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

Step 1: Read the parabola.
The parabola is $y^2=x$. Writing it as $y^2=4ax$ gives $4a=1$, so $a=\tfrac14$, and the vertex is the origin.
Step 2: Write a general tangent.
A tangent of slope $m$ to $y^2=4ax$ is $y=mx+\dfrac{a}{m}$, or $mx-y+\dfrac{a}{m}=0$.
Step 3: Use the distance from the vertex.
The distance from $(0,0)$ to this line is $\dfrac{|a/m|}{\sqrt{m^2+1}}=\dfrac{1}{\sqrt5}$. With $a=\tfrac14$, \[ \frac{1}{4|m|\sqrt{m^2+1}}=\frac{1}{\sqrt5}\implies 16m^2(m^2+1)=5. \]
Step 4: Solve for the slope.
Let $u=m^2$: $16u^2+16u-5=0$, so $u=\dfrac{-16\pm24}{32}$. The positive root is $u=\tfrac14$, hence $m=\pm\tfrac12$.
Step 5: Find the point of contact $Q$.
For $y^2=4ax$ the contact point is $\left(\dfrac{a}{m^2},\dfrac{2a}{m}\right)$. Taking $m=\tfrac12$, $a=\tfrac14$: $x=\dfrac{1/4}{1/4}=1$ and $y=\dfrac{2(1/4)}{1/2}=1$. So $Q=(1,1)$, and indeed $1^2=1$ checks out.
Step 6: Compute $PQ$.
With $P=(5,3)$ and $Q=(1,1)$, \[ PQ=\sqrt{(5-1)^2+(3-1)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt5. \]
\[ \boxed{2\sqrt5} \]