Question

A chord is drawn through the focus of the parabola \(y^2=6x\) such that its perpendicular distance from the vertex is \(\frac{\sqrt5}{2}\). Then its slope can be:

• A. \(\dfrac{2}{\sqrt3}\)
• B. \(\dfrac{2}{\sqrt5}\)
• C. \(\dfrac{\sqrt3}{2}\)
• D. \(\dfrac{\sqrt5}{2}\)

Hint:

For parabola \(y^2=4ax\), always remember: - Focus \(=(a,0)\) - Focal chord with slope \(m\): \(y=m(x-a)\)

Answer 1

The Correct Option is D

Concept: For the parabola: \[ y^2=4ax \] the focus is \((a,0)\). Any chord through the focus having slope \(m\) is called a focal chord and its equation is: \[ y=m(x-a) \] The perpendicular distance of a line: \[ Ax+By+C=0 \] from the origin is: \[ \frac{|C|}{\sqrt{A^2+B^2}} \]

Step 1: Identify the focus of the parabola.
Given parabola: \[ y^2=6x \] Comparing with \(y^2=4ax\), \[ 4a=6 \Rightarrow a=\frac32 \] Hence focus is: \[ \left(\frac32,0\right) \]

Step 2: Equation of chord through the focus.
Equation of line through focus with slope \(m\): \[ y=m\left(x-\frac32\right) \] or \[ mx-y-\frac{3m}{2}=0 \]

Step 3: Use perpendicular distance formula.
Distance from vertex \((0,0)\) is: \[ \frac{\left|-\frac{3m}{2}\right|}{\sqrt{m^2+1}} = \frac{\sqrt5}{2} \] Squaring both sides: \[ \frac{9m^2}{4(m^2+1)}=\frac54 \] \[ 9m^2=5(m^2+1) \] \[ 9m^2=5m^2+5 \] \[ 4m^2=5 \] \[ m^2=\frac54 \] \[ m=\pm\frac{\sqrt5}{2} \] Among the given options: \[ \boxed{\frac{\sqrt5}{2}} \] is correct.