How do you write the hyperbola $36 x^{2} - 9 y^{2} = 324$ $36x^2-9y^2=324$ in standard form?

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Answer 1 · 2016-10-09T13:16:37

$\frac{{(6 x)}^{2}}{18^{2}} - \frac{{(3 y)}^{2}}{18^{2}} = 1$ $(6x)^2/18^2-(3y)^2/18^2=1$

Target form is $\frac{{(y - k)}^{2}}{a^{2}} - \frac{(x - h^{2})}{b^{2}} = 1$ $((y-k)^2)/a^2-((x-h^2))/b^2 =1$

Notice that $36 = 6^{2}; 9 = 3^{2}; 324 = 18^{2}$ $36=6^2"; "9=3^2"; "324=18^2$

So we can write:

$\frac{{(6 x)}^{2}}{a^{2}} - \frac{{(3 y)}^{2}}{b^{2}} = 18^{2}$ $(6x)^2/a^2-(3y)^2/b^2=18^2$

So $h and k = 0 \leftarrow$ $h" and "k=0 larr$ symmetrical about the origin

Divide both sides by $18^{2}$ $18^2$

$\frac{{(6 x)}^{2}}{18^{2}} - \frac{{(3 y)}^{2}}{18^{2}} = 1$ $(6x)^2/18^2-(3y)^2/18^2=1$

Tony B Tony B

How do you write the hyperbola 36x2−9y2=32436x^2-9y^2=324 in standard form?