line through $P(l,3)$ intersect ellipse at $A$ and $D$ and axis at $B$ and $D$. then min. of $|l|$ A line through $P(l,3)$ meets the ellipse $\displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1$ at $A$ and $D$ and meets the  $x$ axis and $y$ axis at $B$ and $C$ so that $PA.PD =PB.PC.$ find minimum value of $|l|$
equation of line through $P(l,3)$ is $y-3=m(x-l)$
intersection with $x$ axis $B(l-\frac{3}{m},0)$ and $y$ axis $C(0,3-ml)$
i wan,t be able to go further, could some help me with this, thanks

Eliminating $y$ from $\frac{x^2}{16}+\frac{y^2}{9}=1$ and $y-3=m(x-l)$ gives
$$(16m^2+9)x^2+16(-2m^2l+6m)x+16(m^2l^2-6ml)=0$$
Letting $\alpha,\beta$ be the $x$-coordinate of $A,D$ respectively, we get
$$\alpha+\beta=-\frac{16(-2m^2l+6m)}{16m^2+9},\quad \alpha\beta=\frac{16(m^2l^2-6ml)}{16m^2+9}\tag1$$
Since we have 
$$PA:PC=PB:PD$$
we get
$$|l-\alpha|:|l-0|=\left|l-\left(l-\frac 3m\right)\right|:|l-\beta|,$$
and so
$$|l^2-(\alpha+\beta)l+\alpha\beta|=\left|\frac{3l}{m}\right|$$
From $(1)$,
$$\left|l^2+\frac{16(-2m^2l+6m)}{16m^2+9}l+\frac{16(m^2l^2-6ml)}{16m^2+9}\right|=\left|\frac{3l}{m}\right|$$
Multiplying the both sides by $16m^2+9$ gives
$$|9l^2|=(16m^2+9)\left|\frac{3l}{m}\right|,$$
i.e.
$$|l|=\frac{16|m|^2+9}{3|m|}$$
Now by AM-GM inequality, we get
$$|l|=\frac{1}{3}\left(16|m|+\frac{9}{|m|}\right)\ge 3\cdot 2\sqrt{16|m|\times\frac{9}{|m|}}=\color{red}{72}.$$
