Question:
<p>Let P(3, 3) be a point on the hyperbola <span class="math-tex">\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)</span>. If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair (a<sup>2</sup>, e<sup>2</sup>) is equal to:</p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{9}{2}, 2\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{3}{2}, 2\right)\)</span></p>
<p style="display:inline">(9, 3)</p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{9}{2}, 3\right)\)</span></p>
Step-by-Step Solution
Key Concept: Determine the values of $a^2$ and $b^2$ by simultaneously using the hyperbola's equation at the point of contact and the intercept properties of the normal line.
<p><span class="math-tex">$\because$</span> The equation of the hyperbola is<br />
<span class="math-tex">$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$</span><br />
<span class="math-tex">$\because$</span> Equation of hyperbola passes through (3, 3)<br />
<span class="math-tex">$\frac{1}{a^{2}}-\frac{1}{b^{2}}=\frac{1}{9}$</span> ...(i)<br />
Equation of normal at point (3, 3) is:<br />
<span class="math-tex">$\frac{x-3}{\frac{1}{a^{2}} \cdot 3}=\frac{y-3}{-\frac{1}{b^{2}} \cdot 3}$</span><br />
<span class="math-tex">$\because$</span> It passes through (9, 0)<br />
<span class="math-tex">$\frac{6}{\frac{1}{a^{2}}}=\frac{-3}{-\frac{1}{b^{2}}}$</span><br />
<span class="math-tex">$\therefore \frac{1}{b^{2}}=\frac{1}{2 a^{2}}$</span> ...(ii)<br />
From equations (i) and (ii),<br />
<span class="math-tex">$a^{2}=\frac{9}{2}$</span>, b<sup>2</sup> = 9<br />
<span class="math-tex">$\because$</span> Eccentricity = e, then <span class="math-tex">$e^{2}=1+\frac{b^{2}}{a^{2}}=3$</span><br />
<span class="math-tex">$\therefore\left(a^{2}, e^{2}\right)=\left(\frac{9}{2}, 3\right)$</span></p>
Correct Answer: D