Let x+y=k be a normal to the parabola ${\mathrm{y}}^2=12\mathrm{x}.$ If P is length of the perpendicular from the focus of the parabola on to this normal then $4\mathrm{k}-2{\mathrm{P}}^2=$

 $\mathrm{Given} \mathrm{parabola} {\mathrm{y}}^2=12\mathrm{x}$

$$\begin{gathered} \Rightarrow 4\mathrm{a}=12 \\ \Rightarrow a=3 \end{gathered}$$

Given normal $\mathrm{x}+\mathrm{y}=\mathrm{k}$

  If $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$ is a normal to ${\mathrm{y}}^2=4\mathrm{ax}$ Then

$$\begin{gathered} {\mathrm{al}}^3+2{\mathrm{alm}}^2+{\mathrm{m}}^2\mathrm{n}=0 \\ Where \mathrm{l}=1, \mathrm{m}=1, \mathrm{n}=-\mathrm{k} \\ 3{\left(1\right)}^3+6\left(1\right) \left(1\right)+1(-\mathrm{k})=0 \\ \Rightarrow k=9 \end{gathered}$$

Focus $(\mathrm{a},0)=(3, 0)$

Normal is $\mathrm{x}+\mathrm{y}-9=0$

$$\begin{gathered} \mathrm{p}=\frac{\left|3+0-9\right|}{\sqrt{2}}=\frac{6}{\sqrt{2}} \\ Now 4\mathrm{k}-2{\mathrm{p}}^2=4\left(9\right)-2\left(\frac{36}{2}\right) \\ =36-36 \\ =0 \end{gathered}$$