Find the equation of the chord of the parabol | vedclass

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(B) The equation of the parabola is $y^2 = 6x$. Comparing this with $y^2 = 4ax$,we get $4a = 6$,so $a = 3/2$.The vertex of the parabola is $(0, 0)$.Th

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$y + 2x = 0$

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(B) The equation of the parabola is $y^2 = 6x$. Comparing this with $y^2 = 4ax$,we get $4a = 6$,so $a = 3/2$.The vertex of the parabola is $(0, 0)$.The coordinates of the ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$.The negative end of the latus rectum is $(a, -2a) = (3/2, -2 	imes 3/2) = (3/2, -3)$.The chord passes through the vertex $(0, 0)$ and the point $(3/2, -3)$.The slope $m$ of the line passing through $(0, 0)$ and $(3/2, -3)$ is $m = \frac{-3 - 0}{3/2 - 0} = \frac{-3}{3/2} = -2$.The equation of the line is $y - 0 = -2(x - 0)$,which simplifies to $y = -2x$ or $y + 2x = 0$.

Find the equation of the chord of the parabola $y^2 = 6x$ passing through the vertex and the negative end of the latus rectum.

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