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

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

Aicharya · Accepted Answer

Let $P\left(t^2, 2 tight)$ 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 $\left(\frac{1}{t^2}, \frac{-2}{t}ight)$. $\left[\because t t^{\prime}=-1ight]$ Given, the chord makes a $	heta$ with positive direction of $x$-axis $\begin{aligned} \Rightarrow \quad 	an 	heta & =\frac{-2 / t-2 t}{1 / t^2-t^2}=\frac{-2}{(1 / t-t)} \ & =(t-1 / t)=2 \cot 	heta \end{aligned}$ Now from Eq. (i) $\begin{aligned} P Q & =(t+1 / t)^2 \ P Q & =(t-1 / t)^2+4 \ & =(2 \cot 	heta)^2+4 \ & =4\left(\cot ^2 	heta+1ight) \ & =4 \operatorname{cosec}^2 	heta \end{aligned}$

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

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