If a chord of the parabola y^2=4x passes thro | vedclass

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(C) Let $P(t^2, 2t)$ be one end of a focal chord $PQ$ of the parabola $y^2=4x$. The coordinates of the other end $Q$ are $(\frac{1}{t^2}, \frac{-2}{t}

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$4 \operatorname{cosec}^2 	heta$

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(C) Let $P(t^2, 2t)$ be one end of a focal chord $PQ$ of the parabola $y^2=4x$. The coordinates of the other end $Q$ are $(\frac{1}{t^2}, \frac{-2}{t})$,since $tt' = -1$. Given that the chord makes an angle $	heta$ with the positive direction of the $X$-axis,the slope of the chord is $	an 	heta$. $	an 	heta = \frac{\frac{-2}{t} - 2t}{\frac{1}{t^2} - t^2} = \frac{-2(t + \frac{1}{t})}{\frac{1-t^4}{t^2}} = \frac{-2(t + \frac{1}{t})}{\frac{(1-t^2)(1+t^2)}{t^2}} = \frac{2t}{t^2-1}$. Alternatively,using the slope formula for a focal chord: $	an 	heta = \frac{2}{t - \frac{1}{t}}$,so $t - \frac{1}{t} = 2 \cot 	heta$. The length of the focal chord $PQ$ is given by $a(t + \frac{1}{t})^2$,where $a=1$. $PQ = (t + \frac{1}{t})^2 = (t - \frac{1}{t})^2 + 4$. Substituting $t - \frac{1}{t} = 2 \cot 	heta$: $PQ = (2 \cot 	heta)^2 + 4 = 4 \cot^2 	heta + 4 = 4(1 + \cot^2 	heta) = 4 \operatorname{cosec}^2 	heta$.

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

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