The $x$-coordinate of the two points $P$ and $Q$ on the parabola $y^2=8x$ are roots of $x^2-17x+11$. 
The $x$-coordinates of the two points $P$ and $Q$ on the parabola $y^2=8x$ are roots of $x^2-17x+11$. If the tangents at $P$ and $Q$ meet at $T$, then find the distance of $T$ from the focus.

The point of intersection of tangents is (GM of abscissa, AM of ordinate)
Hence x coordinate of the point is 
$$x=-\sqrt{\alpha \beta}$$
$$x=-\sqrt {11}$$
And y coordinate will be 
$$y=\frac{y_1+y_2}{2}$$
But $y^2=8 \alpha$ and $y^2=8\beta$
$$y=\frac{2\sqrt {2\alpha} +2\sqrt {2\beta}}{2}$$$$y=\sqrt 2 (\sqrt{\alpha+\beta+2\sqrt {\alpha\beta}})$$
$$y=\sqrt 2 (17+2\sqrt 11)$$
As you may have already realised, I am going wrong. But I can’t seem to pinpoint it. 
 A: Note that
$$x_1+x_2=17,\>\>\>\>\>x_1x_2 =11$$
Then, the intersection coordinates of the two tangents are (GM of abscissa, AM of ordinate),
$$x=\sqrt{x_1x_2} = \sqrt{11}$$
$$y = \frac{y_1+y_2}2=\sqrt2(\sqrt{x_1}+\sqrt{x_2})=\sqrt2\sqrt{x_1+x_2+2\sqrt{x_1x_2} }
=\sqrt{34+4\sqrt{11}}$$
Since the focus of $y^2=8x$ is $(2,0)$, the distance is thus,
$$\sqrt{(x-2)^2+y^2} =\sqrt{(\sqrt{11}-2)^2+34+4\sqrt{11}}=\sqrt{49}=7$$
A: 

Proposition.  Let $\mathcal{P}$ be a parabola on a plane with focus $F$ and directrix $d$.  For two points $A$ and $B$ on $\mathcal{P}$, define $T(A,B)$ to be the intersection of the tangents at $A$ and at $B$ to $\mathcal{P}$.    A straight line $p$ parallel to $d$ meets $\mathcal{P}$ at $P$ and $P'$.  Another straight line $q$ parallel to $d$ meets $\mathcal{P}$ at $Q$ and $Q'$.  Then, the four points $T(P,Q)$, $T(P',Q)$, $T(P,Q')$, and $T(P',Q')$ lie on a circle centered at $F$ with radius $$\sqrt{\text{dist}(d,p)\cdot\text{dist}(d,q)}\,,$$ where $\text{dist}(l,\ell)$ is the distance between two parallel lines $l$ and $\ell$.

Without loss of generality, suppose that $\mathcal{P}$ is given by
$$\mathcal{P}=\big\{(x,y)\in\mathbb{R}^2\,\big|\,y^2=4cx\big\}\,,$$
where $c>0$.  Then, $F=(c,0)$ and the equation for $d$ is $x=-c$.  Suppose that the equation
$$x^2-sx+m=0$$
gives the union of $p$ and $q$.
Now we may assume that $P=\left(a^2,2a\sqrt{c}\right)$ and $Q=\left(b^2,2b\sqrt{c}\right)$ for some real numbers $a$ and $b$.  Note that $$(x-a^2)(x-b^2)=x^2-sx+m$$
so that $a^2+b^2=s$ and $a^2b^2=m$.  
Then, the tangent at $P$ to $\mathcal{P}$ is given by the linear equation
$$y=\frac{\sqrt{c}}{a}x+a\sqrt{c}\,.$$
Similarly, the tangent at $Q$ to $\mathcal{P}$ is given by the linear equation
$$y=\frac{\sqrt{c}}{b}x+b\sqrt{c}\,.$$
Therefore, $T(P,Q)$ has coordinates $\big(ab,(a+b)\sqrt{c}\big)$.  
Ergo, the distance from $T(P,Q)$ to $F$ is
$$\sqrt{(ab-c)^2+\big((a+b)\sqrt{c}\big)^2}=\sqrt{c^2+(a^2+b^2)c+a^2b^2}=\sqrt{c^2+sc+m}.$$
However, since $\text{dist}(d,p)=c+a^2$ and $\text{dist}(d,q)=c+b^2$, we obtain
$$c^2+sc+m=(c+a^2)(c+b^2)=\text{dist}(d,p)\cdot \text{dist}(d,q)\,.$$
The claim follows.

Remark.  Note that the coordinates of $T(P,Q)$ does not quite have the formula the OP claimed
$$T(P,Q)=\left(\sqrt{P_xQ_x},\frac{P_y+Q_y}{2}\right)\,,$$ if $P=(P_x,P_y)$ and $Q=(Q_x,Q_y)$.  Only when both $P$ and $Q$ are in the first quadrant does the formula makes sense.

In the OP's problem $c=2$, $s=17$, and $m=11$.  Hence, the distance to the focus is
$$\sqrt{2^2+17\cdot 2+11}=\sqrt{49}=7\,.$$
