If the locus of the middle points of the chords of the parabola $y^{2} = 2 x$ which touches the circle $x^{2} + y^{2} - 2 x - 4 = 0$ is given by
${(y^{2} + 1 - x)}^{2} = λ (1 + y^{2})$ , then the value of $λ$ is equal to

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Detailed Solution

$y^{2} = 2 x \Rightarrow a = \frac{1}{2} \Rightarrow 2 a = 1 2 h = a (t_{1}^{2} + t_{2}^{2}); k = a (t_{1} + t_{2}) 4 h = t_{1}^{2} + t_{2}^{2}; 2 k = t_{1} + t_{2} 4 k^{2} = 4 h + 2 t_{1} t_{2} t_{1} t_{2} = 2 (k^{2} - h) N o w e q u a t i o n o f c h o r d i s 2 x - (t_{1} + t_{2}) y + t_{1} t_{2} = 0 \dots (i) (1) t o u c h e s t h e c i r c l e x^{2} + y^{2} - 2 x - 4 = 0 u \sin g p = r |\frac{2 + t_{1} t_{2}}{\sqrt{4 + {(t_{1} + t_{2})}^{2}}}| = \sqrt{5} {(2 + t_{1} t_{2})}^{2} = 5 [4 + {(t_{1} + t_{2})}^{2}] 4 {(k^{2} - h + 1)}^{2} = 20 (1 + k^{2}) ({y^{2} - x + 1)}^{2} = 5 (1 + x^{2}) \Rightarrow λ = 5$

$A l i t e r : T = S_{1} a n d u s e p = r T = S_{1} \Rightarrow k y - (x + h) = k^{2} - 2 h \Rightarrow - x + k y + (h - k^{2}) = 0 p = 4 \Rightarrow |\frac{- 1 - k^{2} + h}{\sqrt{1 + k^{2}}}| = \sqrt{5} \Rightarrow l o c u s o f (h, k) i s {(y^{2} - x + 1)}^{2} = 5 (1 + y^{2}) s o λ = 5$