JEE Main 2014 (Online) 19th AprilQuestion 1
A chord is drawn through the focus of the parabola $y^{2}=6x$ such that its distance from the vertex of this parabola is $\frac{\sqrt5}{2}$, then its slope can be
Parabola
Solution
Formulas used. (1) Focus of $y^2=4ax$ is $(a,0)$(2) Distance of $(0,0)$ from $mx-y+c=0$ is $\frac{|c|}{\sqrt{1+m^2}}$Step 1 - Focus. $y^2=6x$ gives $4a=6$, $a=\frac32$; focus $S\big(\frac32,0\big)$. Step 2 - Chord through $S$. With slope $m$: $y=m\big(x-\frac32\big)$, i.e. $mx-y-\frac{3m}{2}=0$. Step 3 - Distance from vertex $O$, use (2). $$\frac{\frac32|m|}{\sqrt{1+m^2}}=\frac{\sqrt5}{2}\Rightarrow 9m^2=5(1+m^2)\Rightarrow 4m^2=5$$ Step 4. $m=\pm\frac{\sqrt5}{2}$; among the options, $\frac{\sqrt5}{2}$. Answer: A