Question

PSQ is a focal chord of the parabola \[ y^2=12x. \] \(A\) and \(B\) are respectively the feet of the perpendiculars drawn from \(P\) and \(Q\) on the directrix of the parabola. If the length of \(AB\) is \(7\sqrt3\) and \[ P=(3t^2,6t), \qquad (0<t<1), \] then \(t=\)

• A. \[ \frac{2}{\sqrt3} \]
• B. \[ \frac{\sqrt3}{2} \]
• C. \[ \frac{\sqrt2}{3} \]
• D. \[ \frac{\sqrt5}{4} \] \bigskip

Hint:

For the parabola \[ y^2=4ax, \] the ends of a focal chord have parameters \(t\) and \(-1/t\). This property is frequently used in JEE and entrance examinations.

Answer 1

The Correct Option is B

Concept: For parabola \(y^2=4ax\), focal chord endpoints have parameters \(t_1,t_2\) with \(t_1t_2=-1\). Use distance between feet of perpendiculars on directrix \(x=-a\) to form equation in \(t\), then solve.

Step 1: \(y^2=12x \Rightarrow 4a=12, a=3\). Focus S(3,0). P has parameter \(t\): \((3t^2,6t)\). Focal chord PSQ: Q has parameter \(-1/t\): \((3/t^2, -6/t)\).

Step 2: Directrix \(x=-3\). Feet A and B: \(A(-3,6t), B(-3,-6/t)\). \(AB = |6t - (-6/t)| = 6|t+1/t|\). Given \(AB=7\sqrt3 \Rightarrow 6(t+1/t)=7\sqrt3\) (since \(t>0\)).

Step 3: \(t+1/t = 7\sqrt3/6 \Rightarrow 6t^2 - 7\sqrt3 t + 6 = 0\). Discriminant: \(147-144=3\). \(t = \frac{7\sqrt3 \pm \sqrt3}{12} = \frac{2\sqrt3}{3}, \frac{\sqrt3}{2}\).

Step 4: Given \(0<t<1\), choose \(t = \frac{\sqrt3}{2}\).

Step 5: Write the final answer. \(\boxed{\frac{\sqrt3}{2}}\)