The line $y-2x+4a=0$ intersects the parabola $y^2=4ax$ at the point $P(ap^2,2ap)$ and the point $Q(aq^2,2aq)$. Find the value of $p+q$ and $pq$.
How do I visualise this on the cartesian plane and approach this problem ?
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How to visualize? You have a line intersecting a parabola. One point of intersection will be to the left of the axis of symmetry, one will be to the right. How to solve. Isolate x in the equation of the line, and substitute into the equation of the parabola. – Doug M Sep 06 '16 at 17:58
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Hint (developed):
Since $\;P\;$ on the line, you get:
$$2ap-2ap^2+4a=0\implies 2a(p-p^2+2)=0\implies\begin{cases}a=0\;,\;\;\text{or}\\{}\\p^2-p-2=0\iff p=-1,2\end{cases}$$
Since we can safely assume $\;a\neq0\;$ (otherwise the "parabola" is just one point...), you already have two options for $\;p\;$...but also for $\;q\;$. Try to finish now.
DonAntonio
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We can rearrange your equations to $f(x)=2x-4a$ and $g(x)=\pm2\sqrt{ax}$. $f$ is a line with positive slope, that intersects the $y$ axis at $-4a$ and $g$ is a parabola on the $y$ axis or, alternatively, a square-root curve above and below the $x$ axis. I'd approach the problem by setting $f=g$ and finding the $x$ co-ordinate of the point then finding the $y$ co-ordinate with $f(x)$. That cool?
Jam
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