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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.

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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).

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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:

the graph

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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 $$m/2+1+1/2\,\sqrt {{m}^{2}-4}$$ $$m=\pm 2$$

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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 $y=-2x+5$.

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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.

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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 $P', Q'$.

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