If the line 5 x-2 y-6=0 is a tangent to the hyperbola 5 x^2-k y^2=12 . then the equation of the normal to this hyperbola at the point ( 6 , p)(p 0) is

Step-by-step solution

Given equation of hyperbola 5 x^2-k y^2=12 x^2 12 5 - y^2 12 k =1 So equation of tangent at (x₁, y₁ ) is aligned & x₁ a^2 x- y₁ b^2 y=1 & x₁ 12 5 x- y₁ 12 k y=1 5 x₁ 12 x- k y₁ 12 y=1 aligned 5 x₁ 2 x- k y₁ 2 y=6 So, 5 x₁ 2 =5 x₁=2, k y₁ 2 =2 y₁= 4 k aligned & Now, 5 4-k 16 k^2 =12 & 20-12= 16 k k=2 aligned Since, ( 6 , p) satisfied the hyperbola 5 6-2 p^2=12 p=-3, p 0 So, equation of normal at ( 6 ,-3) is a^2 x₁ x+ b^2 y₁ y=a^2+b^2 aligned & 12 5 6 x+ 12 2 (-3) y= 12 5 + 12 2 & x 5 6 - y 6 = 1 5 + 1 2 aligned 6 x-