In the parabola y^2 = 6x,the equation of the | 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 $a = \frac{3}{2}$.The vertex of the parabola is $(0

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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 $a = \frac{3}{2}$.The vertex of the parabola is $(0, 0)$.The coordinates of the endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$.The negative end of the latus rectum is $(a, -2a) = (\frac{3}{2}, -2 	imes \frac{3}{2}) = (\frac{3}{2}, -3)$.The equation of the line passing through the vertex $(0, 0)$ and the point $(\frac{3}{2}, -3)$ is given by $y - 0 = \frac{-3 - 0}{3/2 - 0}(x - 0)$.$y = \frac{-3}{3/2}x = -2x$.Therefore,$y + 2x = 0$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

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

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