ a^2\text{ = 18} \\ a\text{ = }\sqrt{18} \\ a\text{ = 3}\sqrt{2} \\ \text{b = 3}\sqrt{2} \end{gathered}[/tex]
Step 4
[tex]\begin{gathered} \text{The standard equation of a parabola is }\frac{x^2}{(3\sqrt{2})^2}\text{ - }\frac{y^2}{(3\sqrt{2})^2}\text{ = 1} \\ or \\ \frac{x^2}{18}\text{ - }\frac{y^2}{18}\text{ = 1} \end{gathered}[/tex]
