Question

The shortest distance between the line $y - x = 1$ and the curve $x = y^2$ is

• A. $\frac{3\sqrt{2}}{8}$
• B. $\frac{2\sqrt{3}}{8}$
• C. $\frac{3\sqrt{2}}{5}$
• D. $\frac{\sqrt{3}}{4}$

Hint:

Parallel tangent method: find where the curve's derivative equals the line's slope, then use the point-to-line distance formula.

Answer 1

The Correct Option is A

Step 1: Concept
Shortest distance occurs along the common normal, where the tangent to $x=y^2$ is parallel to $y-x=1$ (slope $= 1$).

Step 2: Meaning
Curve $x=y^2 \implies \frac{dx}{dy} = 2y \implies \frac{dy}{dx} = \frac{1}{2y}$.

Step 3: Analysis
Set $\frac{1}{2y} = 1 \implies y = 1/2$. Then $x = (1/2)^2 = 1/4$. Point $P(1/4, 1/2)$. Distance from $P(1/4, 1/2)$ to $x - y + 1 = 0$ is $d = \frac{|1/4 - 1/2 + 1|}{\sqrt{1^2 + (-1)^2}}$. $d = \frac{|3/4|}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$.

Step 4: Conclusion
The shortest distance is $\frac{3\sqrt{2}}{8}$. Final Answer: (A)