algebra-precalculusconic sectionsquadraticstangent line
How do I find the two tangents to the parabola: $y=x^2-2x+5$ that go through the point $(1,3)$.
I had tried to find it by creating $2$ unknown linear equations and substituting into the parabola, and then using the discriminant to find the equation (as in the vein of the exercise).
Help is much appreciated.
(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 $(\frac35,\frac15)$.
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
You can do that like that: Line through $P$ is $y= k(x-1)+3$. Plug this in parabola and calculate its discriminant. It must be 0 since equation you get must have 1 solution on $x$. Now you get $k$ (probably two of them).