If the slope of the focal chord of $y^2 = 16x$ is $2,$ then the length of the chord, is

Let $P\left(4t_1^2,8t_1\right)$ and $Q\left(4t_2^2, 8t_2\right)$ be the end-points

of a focal chord of the parabola $y^2=16x.$ Then,

     $PQ=4{\left(t_2-t_1\right)}^2$

Now, Slope of $PQ=2$

$\begin{array}{l}\Rightarrow \frac{8t_2-8t_1}{4t_2^2-4t_1^2}=2\Rightarrow t_2+t_1=1\\ \therefore PQ=4{\left(t_2-t_1\right)}^2=4\left\{{\left(t_2+t_1\right)}^2-4t_1{t}_2\right\}\end{array}$

$\Rightarrow PQ=4\left\{{\left(t_2+t_1\right)}^2+4\right\}=4(1+4)=20.$

A.LITER We know that the length of a focal chord of the parabola $y^2 = 4ax$ making an angle $\theta$ with the axis of the parabola is $4a\cos e{c}^2\theta .$

Here, we have

             $a=4$ and $\tan \theta =2$

$\therefore$ Length of the focal chord $=16\left(1+\frac{1}{4}\right)=20.$