Question

If the tangent drawn from the point \(P(5,3)\) to the parabola \(y^{2}=x\) is at a distance of \(\frac{1}{\sqrt{5}}\) units from the vertex of the parabola and touches the parabola at the point Q, then \(PQ=\)

• A. \(2\sqrt{101}\)
• B. \(25\sqrt{2}\)
• C. \(2\sqrt{5}\)
• D. \(5\sqrt{2}\)

Hint:

For parabola \(y^2=4ax\), the point of tangency for a line with slope \(m\) is always \((a/m^2, 2a/m)\).

Answer 1

The Correct Option is C

Concept: The equation of a tangent to the parabola \(y^2=4ax\) (here \(a=1/4\)) is \(y = mx + a/m\). The distance from the vertex \((0,0)\) to the line \(mx - y + a/m = 0\) is given by \(\frac{|a/m|}{\sqrt{m^2+1}}\).

Step 1: Set up the distance condition.
Given \(a=1/4\), the distance is \(\frac{|1/(4m)|}{\sqrt{m^2+1}} = \frac{1}{\sqrt{5}}\). \[ \frac{1}{4|m|\sqrt{m^2+1}} = \frac{1}{\sqrt{5}} \implies 16m^2(m^2+1) = 5 \implies 16m^4 + 16m^2 - 5 = 0 \] Let \(u = m^2\): \(16u^2 + 16u - 5 = 0\). Using the quadratic formula: \(u = \frac{-16 \pm \sqrt{256 + 320}}{32} = \frac{-16 \pm 24}{32}\). Taking the positive root, \(u = 8/32 = 1/4\). Thus, \(m = \pm 1/2\).

Step 2: Find the point of tangency Q.
Tangent equation \(y = \frac{1}{2}x + \frac{1/4}{1/2} = \frac{1}{2}x + \frac{1}{2} \implies x - 2y + 1 = 0\). Point of tangency for \(y^2=x\) is \((a/m^2, 2a/m) = (1/4 \cdot 4, 2 \cdot 1/4 \cdot 2) = (1, 1)\). So \(Q = (1, 1)\).

Step 3: Calculate the distance \(PQ\).
\(P = (5, 3)\), \(Q = (1, 1)\). \[ PQ = \sqrt{(5-1)^2 + (3-1)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \] center minipage0.3

PQ = 2\(\sqrt{5}\) minipage center