Let P(3,3) be a point on the hyperbola frac{x | vedclass

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(A) Since the point $(3,3)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we have $\frac{9}{a^{2}}-\frac{9}{b^{2}}=1$  $(i)$.The e

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$(\frac{9}{2}, 3)$

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(A) Since the point $(3,3)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we have $\frac{9}{a^{2}}-\frac{9}{b^{2}}=1$  $(i)$.The equation of the normal at $(x_{1}, y_{1})$ to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2} + b^{2}$.Substituting $(x_{1}, y_{1}) = (3,3)$,the normal is $\frac{a^{2}x}{3} + \frac{b^{2}y}{3} = a^{2} + b^{2}$.This normal passes through $(9,0)$,so $\frac{a^{2}(9)}{3} + 0 = a^{2} + b^{2}$ $\Rightarrow 3a^{2} = a^{2} + b^{2}$ $\Rightarrow b^{2} = 2a^{2}$  $(ii)$.Substituting $(ii)$ into $(i)$: $\frac{9}{a^{2}} - \frac{9}{2a^{2}} = 1$ $\Rightarrow \frac{18-9}{2a^{2}} = 1$ $\Rightarrow 9 = 2a^{2}$ $\Rightarrow a^{2} = \frac{9}{2}$.Then $b^{2} = 2(\frac{9}{2}) = 9$.The eccentricity $e$ is given by $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{9}{9/2} = 1 + 2 = 3$.Thus,the ordered pair $(a^{2}, e^{2})$ is $(\frac{9}{2}, 3)$.

Let $P(3,3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If the normal to it at $P$ intersects the $x$-axis at $(9,0)$ and $e$ is its eccentricity,then the ordered pair $(a^{2}, e^{2})$ is equal to

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