Two straight lines rotate about two fixed points `(-a, 0) and (a,o).` If they start from their position of coincidence such that oen rotates at the rate double that of the other, then
The point`(-a,0)` always lies

Options

A. inside the curve
B. outside the curve
C. on the curve
D. none of these

Answer

To solve the problem step by step, we need to analyze the motion of the two lines rotating about the points \((-a, 0)\) and \((a, 0)\). Here's the detailed solution:

### Step 1: Understand the Rotation of the Lines
We have two lines rotating about the points \((-a, 0)\) and \((a, 0)\). One line rotates at an angle \(\theta\) and the other rotates at an angle \(2\theta\) (double the rate).

### Step 2: Determine the Slopes of the Lines
The slope of the line rotating about \((-a, 0)\) at angle \(\theta\) can be expressed as:
\[ m_1 = \tan(\theta) \]
The coordinates of a point on this line can be expressed as:
\[ y = m_1(x + a) \quad \text{(1)} \]

For the line rotating about \((a, 0)\) at angle \(2\theta\), the slope is:
\[ m_2 = \tan(2\theta) \]
The coordinates of a point on this line can be expressed as:
\[ y = m_2(x - a) \quad \text{(2)} \]

### Step 3: Use the Double Angle Formula for Tangent
Using the double angle formula for tangent:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Let \(y = \tan(\theta)(x + a)\) from equation (1). Substituting \(m_1\) into the double angle formula gives:
\[ m_2 = \frac{2y/(x + a)}{1 - (y/(x + a))^2} \]

### Step 4: Set Up the Equations
From the slopes, we can equate the two expressions for \(y\):
\[ \tan(2\theta) = \frac{2y}{x + a} \cdot \frac{(x + a)^2}{(x + a)^2 - y^2} \]
Cross-multiplying gives:
\[ 2y(x + a) = y((x + a)^2 - y^2) \]

### Step 5: Simplify the Equation
Rearranging and simplifying leads to:
\[ y^2 + 2y(x + a) - (x + a)^2 = 0 \]
This is a quadratic equation in \(y\).

### Step 6: Determine the Locus of the Point
To find the locus of the point \((-a, 0)\), substitute \(x = -a\) and \(y = 0\) into the equation:
\[ 0^2 + 2(0)(-a + a) - (-a + a)^2 = 0 \]
This simplifies to:
\[ 0 = 0 \]
indicating that the point \((-a, 0)\) satisfies the locus equation.

### Step 7: Conclusion
Since the point \((-a, 0)\) satisfies the locus equation derived from the rotating lines, it lies on the curve formed by the intersection of the two lines.

### Final Answer
The point \((-a, 0)\) always lies **on the curve**.

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