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.
 A: 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} }$
A: There is a much simpler solution. At the intersection of $y = mx + c$ and $y = x^2$,
$$x^2 - mx - c = 0$$
We are given $|AP - PB| = 1$ and I am taking the case where $P$ is interior to segment $AB$. Just for completeness sakes, if $P$ is exterior to segment $AB$, then we still have $|AP - PB| = 1$ if we consider $AP$ and $PB$ as signed distances or we can say $|AP| + |PB| = 1$. Coming back to the case where $P$ is interior to $AB$, if the roots are $x = \alpha, \beta$,
$ |\alpha + \beta| = |m|$.
But as $\alpha$ and $\beta$ have opposite signs and they are x-coordinates of points $A$ and $B$, $~|\alpha + \beta|~$ is the absolute difference of the horizontal projections of $AP$ and $PB$, which can also be written as $AP |\cos \theta|$ and $PB |\cos \theta|$, where $\tan \theta = m$ is the slope of the line.
$$ |\cos\theta| = \frac{1}{\sqrt{1+m^2}}$$
$$|AP - PB| \cdot |\cos \theta| = |\alpha + \beta| = |m|$$
As $|AP - PB| = 1$,
$$m \sqrt{m^2 + 1} = 1 \implies (m^2 + \frac 12)^2 = \frac 54$$
We get two real solutions for $m$,
$$m = \pm \sqrt{\frac{\sqrt 5 - 1}{2}}$$
You can draw a few lines with equation $$y = \pm \sqrt{\frac{\sqrt 5 - 1}{2}} ~x + c$$ for different values of $c$ with $c \gt 0$ to confirm $|AP - PB| = 1$
As a side note, for values of $c \lt 0$ as long as the line intersects the parabola at two points, $|AP| + |PB| = 1$
A: Since you assume that $m > 0,$ this result of your calculations is good:
$$ AP = \frac{m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}. \tag1$$
Here's where you get in a bit of trouble:
$$ BP \stackrel?= \frac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}. \tag2$$
You want $AP - BP = 1,$ and I think the best interpretation of the problem statement  interprets $AP - BP$ as the difference of two positive lengths (rather than a negative length subtracted from a positive length). Moreover, $AP$ must be the greater of the two lengths in order for the difference to be positive.
The problem with Equation $(2)$ is that if $c > 0$ then the expression on the right side of the equation is negative. A better equation is:
$$ BP = \left\lvert \frac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}\right\rvert.$$
A more useful correct equation is
$$ BP = \begin{cases}
 \dfrac{-m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} & c \geq 0, \\[1ex]
 \dfrac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} & c < 0.
\end{cases} \tag3$$
The $c < 0$ case still looks shaky because of the (apparent) possibility that
$m^2 + 4c < 0,$ which would make the square root undefined, but what actually happens is that for very large negative $c$ the value of $m$ also will be large.
Equation $(1)$, on the other hand, is good because with $m > 0$ you are guaranteed that the expression on the right-hand side of the equation is positive,
and because the expression on the right-hand side is larger than either of the two expressions on the right-hand side of Equation $(3)$,
so you have chosen the correct expression for $AP$ in either case.
The two cases in Equation $(3)$ can (and I think should) be considered separately.
In the $c \geq 0$ case we have
\begin{align}
1 &= AP - BP \\
&= \frac{m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}
 - \frac{-m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} \\
&= m \sqrt{m^2 + 1} 
\end{align}
and therefore
$$ m^4 + m^2 - 1 = 0, $$
for which the only solution (since $m^2$ must be positive) is
$$ m^2 = \frac12(\sqrt5 - 1). $$
In the $c < 0$ case, on the other hand, your further calculations are correct, and $m^2$ is the positive root $v$ of the quadratic equation
$$ v^2 + (4c + 1) v + 4c - 1 = 0, $$
that is,
\begin{align}
 m^2 &= \frac{-(4c + 1) + \sqrt{(4c + 1)^2 - 4(4c - 1)}}{2} \\
     &= \frac{-4c - 1 + \sqrt{(4c - 1)^2 + 4}}{2}.
\end{align}
You cannot eliminate $c$ from the solution in this case because the slope of the line actually does depend on how negative $c$ is.
With a $y$-intercept very far down the negative $y$ axis you need a steep slope in order to intersect the parabola.
