Question

If a chord of the parabola $y^{2}=4 x$ 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. $4 \operatorname{cosec}^{2} \theta$
• D. $4 \sec ^{2} \theta$

Answer 1

Answer: (C)

Let $P\left(t^{2}, 2 t\right)$ be the one end of a focal chord $P Q$ of the parabola $y^{2}=4 x$, the coordinate of the other end $Q$ are $\left(\frac{1}{t^{2}}, \frac{-2}{t}\right).\,\,\,\,\left[\because t t^{'}=-1\right]$
$\therefore P Q =\sqrt{\left(t^{2}-\frac{1}{t^{2}}\right)^{2}+\left(2 t+\frac{2}{t}\right)^{2}} $
$=\left(t+\frac{1}{t}\right) \sqrt{\left(t-\frac{1}{t}\right)^{2}+4} $
$P Q =\left(t+\frac{1}{t}\right)^{2}\,\,\,...(i)$
Given, the chord makes a $\theta$ with positive direction of $x$ -axis
$\Rightarrow \tan \theta =\frac{-2 / t-2 t}{1 / t^{2}-t^{2}}=\frac{-2}{(1 / t-t)} $
$=(t-1 / t)=2 \cot \theta$
Now from Eq. (i)
$P Q =(t+1 / t)^{2} $
$P Q =(t-1 / t)^{2}+4 $
$=(2 \cot \theta)^{2}+4 $
$=4\left(\cot ^{2} \theta+1\right) $
$=4\, \text{cosec}^{2} \theta$