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.
$\endgroup$ 43 Answers
$\begingroup$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, $|AP| + |PB| = 1$
$\endgroup$ 2 $\begingroup$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} }$
$\endgroup$ 1 $\begingroup$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 $B.$
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