JEE Main 2019 — Parabola Question with Solution

The given parabola $y^2=16x$ is of the form $y^2=4ax,$ hence $a=4$ and the focus of the parabola is $\left(a, 0\right)=\left(4, 0\right).$

The equation of a line joining the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right).$

Therefore, equation of chord joining $P\left(1, 4\right)$ to focus $S\left(4, 0\right)$ is

$y-0=-\frac{4}{3}\left(x-4\right)$

$\Rightarrow 3y=-4x+16$

$\Rightarrow 4x +3y-16=0$

To find the points where this line will cut the parabola put $x=\frac{y^2}{16}$ from the parabola into the line, to get

$4\left(\frac{y^2}{16}\right)+ 3y-16=0$

$\Rightarrow y^2+ 12y-64=0$

$\Rightarrow \left(y+16\right)\left(y-4\right)=0$

$\Rightarrow y=-16$ or $y=4$

And $x=\frac{y^2}{16}$

$\Rightarrow x=16$ or $x=1$

Thus, the points are $\left(1, 4\right)$ and $\left(16,-16\right),$ but $\left(1, 4\right)$ is the given point $P.$

$\Rightarrow Q=\left(16,-16\right)$

The distance between the points $\left(x_1, y_1\right) \& \left(x_2, y_2\right)$ is $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

So, the length of focal chord $PQ$ is

$PQ=\sqrt{{\left(16-1\right)}^2+{\left(-16-4\right)}^2}$

$=\sqrt{225+400}=\sqrt{625}=25$ units.