Question

For parabola \(y^2=5x\), normal at P meets x-axis at Q. If PQ subtends \(60^\circ\) at vertex, slope of normal is

• A. \(\pm2\sqrt3\)
• B. \(\pm\sqrt2\)
• C. \(\pm\frac{2}{\sqrt3}\)
• D. \(\pm2\sqrt2\) \bigskip

Hint:

For parabola normals, parameter method is fastest.

Answer 1

The Correct Option is C

Concept: For \(y^2=4ax\), slope of normal at parameter \(t\): \[ m_n=-t \]

Step 1: Rewrite parabola.
\[ y^2=5x \Rightarrow 4a=5 \Rightarrow a=\frac54 \]

Step 2: Use parametric point.
\[ P(at^2,2at) \] Slope of normal: \[ m=-t \]

Step 3: Angle condition gives \(t\).
Using geometry: \[ t=\frac{2}{\sqrt3} \]

Step 4: Final slope.
\[ m=\pm\frac{2}{\sqrt3} \] \[ \boxed{(C)} \]