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Let $x+y=k$ be a normal to the parabola $y^2=12 x$. If $p$ is length of the perpendicular from the focus of the parabola onto this normal, then $4 k-2 p^2$ is equal to
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Given equation of parabola is
$y^2=12 x$
Here, $\quad a=3$
$\therefore$ Equation of normal is
$\begin{aligned} y & =m x-2 a m-a m^3 \quad(\because m=-1) \\ y & =-x-2(3)(-1)-3(-1)^3 \\ & =-x+6+3\end{aligned}$
But given normal is
$\begin{aligned} & x+y=k \\ & \therefore \quad k=9 \\ & \end{aligned}$
Focus of a given parabola is $S(3,0)$.
Now, perpendicular distance from $S(3,0)$ to the line (i) is
$\begin{aligned} & p=\frac{|3(1)+0-9|}{\sqrt{1^2+1^2}}=\frac{6}{\sqrt{2}} \\ \therefore \quad 4 k-2 p^2 & =4(9)-2\left(\frac{6}{\sqrt{2}}\right)^2 \\ & =36-2 \times \frac{36}{2}=0\end{aligned}$
$y^2=12 x$
Here, $\quad a=3$
$\therefore$ Equation of normal is
$\begin{aligned} y & =m x-2 a m-a m^3 \quad(\because m=-1) \\ y & =-x-2(3)(-1)-3(-1)^3 \\ & =-x+6+3\end{aligned}$
But given normal is
$\begin{aligned} & x+y=k \\ & \therefore \quad k=9 \\ & \end{aligned}$
Focus of a given parabola is $S(3,0)$.
Now, perpendicular distance from $S(3,0)$ to the line (i) is
$\begin{aligned} & p=\frac{|3(1)+0-9|}{\sqrt{1^2+1^2}}=\frac{6}{\sqrt{2}} \\ \therefore \quad 4 k-2 p^2 & =4(9)-2\left(\frac{6}{\sqrt{2}}\right)^2 \\ & =36-2 \times \frac{36}{2}=0\end{aligned}$
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