Figure shows the elipticla path of a planet about the sun. The two shaded parts have equal area. If `t_1 and t_2` be the tim taken by the planet to go from a to b and from c to d respectively

Options

A. `t_1ltt_2`
B. `t_1=t_2`
C. `t_1gtt_2`
D. insufficient information to deduce the relation between `t_1 and t_2`

Answer

To solve the problem, we need to apply Kepler's Second Law of Planetary Motion, which states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

### Step-by-Step Solution:

1. **Identify the Areas**:
- We have two shaded areas in the elliptical path of the planet: Area A1 (from point A to point B) and Area A2 (from point C to point D). According to the problem, these two areas are equal: A1 = A2.

2. **Apply Kepler's Second Law**:
- According to Kepler's Second Law, the area swept out by the planet in a given time interval is constant. This can be mathematically expressed as:
\[ \frac{dA_1}{dt_1} = \frac{dA_2}{dt_2} \]
- Here, \(dA_1\) is the area swept out from A to B in time \(t_1\), and \(dA_2\) is the area swept out from C to D in time \(t_2\).

3. **Set Up the Equation**:
- Since A1 = A2, we can substitute this into our equation:
\[ \frac{A_1}{t_1} = \frac{A_2}{t_2} \]
- This simplifies to:
\[ \frac{A_1}{t_1} = \frac{A_1}{t_2} \]

4. **Cancel the Areas**:
- Since A1 is equal to A2, we can cancel them out from both sides of the equation:
\[ \frac{1}{t_1} = \frac{1}{t_2} \]

5. **Conclude the Relationship**:
- From the equation above, we can conclude that:
\[ t_1 = t_2 \]
- This indicates that the time taken by the planet to travel from A to B (t1) is equal to the time taken to travel from C to D (t2).

### Final Answer:
Thus, the correct conclusion is that \(t_1 = t_2\).

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