Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$. If 

$\mathrm{P}(1,\beta ),\beta >0\text{ is a point on this ellipse, then the equation of the normal to it at }\mathrm{P}\text{ is: }$$4x-2y=k then k is$

$\begin{array}{c}\text{The ecentricity is } e=\frac{1}{2} \\ \text{Given that the directrix is }x=4\\ \text{Hence }\frac{a}{e}=4\\ \text{it gives }a=2, by substituting the eccentricity value\\ b^2=a^2\left(1-e^2\right)\\ =4\left(1-\frac{1}{4}\right)\\ =3\\ \text{Equation of the ellipse is }\frac{x^2}{4}+\frac{y^2}{3}=1\end{array}$

$\begin{array}{l}\text{Since }P\left(1,\beta \right)\text{\hspace{0.17em}is a point on the ellipse}\\ \text{Substitute the point in the equation of the ellipse}\\ \text{Hence, }P\left(1,\frac{3}{2}\right)\text{\hspace{0.17em}is given point}\\ \text{The equation of the normal is }\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2{e}^2\\ \text{Substitute the appropriate values, we get }4x-2y=1\end{array}$