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 :

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

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 :

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