<s>[INST] Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \sqrt{m^2 + 4c}}{2}$, $B_x = \frac{m - \sqrt{m^2 + 4c}}{2} $
$A_y = \frac{m^2 + m\sqrt{m^2 + 4c}}{2} + c$, $B_y = \frac{m^2 - m\sqrt{m^2 + 4c}}{2} + c$
using distance formula(not showing all steps)
$AP = \frac{m + \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$BP = \frac{m - \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$AP - BP = 1$
$(\sqrt{m^2 + 4c})(\sqrt{m^2 + 1}) = 1$
$m^4 + m^2(4c + 1) + 4c - 1 = 0$
well I could manipulate this into quadratic but that doesn't really help me with coefficient with c.
 [/INST] The line $y = m x + c$ has the parametric equation
$ (x, y) = (0, c) + t (\cos \theta, \sin \theta) $
where $\theta$ is the angle between the line and the positive $x$-axis.
and $m = \tan \theta $
Intersecting this line with the parabola $y = x^2$ yields
$ c + t \sin \theta = t^2 \cos^2 \theta $
which has two solutions
$t = \dfrac{1}{2 \cos^2 \theta } ( \sin \theta \pm \sqrt{ \sin^2 \theta + 4 c \cos^2 \theta} ) $
The absolute values of $t$ are (assuming $c \ge 0 $ )
$ | t_1 | = \dfrac{1}{2 \cos^2 \theta } ( \sin \theta + \sqrt{ \sin^2 \theta + 4 c \cos^2 \theta} ) $
$ | t_2 | = \dfrac{1}{2 \cos^2 \theta } ( \sqrt{ \sin^2 \theta + 4 c \cos^2 \theta} - \sin \theta ) $
The difference between these two absolute values is equal to $1$, hence,
$ \dfrac{ |\sin \theta | }{ \cos^2 \theta } = 1 $
So that
$ \cos^2 \theta = 1 - \sin^2 \theta = | \sin \theta |  $
which becomes
$ 1 - | \sin \theta |^2 = | \sin \theta | $
Therefore,
$ |  \sin \theta | = \frac{1}{2} ( -1 + \sqrt{5} ) $
Therefore,
$ \theta = \pm \sin^{-1} \left( \dfrac{ -1 + \sqrt{5}}{2} \right) $
Now,
$\cos^2 \theta = 1 - \sin^2 \theta = | \sin \theta | =  \dfrac{-1 +  \sqrt{5}}{2}$
Hence,
$\begin{equation} \begin{split}
m^2 &= \tan^2 \theta = \sec^2 \theta - 1  \\
&= \dfrac{1}{\cos^2\theta} - 1  \\
&= \dfrac{2}{-1 + \sqrt{5}} - 1 = \dfrac{\sqrt{5} + 1 }{2} - 1 \\
&= \boxed{\dfrac{\sqrt{5} - 1}{2}} \end{split}\end{equation}$
But, what if $ c \lt 0 $ ?
Clearly, (by graphing the situation), both $t_1$ and $t_2$ will be positive, or both negative.  Their positive difference is
$ \Delta t = 1 = \dfrac{\sqrt{ \sin^2 \theta + 4 c \cos^2 \theta }} { \cos^2 \theta } $
Hence, we must have
$ \sin^2 \theta + 4 c \cos^2 \theta = \cos^4 \theta $
or
$ \cos^4 \theta + (1 - 4 c) \cos^2 \theta - 1 = 0 $
Solving for $\cos^2 \theta$ from this equation
$ \cos^2 \theta = \dfrac{1}{2} ( 4 c - 1 + \sqrt{ (4 c - 1)^2 + 4 } ) $
Hence,
$ \begin{equation} \begin{split}
 m^2 &= \tan^2 \theta = \dfrac{1}{\cos^2 \theta} - 1 \\
&= \dfrac{2}{ 4 c - 1 + \sqrt{ (4 c - 1)^2 + 4 }} - 1 \\
 \end{split}\end{equation}$
i.e.
$m^2 =  \dfrac{ 3 - 4 c - \sqrt{ (4 c - 1)^2 + 4 }}{ 4 c - 1 + \sqrt{(4c - 1)^2 + 4 }} $
simplifying further, by eliminating the surd in the denominator,
$ m^2 = \boxed{ \dfrac{ -1 - 4 c +  \sqrt{ (4 c - 1)^2 + 4 } }{2} }$
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