Question

If a chord of the parabola $y^2=4x$ passes through its focus and makes an angle $\theta$ with the $X$ -axis, then its length is

• A. $4{\cos }^2\theta$
• B. $4{\sin }^2\theta$
• C. $4{\mathrm{cosec}}^2\theta$
• D. $4{\sec }^2\theta$

Answer 1

Answer: (C)

Let $P\left(t^2,2t\right)$ be the one end of a focal chord $PQ$ of the parabola $y^2=4x$, the coordinate of the other end $Q$ are $\left(\frac{1}{t^2},\frac{-2}{t}\right).\left[\because t{t}^{{}^{\prime }}=-1\right]$
$\therefore PQ=\sqrt{{\left(t^2-\frac{1}{t^2}\right)}^2+{\left(2t+\frac{2}{t}\right)}^2}$
$=\left(t+\frac{1}{t}\right)\sqrt{{\left(t-\frac{1}{t}\right)}^2+4}$
$PQ={\left(t+\frac{1}{t}\right)}^2...(i)$
Given, the chord makes a $\theta$ with positive direction of $x$ -axis
$\Rightarrow \tan \theta =\frac{-2/t-2t}{1/t^2-t^2}=\frac{-2}{(1/t-t)}$
$=(t-1/t)=2\cot \theta$
Now from Eq. (i)
$PQ=(t+1/t{)}^2$
$PQ=(t-1/t{)}^2+4$
$=(2\cot \theta {)}^2+4$
$=4\left({\cot }^2\theta +1\right)$
$=4{\text{cosec}}^2\theta$