<s>[INST] Find the point $T(a, b)$ on the curve $y = x^2 + 1$ whose tangent passes through the point $P(1, 0)$ from the left. Find the point $T(a, b)$ on the curve $y = x^2 + 1$ whose tangent passes through the point $P(1, 0)$ from the left.
$$
\\ \begin{align}
\\
\\ f(x) &= x^2 + 1
\\ f(a) &= a^2 + 1 = b
\\
\\ T(a, f(a))
\\
\\ f'(x) &= 2x
\\ f'(a) &= 2a = m_{tan}
\\
\\ x_1 &= 1
\\ y_1 &= 0
\\
\\ P(x_1, y_1)
\\
\\ y_1 - b &= m_{tan}(x_1 - a)
\\ 0 - a^2 - 1 &= 2a(1 - a)
\\ -a^2 - 1 &= 2a - 2a^2
\\ a^2 - 2a - 1 &= 0
\\ (a - 1)(a - 1) &= 0
\\ a &= 1
\\ b &= 1^2 + 1 = 2
\\
\\ T(1, 2)
\\ P(1, 0)
\\ \end{align}
$$
At this point, I see that the line that goes through the points $T$ and $P$ would have to have a vertical slope.
I've made an error somewhere along the line, but I don't quite see it. Perhaps I'm going about this all wrong?
 [/INST] 
Find the point $T(a,b)$ on the curve $y=x^2+1$ whose tangent passes through the point $P(1,0)$ from the left.

Parametrize $(t,t^2+1)$, $y'=2x=2t$, Eqn. of tangent, $y-(t^2+1)=2t(x-t)\implies y=1+2tx-t^2$,  at $y=0$, $x=\frac{t^2-1}{2t}<1\implies t\in(-\infty,1-\sqrt2)\cup(0,1+\sqrt2)$
So, $$x\in(\infty,1-\sqrt2)\cup(0,1+\sqrt2)$$

More shorter method:
Write Chord of Contact (while keeping the graph of $x^2+1$ in mind) and find intersection with curve:
$$\frac12(y+0)=x.1+1\implies y=2x+2=x^2+1\implies x=1-\sqrt2,1+\sqrt2$$
It's fairly easy to proceed now.
</s>