Thank you for reporting, we will resolve it shortly
Q.
In the parabola $y^2=6 x$, the equation of the chord through vertex and negative end of latus rectum, is
Conic Sections
Solution:
$4 a=6 \Rightarrow a=\frac{3}{2}$
Negative end of latus rectum is $(a,-2 a)$, i.e. $\left(\frac{3}{2},-3\right)$, vertex is $(0,0)$ Line through these two points is $\frac{y-0}{x-0}=\left(\frac{-3-0}{\frac{3}{2}-0}\right)$ or $2 x+y=0$