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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
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Verified Answer
The correct answer is:
$4 \operatorname{cosec}^2 \theta$
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 $\left(\frac{1}{t^2}, \frac{-2}{t}\right)$.
$\left[\because t t^{\prime}=-1\right]$
Given, the chord makes a $\theta$ with positive direction of $x$-axis
$\begin{aligned}
\Rightarrow \quad \tan \theta & =\frac{-2 / t-2 t}{1 / t^2-t^2}=\frac{-2}{(1 / t-t)} \\
& =(t-1 / t)=2 \cot \theta
\end{aligned}$
Now from Eq. (i)
$\begin{aligned}
P Q & =(t+1 / t)^2 \\
P Q & =(t-1 / t)^2+4 \\
& =(2 \cot \theta)^2+4 \\
& =4\left(\cot ^2 \theta+1\right) \\
& =4 \operatorname{cosec}^2 \theta
\end{aligned}$
$\left[\because t t^{\prime}=-1\right]$
Given, the chord makes a $\theta$ with positive direction of $x$-axis
$\begin{aligned}
\Rightarrow \quad \tan \theta & =\frac{-2 / t-2 t}{1 / t^2-t^2}=\frac{-2}{(1 / t-t)} \\
& =(t-1 / t)=2 \cot \theta
\end{aligned}$
Now from Eq. (i)
$\begin{aligned}
P Q & =(t+1 / t)^2 \\
P Q & =(t-1 / t)^2+4 \\
& =(2 \cot \theta)^2+4 \\
& =4\left(\cot ^2 \theta+1\right) \\
& =4 \operatorname{cosec}^2 \theta
\end{aligned}$
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