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 the parabola \[ y^2=4ax, \] the length of the double ordinate at \(x=k\) is \[ 4\sqrt{ak}. \] Also, if a parabola has vertex \((h,0)\) and focus \((h+a,0)\), then its equation is \[ y^2=4a(x-h). \]

Step 1: Find the value of \(k\). Given that the double ordinate at \[ x=k \] has length \[ 2\sqrt6\,a. \] Using \[ 4\sqrt{ak}=2\sqrt6\,a, \] \[ 2\sqrt{ak}=\sqrt6\,a. \] Squaring, \[ 4ak=6a^2. \] \[ k=\frac{3a}{2}. \]

Step 2: Find the new parabola. For the original parabola, \[ O=(0,0), \qquad S=(a,0). \] The line \[ x=k \] meets the \(X\)-axis at \[ P=\left(\frac{3a}{2},0\right). \] The new parabola has \[ \text{vertex}=P=\left(\frac{3a}{2},0\right) \] and \[ \text{focus}=S=(a,0). \] Hence its parameter is \[ a_1 = PS = \frac{3a}{2}-a = \frac{a}{2}. \] Since the focus lies to the left of the vertex, its equation is \[ y^2=-4a_1\left(x-\frac{3a}{2}\right) = -2a\left(x-\frac{3a}{2}\right). \]

Step 3: Find the double ordinate through \(O\). At the point \[ O=(0,0), \] substitute \(x=0\) into the new parabola: \[ y^2 = -2a\left(-\frac{3a}{2}\right). \] \[ y^2=3a^2. \] \[ y=\pm \sqrt3\,a. \] Therefore the double ordinate through \(O\) has length \[ 2(\sqrt3\,a) = 2\sqrt3\,a. \]

Step 4: Write the final answer. \[ \boxed{2\sqrt3\,a} \]