Question

A, B are fixed points with coordinates (0,a) and (0,b) (a>0,b>0). P is a variable point (x,0) referred to as the rectangular axis. If the angle \(\angle\)APB is maximum, then

• A. x²=ab
• B. x²=a+b
• C. x=\(\frac{1}{ab}\)
• D. x=\(\frac{a+b}{2}\)

Answer 1

The Correct Option is A

Given:
Points \( A(0, a) \), \( B(0, b) \), with \( a > 0 \), \( b > 0 \), and point \( P(x, 0) \) on the x-axis.

We want to maximize angle \( \angle APB \). 

Step 1: Slopes of lines AP and BP
\[ \text{slope of } AP = -\frac{a}{x}, \quad \text{slope of } BP = -\frac{b}{x} \]
Step 2: Use angle between lines formula
\[ \tan(\angle APB) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{b-a}{x}}{1 + \frac{ab}{x^2}} \right| = \left| \frac{(b-a)x}{x^2 + ab} \right| \]
Step 3: Maximize the expression
Let: \[ f(x) = \frac{(b-a)x}{x^2 + ab} \]
Using the quotient rule: \[ f'(x) = \frac{(b-a)(ab - x^2)}{(x^2 + ab)^2} \]
Set \( f'(x) = 0 \) to find maximum: \[ (b - a)(ab - x^2) = 0 \Rightarrow x^2 = ab \Rightarrow x = \sqrt{ab} \]
Conclusion:
The angle \( \angle APB \) is maximum when: \[ \boxed{x = \sqrt{ab}} \]
Correct answer: Option (A)