Tangents are drawn from point (1,1) to the ellipse S ^2+4 y^2-2 x+8 y+1=0 . If m₁, m₂ (m₁>m₂ ) are the slopes of these tangents, then with respect to the given ellipse the point P ( m ₁, ~m ₂ )

Step-by-step solution

Given ellipse x^2+4 y^2-2 x+8 y+1=0 (x-1)^2 4 + (y+1)^2 1 =1 ...(i) ...(i) Let m be slope of tangent passing through (1,1) on equation (i) Hence y-1=m(x-1) y=m x+(m-1) ...(ii) When y=m x+c is a tangent to x^2 a^2 + y^2 b^2 =1 then c^2=a^2 m^2+b^2 (1-m)^2=4 m^2+1 m=0 and m= 2 3 Hence there will be two tangents from (1,1) m₁=0, m₂= 2 3 P (m₁, m₂ ) will shift to P (m₁+1, m₂-1 ) because given ellipse have centre at (1,-1) Hence P (m₁+1,, m₂-1 )= (1, -1 3 ) Given S= (x-1)^2 4 + (y+1)^2 1 -1 Now S_ (1,- 1 3 ) =0+ (1- 1 3