The length of the chord of the parabola y^{2} | vedclass

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Question

(B) The vertex of the parabola $y^{2}=4ax$ is at the origin $O(0,0)$.The line passing through the vertex $O$ making an angle $\alpha$ with the $x$-axi

Vedclass · Accepted Answer

$4a \cot \alpha \operatorname{cosec} \alpha$

Vedclass · Answer

(B) The vertex of the parabola $y^{2}=4ax$ is at the origin $O(0,0)$.The line passing through the vertex $O$ making an angle $\alpha$ with the $x$-axis is $y = x 	an \alpha$.To find the intersection point $P$ of this line and the parabola,substitute $y = x 	an \alpha$ into $y^{2}=4ax$:$(x 	an \alpha)^{2} = 4ax$$x^{2} 	an^{2} \alpha = 4ax$Since $P$ is not the origin,$x 
eq 0$,so $x 	an^{2} \alpha = 4a$,which gives $x = 4a \cot^{2} \alpha$.Then $y = (4a \cot^{2} \alpha) 	an \alpha = 4a \cot \alpha$.Thus,the coordinates of $P$ are $(4a \cot^{2} \alpha, 4a \cot \alpha)$.The length of the chord $OP$ is the distance from $(0,0)$ to $(4a \cot^{2} \alpha, 4a \cot \alpha)$:$OP = \sqrt{(4a \cot^{2} \alpha)^{2} + (4a \cot \alpha)^{2}}$$OP = \sqrt{16a^{2} \cot^{4} \alpha + 16a^{2} \cot^{2} \alpha}$$OP = 4a \cot \alpha \sqrt{\cot^{2} \alpha + 1}$$OP = 4a \cot \alpha \operatorname{cosec} \alpha$ (since $0  0$ and $\operatorname{cosec} \alpha > 0$).

The length of the chord of the parabola $y^{2}=4ax$ $(a>0)$ which passes through the vertex and makes an acute angle $\alpha$ with the axis of the parabola is

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