My hunch is that you were supposed to solve the case $c \geq 0.$
This could have been stated explicitly, or it could have been implied by stating that $P$ is between $A$ and $B.$
A: We may also work from the disposition of the points of interest toward obtaining the slope of the line.  We then need to treat this question as two separate cases.
The "easier" of the two to set up geometrically occurs for $ \ c \ < \ 0 \ \ , \ $ in which $ \ P \ , \ $ the $ \ y-$intercept of the line $ \ y \ = \ mx + c \ $ lies "below" the vertex of the parabola $ \ y \ = \ x^2 \ \ . \ $  Both intersection points $ \ A \ $ and $ \ B \ $ of the line with the parabola lie on the same side of the $ \ y-$axis (since we want $ \ m \ > \ 0 \ ) \ , \ $ so determining the circumstance for which $ \ AP - BP \ = \ 1 \ $ involves requiring the distance between $ \ A \ $ and $ B \ $ to equal $ \ 1 \ \ . \ $  Since the slope $ \ m \ $ of the line is the tangent of the angle $ \ \theta \ $ that the line makes to the "direction" of the positive $ \ x-$axis, then if we call $ \ \Delta \ $ the difference in the $ \ x-$coordinates of the intersection points, we have $ \ B \ (X \ , \ X^2) \ $ and $ \ A \ (X + \Delta \ , \ [X + \Delta]^2) \ \ , \ $ where we may also write
the $ \ y-$coordinate of $ \ A \ $ as $  \ X^2 + \Delta·\tan \theta \ = \ X^2 + m·\Delta \ \ . \ $  The requirement for the separation between the intersection points then gives us
$$ \Delta^2 \ + \ (m·\Delta)^2 \ \ = \ \ 1 \ \ \Rightarrow \ \ \Delta \ \ = \ \ \frac{1}{\sqrt{1 \ + \ m^2}} \ \ . \ $$
If we solve directly for the intersection points (if they exist), they must satisfy $ \ x^2 \ = \ m·x + c \ \Rightarrow \ x^2 - m·x - c \ = \ 0 \ \ , \ $ which has the discriminant $ \ \mathcal{D} \ = \ m^2 + 4c \ \ . \ $  The line will be just tangent to the parabola when this discriminant is zero; since $ \ c \ < \ 0 \ \ , \ $ the slope of the tangent line $ \ m \ = \ 2·\sqrt{|c|} \ \ . \ $
Thus, we must have $ \ m \ > \ 2·\sqrt{|c|} \ $ in order to have two intersection points on the parabola, these lying at $ \ x \ = \ \frac{m}{2} \ \pm \ \frac{\sqrt{\mathcal{D}}}{2} \ \ . \ $  This leads to
$$ \sqrt{\mathcal{D}} \ \ = \ \ \Delta \ \ \Rightarrow \ \ \sqrt{m^2 - 4·|c|}  \ \ = \ \  \frac{1}{\sqrt{1 \ + \ m^2}}  $$
and the biquadratic equation $ \ m^4 \ + \ (1  -   4·|c|)·m^2 \ - \ (1  +  4·|c|)  \ = \ 0 \ \ . $  (It becomes clear why the problem asks for $ \ m^2 \ ) \ . \ $  The slope solution that guarantees intersection points between the line and the parabola for this case is then
$$ m^2 \ \ = \ \  \frac{4·|c| \ - \ 1}{2} \ + \ \frac{\sqrt{ \ (1 - 4·|c|)^2 \ + \  16·|c| + 4} }{2} $$ $$  = \ \ \frac{4·|c| \ - \ 1}{2} \ + \ \frac{\sqrt{ \ (1 + 4·|c|)^2 \ +  \ 4}}{2} \ \ > \ \ 4·|c| \ \ . $$
Using this for $ \ c \ = \ -4 \ \ , \ $ we find that $ \ m^2 \ = \ \frac{15}{2} + \frac{\sqrt{293}}{2} \ \approx \ 16.0586 \ \Rightarrow \ m \ \approx \ 4.0073 \ \ . \ $  With $ \ \mathcal{D} \ \approx \ 0.0586 \ , \ $ the intersection points are then $ \ B \ (1.883 \ , \ 3.544) \ $ and $ \ A \ (2.125 \ , \ 4.514) \ \ ; \ $ the distance separating them is indeed $ \ 1 \ $ .  [Alternatively, with $ \ P \ (0 \ , \ -4) \ \ , \ $ we can verify that $ \ AP \ \approx \ 8.775 \ $ and $ \ BP \ \approx \ 7.775 \ \ . \ ] $

