JEE Main 202226 Jun, Evening shiftMathematicsEasyCoordinate Geometry478× since 2002The 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 :AOption A. (15,−23)\left( {15, - 2\sqrt 3 } \right)(15,−23)BOption B. (9,23)\left( {9,2\sqrt 3 } \right)(9,23)COption C. (−1,93)\left( { - 1,9\sqrt 3 } \right)(−1,93)DOption D. (−1,63)\left( { - 1,6\sqrt 3 } \right)(−1,63)CheckSkip · show solution
