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

Step 1: Find the focus of the parabola.
Compare $y^2=6x$ with $y^2=4ax$. Then $4a=6$, so $a=\dfrac32$ and the focus is $\left(\dfrac32,0\right)$.
Step 2: Write the focal chord.
A chord through the focus with slope $m$ is $y=m\left(x-\dfrac32\right)$, or $mx-y-\dfrac{3m}{2}=0$.
Step 3: Use the distance from the vertex.
The vertex is the origin. Its distance from the chord is \[ \frac{\left|-\frac{3m}{2}\right|}{\sqrt{m^2+1}}. \] We are told this equals $\dfrac{\sqrt5}{2}$.
Step 4: Square both sides.
\[ \frac{9m^2}{4(m^2+1)}=\frac54. \]
Step 5: Clear the fractions.
$9m^2=5(m^2+1)$, so $9m^2=5m^2+5$, giving $4m^2=5$ and $m^2=\dfrac54$.
Step 6: Solve and pick the listed value.
$m=\pm\dfrac{\sqrt5}{2}$. The matching option is the positive one. \[ \boxed{\frac{\sqrt5}{2}} \]