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?

the tangent to $y = x^2 + 1$ at $(a, a^2 + 1$ is $$y - a^2 - 1 = 2a(x-a)$$ if the point $x = 1, y = 0$ is on it, then $$-a^2 - 1 = 2a(1-a) $$ this is a quadratic equation, you can find two values for $a$ by solving it.
you get $$a^2 - 2a = 1 \to (a-1)^2=2 \to a = 1 \pm \sqrt 2  $$
