# Tangents to Parabola through given points

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.

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).

Suppose that $\ell(x) = mx + b$ is the equation for the line tangent to $f(x) = x^2 - 2x + 5$ passing through the point $(1,3)$. Notice that the graph of the function \begin{align} (f-\ell)(x) &= (x^2-2x+5) - (mx + b) \\ &= x^2 + (-2-m)x + (5-b) \end{align} is also a parabola. Moreover, since the graphs of $\ell$ and $f$ are tangent at $x=1$, it follows that $1$ is a double root of this function. Therefore there is some constant $C$ such that \begin{align} (f-\ell)(x) &= C(x-1)^2\\ &= Cx^2 - 2Cx + C.\end{align} Equating the coefficients in the two formulae for $f-\ell$, we obtain $$\begin{cases} 1 = C \\ -2-m = -2C \\ 5-b = C. \end{cases}$$ Solving for $m$ and $b$ (which is relatively easy, since $C=1$), we get $b=4$ and $m=0$. Thus the equation of the tangent line is given by $$\ell(x) = 4.$$ You can see this at Desmos, where we can get the following graph:

making the Ansatz $$y=mx+n$$ and since $$P(1;3)$$ is situated on the given line we get $$y=m(x-1)+3$$ the tangentline intersects the given parabola only in one Point, thus $$m(x-1)+3=x^2-2x+5$$ must have only one solution. Can you finish? from the solution we get $$m/2+1+1/2\,\sqrt {{m}^{2}-4}$$ $$m=\pm 2$$

Let $(t,t^2-2t+5)$ be a touching point.

Thus, $2t-2$ is a slope of the tangent line and we get an equation of the tangent line: $$y-(t^2-2t+5)=(2t-2)(x-t).$$

Now, take $y=3$ and $x=1$.

Thus, $$3-(t^2-2t+5)=(2t-2)(1-t),$$ which gives $t\in\{0,2\}$.

Finally, I got $y=2x+1$ or $y=-2x+5$.

You can do also like this (a little more sofisticated):

Given parabola $y-4 = (x-1)^2$ has parameter $p={1\over 2}$ so it focus is $F(1,{17\over 4})$ and directrix $d: y={15\over 4}$. Then the reflection $F'$ across tangent line $\ell$ through $P$ must lie on $d$. So $F'(x',{15\over 4})$. But $\ell$ is also perpendicular bisetric of $FF'$ so $FP=F'P$. So $PF' = {5\over 4}$ and thus

$$(x'-1)^2+ ({15\over 4}- 3)^2 = \Big({5\over 4}\Big)^2$$

so $(x'-1)^2= 1$ and thus $x' = 2$ or $x'=0$. Finally since touching point is having the same $x$ coordinate as $F'$ we can calculate the touching points $T_1(0,5)$ and $T_2(2,5)$ and then write the equation of tangents.

Consider the parabola $y=x^2$.

At point $P(t,t^2) \;(t>0)$, slope is $2t$ and equation of tangent is $y=2tx-t^2$, which crosses the $y$-axis at $C(0,-t^2)$. By symmetry, the tangent at $Q(-t, t^2)$ also has the same $y$-intercept.

Putting $t=1$ gives $P(1,1), Q(-1,1), C(0,-1)$.

Translate the parabola by $(+1, +4)$.
Equation of translated parabola is $y=(x-1)^2+4$ or $y=x^2-2x+5$. The translated points (indicated by $'$) are $P'(2,5), Q'(0,5), C'(1,3)$.

Hence, points on the parabola whose tangents the $y$-axis at $C'(1,3)$ are $P', Q'$.