For the case with $ \ c \ > \ 0 \ \ , \ $ the intersection points are on opposite sides of the $ \ y-$axis, with $ \ x_A \ > \ 0 \ $ and $ \ x_B \ < \ 0 \ \ ; \ $  because $ \ m \ > \ 0 \ \ ,  \ $ we will have $ \ AP \ > \ BP \ \ . \ $  We can form two similar triangles with vertices $ \ A \ , \ P \ , \ (x_A \ , \ c) \ $ and $ \ B \ , \ P \ , \ (x_B \ , \ c) \ \ ; \ $ if we arrange the smaller triangle "within" the larger one, the requirement that $ \ AP - BP \ = \ 1 \ $ again is equivalent to having the separation between $ \ A \ (x_A \ , \ x_A^2 - c) \ $ and $ \ B' \ (-x_B \ , \ 2c - x_B^2) \ $ be equal to $ \ 1 \ \ . $  Constructing the small triangle $ \ \triangle AB'C \ $ with $ \ C \ (x_A \ , \ 2c - x_B^2) \ \ , \ $ we have $ \ AC \ = \ m·B'C \ \ , \ $ and so $ \ \Delta^2 \ + \ (m·\Delta)^2 \ \ = \ \ 1 \ \ . \ $  In this way, we can again use the relation $ \ \Delta \  = \  \frac{1}{\sqrt{1 \ + \ m^2}} \ \ , \ $ with this quantity now being $ \  x_A - (-x_B) \ = \ x_A + x_B  \ \ . $
The intersection points are also found in the same way as before, except there is no "tangency" restriction for $ \ m \ \ . \ $ We now obtain
$$ \Delta \ \ = \ \  \left( \  \frac{m}{2} \ + \ \frac{\sqrt{\mathcal{D}}}{2} \ \right) \ + \ \left( \ \frac{m}{2} \ - \ \frac{\sqrt{\mathcal{D}}}{2} \ \right) \ \ = \ \ m $$
$$ \Rightarrow \ \  \Delta \ \ = \  \frac{1}{\sqrt{1 \ + \ m^2}} \ \ = \ \ m \ \  \Rightarrow \ \ m^2 \ · \ (1 \ + \ m^2) \ \ = \ \ 1 \ \ \Rightarrow \ \ m^4 \ + \ m^2 \ - \ 1 \ \ = \ \ 0 \ \ , $$
for which the permissible solution is $ \ m^2 \ = \ -\frac12 + \frac{\sqrt5}{2} \ \ . \ $  Perhaps surprisingly, the condition $ \ AP - BP \ = \ 1 \ \ $ is satisfied by only one slope for the line, regardless  (!) of the value of $ \ c \ > \ 0 \  \ . \ $  [The "Golden Ratio" emerges stealthily, in that we have $ \ m^2 \ = \ \frac{1}{\phi} \ \Rightarrow \ m \ = \ \frac{1}{\sqrt{\phi}} \ \ . \ ] $
We may also note that the two cases meld seamlessly at $ \ c \ = \ 0 \ \ : \ $ applying our slope formula for $ \ c \ < \ 0 \ $ produces
$$ m^2 \ \  = \ \ \frac{4·0 \ - \ 1}{2} \ + \ \frac{\sqrt{ \ (1 + 4·0)^2 \ +  \ 4}}{2} \ \ = \ \  -\frac{1}{2} \ + \ \frac{\sqrt{ 1 \ + \ 4}}{2} \ \ = \ \ -\frac{1}{2} \ + \ \frac{\sqrt{5}}{2} \ \ . $$  The point $ \ B \ $ becomes "identified" with $ \ P \ \ , $ now at the origin, and the intersection point $ \ A \ $ of the line $ \ y \ = \ \frac{1}{\sqrt{\phi}}·x \ $ with the parabola is located at $ \ \left(\frac{1}{\sqrt{\phi}} \ , \ \frac{1}{\phi} \right) \ \ ; \ $ its distance from $ \ P \ $ is indeed $$ \ \left(\frac{1}{\sqrt{\phi}} \right)^2 \ + \ \left(\frac{1}{\phi} \right)^2 \ \ = \ \ \frac{1}{\phi}  \ + \ \frac{1}{\phi^2} \ \ = \ \ \frac{\phi \ + \ 1}{\phi^2} \ \ = \ \ \frac{\phi^2}{\phi^2} \ \ = \ \ 1 \ \ . $$
