Question

Let $L{L}^{\prime }$ be the latus rectum of $y^2=4ax$ and $P{P}^{\prime }$ be a double ordinate drawn between the vertex and the latus rectum. If the area of trapezium $P{P}^{\prime }L{L}^{\prime }$ is maximum when the distance $P{P}^{\prime }$ from the vertex is $\frac{a}{k}$, then the value of $k$ is

Answer 1

Given : $y^2=4ax$
$L{L}^{\prime }=4a$
Let $P\equiv (a{t}^2,2at)$


Here $P{P}^{\prime }$ is double ordinate
$\Rightarrow OM=a{t}^2\Rightarrow MN=ON-OM=a-a{t}^{2}P{P}^{\prime }=2PM=4at$

Now the area of trapezium $P{P}^{\prime }{L}^{\prime }L$
$A=\frac{1}{2}(P{P}^{\prime }+L{L}^{\prime })\times MN\Rightarrow A=\frac{1}{2}(4at+4a)(a-a{t}^2)\Rightarrow A=-2a^2(t^3+t^2-t-1)\Rightarrow \frac{dA}{dt}=-2a^2(3t^2+2t-1)\Rightarrow \frac{dA}{dt}=-2a^2(3t-1)(t+1)\Rightarrow \frac{d^{2}t}{d{t}^2}=-2a^2(6t+2)$

For maxima and minima
$\frac{dA}{dt}=0\Rightarrow t=-1,\frac{1}{3}$

When $t=-1$
$\frac{d^{2}A}{d{t}^2}=8a^2>0$
When $t=\frac{1}{3}$
$\frac{d^{2}A}{d{t}^2}=-4a^2<0$
So, $A$ is maximum when $t=\frac{1}{3}$

Therefore, distance of $P{P}^{\prime }$ from vertex is
$OM=\frac{a}{9}\therefore k=9$