Question

Let \(O\) and \(S\) be the vertex and focus of the parabola \[ y^2=4ax \] respectively and \(x=k\) be its double ordinate of length \(2\sqrt6\,a\). If the line \(x=k\) cuts the \(X\)-axis at \(P\), then the length of the double ordinate drawn through \(O\) to the parabola having \(P\) and \(S\) as vertex and focus is

• A. \(4\sqrt6\,a\)
• B. \(4\sqrt3\,a\)
• C. \(2\sqrt2\,a\)
• D. \(2\sqrt3\,a\) \bigskip

Hint:

For the parabola \[ y^2=4ax, \] the length of the double ordinate at \(x=k\) is \[ 4\sqrt{ak}. \] Always use this formula first to determine the ordinate position before forming the new parabola.

Answer 1

The Correct Option is D

Concept: For a parabola, double ordinate at \(x=k\) has length \(4\sqrt{ak}\). Use this to find \(k\), then translate the parabola to have vertex at that ordinate's foot and focus at the original focus. Find the new equation and evaluate the double ordinate through the origin.

Step 1: Original \(y^2=4ax\). Double ordinate length at \(x=k\) is \(2\sqrt6 a \Rightarrow 4\sqrt{ak}=2\sqrt6 a \Rightarrow 2\sqrt{ak}=\sqrt6 a \Rightarrow 4ak=6a^2 \Rightarrow k=3a/2\).

Step 2: New parabola: vertex at \((3a/2,0)\), focus at S(a,0). Distance vertex to focus = \(a/2\) to the left, so equation: \(y^2 = -4(a/2)(x - 3a/2) = -2a(x - 3a/2)\).

Step 3: At origin O(0,0): \(y^2 = -2a(0 - 3a/2) = 3a^2 \Rightarrow y = \pm \sqrt3 a\). Double ordinate length = \(2\sqrt3 a\).

Step 4: Write the final answer. \(\boxed{2\sqrt3 a}\)