Let P(3, 3) be a point on the hyperbola,
${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$. If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair (a², e²) is equal to :

Given hyperbola, ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$<br><br>Point P (3, 3) is on the parabola<br><br>$\therefore$ ${9 \over {{a^2}}} - {9 \over {{b^2}}} = 1$ ...(1)<br><br>Equation of normal at (x₁, y₁),<br><br>${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2}{e^2}$<br><br>Normal at p(3, 3),<br><br>${{{a^2}x} \over 3} - {{{b^2}y} \over 3} = {a^2}{e^2}$ ... (2)<br><br>It intersect x-axis at (9, 0),<br><br>Putting in equation (2),<br><br>${{9{a^2}} \over 3} - 0 = {a^2}{e^2}$<br><br>$\Rightarrow 3{a^2} = {a^2}{e^2}$<br><br>$\Rightarrow {e^2} = 3$<br><br>Also, ${e^2} = 1 + {{{b^2}} \over {{a^2}}}$<br><br>$\Rightarrow 3 = 1 + {{{b^2}} \over {{a^2}}}$<br><br>$\Rightarrow {b^2} = 2{a^2}$<br><br>Putting the value of b² in equation (1),<br><br>${9 \over {{a^2}}} - {9 \over {2{a^2}}} = 1$<br><br>$\Rightarrow {a^2} = {9 \over 2}$<br><br>$\therefore$ $({a^2},{e^2}) = \left( {{9 \over 2},3} \right)$