If the locus of the middle points of the chords of the parabola $y^2=2x$ which touches the circle $x^2+y^2-2x-4=0$ is given by 

${\left(y^2+1-x\right)}^2=\lambda (1+y^2)$, then the value of $\lambda$ is equal to

$$\begin{gathered} y^2=2x\Rightarrow a=\frac{1}{2}\Rightarrow 2a=1 \\ 2h=a\left(t_1^2+t_2^2\right) ; k=a(t_1+t_2) \\ 4h=t_1^2+t_2^2 ; 2k=t_1+t_2 \\ 4k^2=4h+2t_1{t}_2 \\ {t}_1{t}_2=2(k^2-h) \\ Now equation of chord is \\ 2x-\left(t_1+t_2\right)y+t_1{t}_2=0 \dots \left(i\right) \\ \left(1\right) touches the circle \\ {x}^2+y^2-2x-4=0 \\ u\sin g p = r \\ \left|\frac{2+t_1{t}_2}{\sqrt{4+{\left(t_1+t_2\right)}^2}}\right|=\sqrt{5} \\ {\left(2+t_1{t}_2\right)}^2=5\left[4+{\left(t_1+t_2\right)}^2\right] \\ 4{(k^2-h+1)}^2=20(1+k^2) \\ ({y^2-x+1)}^2=5(1+x^2)\Rightarrow \lambda =5 \end{gathered}$$

$$\begin{gathered} Aliter : T=S_1 and use p=r \\ T=S_1\Rightarrow ky-(x+h)=k^2-2h \\ \Rightarrow -x+ky+(h-k^2)=0 \\ p=4\Rightarrow \left|\frac{-1-k^2+h}{\sqrt{1+k^2}}\right|=\sqrt{5} \\ \Rightarrow locus of (h, k) is {(y^2-x+1)}^2=5(1+y^2) \\ so \lambda =5 \end{gathered}$$