Question

The minimum radius vector of the curve \( \frac{a^2}{x^2} + \frac{b^2}{y^2} = 1 \) is of length

• A. $a - b$
• B. $a + b$
• C. $2a + b$
• D. None of these

Hint:

For minimum distance problems, minimize $r^2$ instead of $r$ for easier calculations.

Answer 1

The Correct Option is D

Concept: Minimize distance from origin: \[ r^2 = x^2 + y^2 \]

Step 1: Use given constraint.
\[ \frac{a^2}{x^2} + \frac{b^2}{y^2} = 1 \]

Step 2: Apply substitution or Lagrange multipliers.
Minimize: \[ x^2 + y^2 \] subject to constraint.

Step 3: Use symmetry approach.
Let: \[ \frac{a^2}{x^2} = \frac{b^2}{y^2} \]

Step 4: Solve relation.
\[ \frac{x^2}{y^2} = \frac{a^2}{b^2} \Rightarrow x = \frac{a}{b}y \]

Step 5: Substitute back.
\[ \frac{a^2}{\frac{a^2}{b^2}y^2} + \frac{b^2}{y^2} = 1 \] \[ \Rightarrow \frac{b^2}{y^2} + \frac{b^2}{y^2} = 1 \Rightarrow \frac{2b^2}{y^2} = 1 \] \[ y^2 = 2b^2,\quad x^2 = 2a^2 \]

Step 6: Compute minimum distance.
\[ r^2 = 2a^2 + 2b^2 = 2(a^2 + b^2) \] \[ r = \sqrt{2(a^2 + b^2)} \] Conclusion:
Not in options $\Rightarrow$ None of these