conic sectionsgeometry
A parabola touches x-axis at $(1,0)$ and $y=x$ at $(1,1)$. Find its focus.
My attempt : All I can say is that as angle subtended by this chord at focus is $90^\circ$ as angle between tangents is $45^\circ$. I can find equation of directrix by taking mirror image of focus in tangents and then use the fact that distance from focus is same as from directrix. But, this will get dirty.
Any help will be appreciated.
PERHAPS THIS HELPS
I've been thinking about your problem for a while. I wouldn't say that I could solve it. However, I have an idea that I would like to share with you.
You certainly remember the Dandelin spheres which may have something to do with your conjecture. The following figure depicts two cones (a black one and a grey one). These cones share the vertical axis and a Dandelin sphere. Tangent to this sphere there is a blue plane whose edge can be seen in the figure:
The blue plane and the black cone intersect in a parabola. Also, the blue plane intersects with the other cone in a hyperbola. The parabola's focus and one of the foci of the hyperbola are common.
My conjecture is that this is how to imagine your couple of parabola and hyperbola with a common focus.
Then, I don't know how to proceed.
The reflection of a parabola's focus in any tangent line gives a point on the directrix. (Why?) Therefore, if you have any two tangents (regardless of the angle they make with one another), then you get two reflected foci, which in turn determine the directrix. With a focus and a directrix, you have a unique parabola. $\square$
Best Answer
(Sketch.) Let $P$ and $Q$ be two points on a parabola, and let $R$ be the point where the respective tangents to the parabola intersect. Let $X$ the midpoint of $PQ$. Then $RX$ is parallel to the axis of symmetry of the parabola (proved below). Draw lines through $P$ and $Q$ parallel to $RX$, and reflect these lines in the respective tangents; the focus $F$ is the intersection of the reflected lines.
To show that $RX$ is parallel to the axis of symmetry: Drop perpendiculars from $P,Q,R$ to the directrix, meeting it at $P',Q',R'$ respectively. As you alluded to, the tangent at $P$ is the perpendicular bisector of the segment $FP'$, and likewise for $Q$ and $FQ'$. So, in $\triangle FP'Q'$, two of the perpendicular bisectors pass through $R$; therefore the third does as well. Since $RR'$ is a line through $R$ and perpendicular to $P'Q'$, it must be the perpendicular bisector, that is, $R'$ is the midpoint of $P'Q'$. By parallels, (the extension of) $RR'$ bisects $PQ$, that is, $RR'$ passes through $X$. So $RX$ is perpendicular to the directrix, as claimed.
Edit: Just for reference, here's what this looks like analytically: The direction of $RX$ is $(2,1)$; reflecting in $RP$ just means exchanging $x$ and $y$ coordinates, so the direction of $PF$ is $(1,2)$, and the line through $P$ in that direction is $2x-y=1$. Reflecting in $RQ$ means negating the $y$-coordinate, so the direction of $QF$ is $(2,-1)$, and the line through $Q$ in that direction is $x+2y=1$. The intersection of these lines is $(\frac35,\frac15)$.