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.

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 $$ \ \left(\frac{1}{\sqrt{\phi}} \right)^2 \ + \ \left(\frac{1}{\phi} \right)^2 \ \ = \ \ \frac{1}{\phi}  \ + \ \frac{1}{\phi^2} \ \ = \ \ \frac{\phi \ + \ 1}{\phi^2} \ \ = \ \ \frac{\phi^2}{\phi^2} \ \ = \ \ 1 \ \ . $$
