Curve tangent line A curve with the equation $x^2-x+1$ has two tangent lines $a$ and $b$ that intersects at $x=1$, what is $y$? can I determine $y$ when $a$ and $b$ are perpendicular, or the gradient of $a$ and $b$ are $-1$
I tried using the second point, but knowing that they have the same gradient means they won't intersect, so it cannot be determined from that
from the first point, I know that the gradient of the tangent lines must be $2x-1$, and the equation for line is $y=mx+b$
if both are perpendicular, then $m_a × m_b = -1$. so does that mean I could make it $(2x-1)(2x-1)=-1$? because both lines have the same gradient?
I tried by making the gradient for line $a = (2x-1)$ and gradient for line $b -1/(2x-1)$. but they intersect at $x=1$
I'm stuck right there, is there a way of using either one to determine the y which both lines intersect?

Let $A$ be the $x$-coordinate of the point of tangency of the line $a$, then its equation is
$$y=(2A-1)(x-A)+A^2-A+1.$$
Similarly for $b$,
$$y=(2B-1)(x-B)+B^2-B+1.$$
Therefore for $x=1$ we have
$$(2A-1)(1-A)+A^2-A+1=y=(2B-1)(1-B)+B^2-B+1$$
and we finally find
$$(A-B)(A+B-2)=0.$$
It follows that if $a$ and $b$ are distinct lines then $A+B=2$.
Therefore we have infinite pairs of lines $a$ and $b$ that satisfy the given conditions. The $y$-coordinate of their intersection point is $A(2-A)=AB$ (it depends on the pair of lines).

