A line always passing the point

Since $C: { y }^{ 2 }=2px$ , $F=\left( \frac { p }{ 2 } ,0 \right)$ . Then let $l:k\cdot y=x-\frac { p }{ 2 }$ .

Let's find the intersections, $l\times C:$

${ \left( x-\frac { p }{ 2 } \right) }^{ 2 }=2{ k }^{ 2 }\cdot p\cdot x\\ { x }^{ 2 }-p\cdot x+\frac { { p }^{ 2 } }{ 4 } =2{ k }^{ 2 }\cdot p\cdot x\\ { x }^{ 2 }-\left( 1+2{ k }^{ 2 } \right) \cdot p\cdot x+\frac { { p }^{ 2 } }{ 4 } =0$

${ x }_{ 1,2 }=\left( 1+2{ k }^{ 2 } \right) \cdot \frac { p }{ 2 } \pm \sqrt { \frac { { \left( 1+2{ k }^{ 2 } \right) }^{ 2 }\cdot { p }^{ 2 } }{ 4 } -\frac { { p }^{ 2 } }{ 4 } } \\ { x }_{ 1,2 }=\left( 1+2{ k }^{ 2 } \right) \cdot \frac { p }{ 2 } \pm \sqrt { { \left( 1+2{ k }^{ 2 } \right) }^{ 2 }-1 } \cdot \frac { p }{ 2 } \\ { x }_{ 1,2 }=\left( 1+2{ k }^{ 2 } \right) \cdot \frac { p }{ 2 } \pm \sqrt { 4{ k }^{ 2 }+4{ k }^{ 4 } } \cdot \frac { p }{ 2 } \\ { x }_{ 1,2 }=\left( 1+2{ k }^{ 2 }\pm 2k\sqrt { 1+{ k }^{ 2 } } \right) \cdot \frac { p }{ 2 }$

Knowing ${ x }_{ 1,2 }$ , we can derive ${ y }_{ 1,2 }$ :

$y=\frac { x-\frac { p }{ 2 } }{ k } \\ { y }_{ 1,2 }=\left( k\pm \sqrt { 1+{ k }^{ 2 } } \right) \cdot p$

Thus we got points $A$ and $B$ :

$A=\left( \left( 1+2{ k }^{ 2 }+2k\sqrt { 1+{ k }^{ 2 } } \right) \cdot \frac { p }{ 2 } ,\left( k+\sqrt { 1+{ k }^{ 2 } } \right) \cdot p \right) \\ B=\left( \left( 1+2{ k }^{ 2 }-2k\sqrt { 1+{ k }^{ 2 } } \right) \cdot \frac { p }{ 2 } ,\left( k-\sqrt { 1+{ k }^{ 2 } } \right) \cdot p \right)$

Since point $D$ is the symmetry point about the x-axis of point $A$ :

$D=\left( \left( 1+2{ k }^{ 2 }+2k\sqrt { 1+{ k }^{ 2 } } \right) \cdot \frac { p }{ 2 } ,\left( -k-\sqrt { 1+{ k }^{ 2 } } \right) \cdot p \right)$

Now we can consider the line $BD$ :

$\Delta x={ x }_{ D }-{ x }_{ B }=2k\sqrt { 1+{ k }^{ 2 } } \cdot p\\ \Delta y=-2k\cdot p$

Then, $BD:\frac { x-{ x }_{ B } }{ \Delta x } =\frac { y-{ y }_{ B } }{ \Delta y }$

Let $x=\lambda \cdot p,y=0$ :

$\frac { \lambda \cdot p-\left( 1+2{ k }^{ 2 }-2k\sqrt { 1+{ k }^{ 2 } } \right) \cdot \frac { p }{ 2 } }{ 2k\sqrt { 1+{ k }^{ 2 } } \cdot p } =\frac { 0-\left( k-\sqrt { 1+{ k }^{ 2 } } \right) \cdot p }{ -2k\cdot p } \\ \frac { \lambda -\left( 1+2{ k }^{ 2 }-2k\sqrt { 1+{ k }^{ 2 } } \right) \cdot \frac { 1 }{ 2 } }{ 2k\sqrt { 1+{ k }^{ 2 } } } =\frac { k-\sqrt { 1+{ k }^{ 2 } } }{ 2k } \\ \frac { \lambda -\left( 1+2{ k }^{ 2 }-2k\sqrt { 1+{ k }^{ 2 } } \right) \cdot \frac { 1 }{ 2 } }{ \sqrt { 1+{ k }^{ 2 } } } =k-\sqrt { 1+{ k }^{ 2 } } \\ \lambda -\left( 1+2{ k }^{ 2 }-2k\sqrt { 1+{ k }^{ 2 } } \right) \cdot \frac { 1 }{ 2 } =k\sqrt { 1+{ k }^{ 2 } } -1-{ k }^{ 2 }\\ \lambda -\frac { 1 }{ 2 } -{ k }^{ 2 }+k\sqrt { 1+{ k }^{ 2 } } =k\sqrt { 1+{ k }^{ 2 } } -1-{ k }^{ 2 }\\ \lambda =\boxed { -\frac { 1 }{ 2 } }$

Alternative solution:

Let $d$ be the direttrice of the parabola and $F', A', B', D'$ be the projections of the corresponding points onto $d$ . Then $A'A=AF=a,B'B=BF=b$ and $AO=\alpha ,BO=\beta$ . Since $A'A$ and $B'B$ are perpendicular to $d$ , they are parallel to x-axis. That means $\frac { AF }{ A'F' } =\frac { BF }{ B'F' }$ and $\frac { a }{ \alpha } =\frac { b }{ \beta }$ .

Since point $D$ is the symmetry point about the x-axis of point $A$ : $D'D=DF=a$ and $D'F'=\alpha$ . Thus $\frac { D'D }{ D'F' } =\frac { a }{ \alpha } ,\frac { B'B }{ B'F' } =\frac { b }{ \beta }$ . Since $\frac { D'D }{ D'F' } =\frac { B'B }{ B'F' }$ and $D'D\bot D'F',B'B\bot B'F'$ , we have two similar triangles: $\Delta F'D'D=\Delta F'B'B$ . That means $\angle DF'D'=\angle BF'B'$ , so points $F',B,D$ lay on one line $BD$ . That means $BD$ intersects with x-axis at point $F'$ .

Point $F'$ is a projection of the focus on the direttrice, so $F'=\left( -\frac { p }{ 2 } ,0 \right)$ .