The product of the perpendicular distances drawn from the points (3,0) and (-3,0) to the tangent of the ellipse x^2 36 + y^2 27 =1 at (3, 9 2 ) is

Step-by-step solution

Let P(6 , 3 3 ) be any point on the ellipse x^2 36 + y^2 27 =1 . The equation of the tangent at P(6 , 3 3 ) is x 6 + y 3 3 =1 . 3 3 x +6 y -18 3 =0 The product of the lengths of the perpendiculars from (3,0) and (-3,0) an Eq. (i) is given by aligned & P= | 3 3 3 -18 3 27 ^2 +36 ^2 | | 3 3 3 +18 3 27 ^2 +36 ^2 | & = 36 27-9 27 ^2 36 ^2 +27 ^2 & = 9 27 (4- ^2 ) 36 (1- ^2 )+27 ^2 & = 9 27 (4- ^2 ) 36-9 ^2 & = 9 27 (4- ^2 ) 9 (4- ^2 ) =27 . & aligned