The mid point of chord $2x+y–5 = 0$ of the parabola $y^{2 }= 4x$is

eqn. of mid point of chord is $S_1=S_{11}$

Suppose that $\left(x_1,y_1\right)$ is any point on the locus.

Write the equation of midpoint chord and compare with the given equation

Sollving given two equations

$$\begin{gathered} 2\mathrm{h}+\mathrm{k}=5\dots ..\left(1\right)\text{ and }{\mathrm{k}}^2=4\mathrm{h}\dots \dots .\left(2\right) \\ \begin{array}{r}{\mathrm{k}}^2=2(5-\mathrm{k})\\ \text{ or, }{\mathrm{k}}^2+2\mathrm{k}-10=0\\ \Rightarrow \mathrm{k}=\frac{-2\pm \sqrt{4+40}}{2}=-1\pm \sqrt{11}\end{array} \\ \text{ For, }\mathrm{k}=-1+\sqrt{11},\mathrm{h}=3-\frac{\sqrt{11}}{2}\text{ and for }\mathrm{k}=-1-\sqrt{11},\mathrm{h}=3+\frac{\sqrt{11}}{2}\text{. } \end{gathered}$$

$$\begin{gathered} \text{ End points of the chord of the parabola are }\left(3-\frac{\sqrt{11}}{2},-1+\sqrt{11}\right)\text{ and} \\ (3+\left.\frac{\sqrt{11}}{2},-1-\sqrt{11}\right)\text{. } \\ \therefore \text{ the mid point of the chord is }(3,-1)\text{. } \end{gathered}$$