The $x$-coordinate of the two points $P$ and $Q$ on the parabola $y^2=8x$ are roots of $x^2-17x+11$.

Question

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.

Answer 1

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,

$$\sqrt{(x-2)^2+y^2} =\sqrt{(\sqrt{11}-2)^2+34+4\sqrt{11}}=\sqrt{49}=7$$

Answer 2

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.

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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.

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