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In the parabola $y^2=6 x$, the equation of the chord through vertex and negative end of latus rectum, is
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The correct answer is:
$y+2 x=0$
Vertex $\equiv(0,0)$. End points of latus rectum are $(a, \pm 2 a)$. Here $\quad a=\frac{6}{4}$
Therefore, $-v e$ end of latus rectum is $\left(\frac{3}{2},-3\right)$
Line through the point is $y=\frac{-3}{3 / 2} x$ or $y+2 x=0$.
Therefore, $-v e$ end of latus rectum is $\left(\frac{3}{2},-3\right)$
Line through the point is $y=\frac{-3}{3 / 2} x$ or $y+2 x=0$.
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