Find the length of the chord of the parabola | vedclass

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(A) The equation of the parabola is $y^2 = 4x$. Here, $a = 1$. The vertex of the parabola is $(0, 0)$. The chord passes through the vertex $(0, 0)$ an

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$8$

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(A) The equation of the parabola is $y^2 = 4x$. Here, $a = 1$. The vertex of the parabola is $(0, 0)$. The chord passes through the vertex $(0, 0)$ and makes an angle $	heta = 30^{\circ}$ with the $x$-axis. The equation of the line passing through the origin with slope $m = 	an(30^{\circ}) = \frac{1}{\sqrt{3}}$ is $y = \frac{1}{\sqrt{3}}x$, or $x = \sqrt{3}y$. Substitute $x = \sqrt{3}y$ into the parabola equation: $y^2 = 4(\sqrt{3}y) \implies y^2 - 4\sqrt{3}y = 0$. $y(y - 4\sqrt{3}) = 0$. So, $y = 0$ or $y = 4\sqrt{3}$. If $y = 0$, $x = 0$. If $y = 4\sqrt{3}$, $x = \sqrt{3}(4\sqrt{3}) = 12$. The endpoints of the chord are $(0, 0)$ and $(12, 4\sqrt{3})$. The length of the chord $L$ is $\sqrt{(12 - 0)^2 + (4\sqrt{3} - 0)^2} = \sqrt{144 + 16(3)} = \sqrt{144 + 48} = \sqrt{192}$. $\sqrt{192} = \sqrt{64 	imes 3} = 8\sqrt{3}$.

Find the length of the chord of the parabola $y^2 = 4x$ which passes through the vertex and makes an angle of $30^{\circ}$ with the $x$-axis. (in $\sqrt{3}$)

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