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

The angle ∠APB is maximum when point P is on the x-axis, so x = 0.

The correct answer is option (B): x²=a+b

Answer 2

Given points \( A(0, a) \), \( B(0, b) \) with \( a > 0 \) and \( b > 0 \), and \( P(x, 0) \) is a variable point on the x-axis. We need to find the position of \( P \) such that the angle \( \angle APB \) is maximized.
To solve this, we will use trigonometric principles and the concept of maximization of angles in a triangle.
Step 1: Find the Slopes of Lines AP and BP
The slopes of lines \( AP \) and \( BP \) are given by:
\[ \text{slope of } AP = \frac{0 - a}{x - 0} = \frac{-a}{x} = -\frac{a}{x} \]
\[ \text{slope of } BP = \frac{0 - b}{x - 0} = \frac{-b}{x} = -\frac{b}{x} \]
Step 2: Find the Tangent of the Angle \( \angle APB \)
The tangent of the angle between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan(\angle APB) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Here, \( m_1 = -\frac{a}{x} \) and \( m_2 = -\frac{b}{x} \), so:
\[ \tan(\angle APB) = \left| \frac{-\frac{a}{x} - \left(-\frac{b}{x}\right)}{1 + \left(-\frac{a}{x}\right)\left(-\frac{b}{x}\right)} \right| \]
\[ = \left| \frac{-\frac{a}{x} + \frac{b}{x}}{1 + \frac{ab}{x^2}} \right| \]
\[ = \left| \frac{\frac{b - a}{x}}{1 + \frac{ab}{x^2}} \right| \]
\[ = \left| \frac{b - a}{x} \cdot \frac{x^2}{x^2 + ab} \right| \]
\[ = \left| \frac{(b - a)x}{x^2 + ab} \right| \]
Step 3: Maximize the Tangent Expression
To maximize \( \tan(\angle APB) \), we need to maximize:
\[ \left| \frac{(b - a)x}{x^2 + ab} \right| \]
Taking the derivative of the expression inside the absolute value with respect to \( x \):
\[ f(x) = \frac{(b - a)x}{x^2 + ab} \]
Step 4: Use Calculus to Find the Maximum
Using the quotient rule for derivatives:
\[ f'(x) = \frac{(b - a)(x^2 + ab) - (b - a)x(2x)}{(x^2 + ab)^2} \]
\[ = \frac{(b - a)x^2 + (b - a)ab - 2(b - a)x^2}{(x^2 + ab)^2} \]
\[ = \frac{(b - a)ab - (b - a)x^2}{(x^2 + ab)^2} \]
\[ = \frac{(b - a)(ab - x^2)}{(x^2 + ab)^2} \]
Setting the derivative equal to zero for maximization:
\[ (b - a)(ab - x^2) = 0 \]
Since \( b \neq a \) (given \( a > 0 \) and \( b > 0 \)):
\[ ab - x^2 = 0 \]
\[ x^2 = ab \]
\[ x = \sqrt{ab} \] Step 5: Conclusion
The angle \( \angle APB \) is maximized when \( P \) is at the position \( x = \sqrt{ab} \).
Therefore, the position of \( P \) that maximizes \( \angle APB \) is \( \boxed{(\sqrt{ab}, 0)} \).
so the correct Answer is Option A