Let the equation of the tangent at a point P on the parabola x^2-4 x-4 y+16=0 be 2 x-y-5=0 . If the equation of the normal drawn at P to this parabola is ax + y + c =0 , then ac = .0

Step-by-step solution

x^2-4 x-4 y+16=0 2 x-4-4 d y d x =0 d y d x = 4-2 x -4 =-1+ 1 2 x 2 x-y-5=0 is tangent on the parabola. aligned & m=2 & -1+ 1 2 x=2 m= d y d x & 1 2 x=3 x=6 aligned So, 2 6-y-5=0 y=7 . P (6,7) Now, equation of normal at P(6,7) is aligned & (y-7)=- 1 2 (x-6) & 2 y-14=-x+6 x+2 y-20=0 . & 1 2 x+y-10=0...(1) aligned The given equation normal is a x+y+c=0...(2) from (1) and (2) a= 1 2 and c=-10 a c= 1 2 (-10)=-5