Question

The lines joining the origin to the points of intersection of the line lx + my = 1 and y²= 4ax are at right angles. Show that 4al = 1.

Answer 1

The points of intersection of $lx+my=1$ and $y^2=4ax$ are:

$lx+my=1\Rightarrow my=1-lx\Rightarrow y=\dfrac{1-lx}m$

$\left( \dfrac{1-lx}m\right)^2=4ax\Rightarrow \dfrac{1-2lx+l^2x^2}{m^2}=4ax\Rightarrow$

$l^2x^2+(-2l-4am^2)x+1=0\Rightarrow x=\dfrac{l+2am^2\pm 2m\sqrt{a(l+am^2)}}{l^2}$

$\Rightarrow y=-2am\mp 2\sqrt{a(l+am^2)}$

The points of intersection are:

$P\left( \dfrac{l+2am^2+2m\sqrt{a(l+am^2)}}{l^2}, -2am-2\sqrt{a(l+am^2)}\right)=\overrightarrow{OP}$

$Q\left( \dfrac{l+2am^2-2m\sqrt{a(l+am^2)}}{l^2}, -2am+2\sqrt{a(l+am^2)}\right)=\overrightarrow{OQ}$

We know that the lines joining the origin $O$ are at right angles: $\overrightarrow{OP}\cdot \overrightarrow{OQ}=0:$

$\dfrac{(l+2am^2+2m\sqrt{a(l+am^2)})(l+2am^2-2m\sqrt{a(l+am^2)})}{l^2\cdot l^2}+$

$+(-2am-2\sqrt{a(l+am^2)})(-2am+2\sqrt{a(l+am^2)})=$

$=\dfrac{1}{l^2}-4la=0 \Rightarrow \dfrac{1}{l^2}=4la \Rightarrow 1=4lal^2\Rightarrow \boxed{1=4l^3a}$

We see that the problem statement is incorrect. It is correct that $4al^3=1$ is satisfied.