MathematicsEasy57×since 2003Q3954The normal to the hyperbola x2a2−y29=1{{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1a2x2−9y2=1 at the point (8,33)\left( {8,3\sqrt 3 } \right)(8,33) on it passes through the point :A(15,−23)\left( {15, - 2\sqrt 3 } \right)(15,−23)B(9,23)\left( {9,2\sqrt 3 } \right)(9,23)C(−1,93)\left( { - 1,9\sqrt 3 } \right)(−1,93)D(−1,63)\left( { - 1,6\sqrt 3 } \right)(−1,63)Check answerSkip