The eccentricity of the ellipse `3x^2+4y^2=12` is decreasing at the rate of 0.1 per sec.The time at which it will coincide with auxiliary circle is:

Options

A. 2 seconds
B. 3 seconds
C. 5 seconds
D. 6 seconds

Explanation

To solve the problem, we need to find the time at which the eccentricity of the ellipse coincides with that of the auxiliary circle. Let's break down the solution step by step.

### Step 1: Write the equation of the ellipse
The given equation of the ellipse is:
\[ 3x^2 + 4y^2 = 12 \]

### Step 2: Standard form of the ellipse
To convert this into the standard form, we divide the entire equation by 12:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
This shows that \( a^2 = 4 \) and \( b^2 = 3 \).

### Step 3: Find the eccentricity of the ellipse
The eccentricity \( e \) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 4: Determine the rate of change of eccentricity
We are given that the eccentricity is decreasing at the rate of \( \frac{dE}{dt} = -0.1 \) per second.

### Step 5: Set up the equation for eccentricity over time
Let \( E(t) \) be the eccentricity at time \( t \). The equation can be expressed as:
\[ E(t) = \frac{1}{2} - 0.1t \]

### Step 6: Find when the eccentricity equals that of the auxiliary circle
The eccentricity of the auxiliary circle is 1. We need to find the time \( t \) when:
\[ E(t) = 1 \]
Setting the equation equal to 1:
\[ \frac{1}{2} - 0.1t = 1 \]

### Step 7: Solve for \( t \)
Rearranging the equation:
\[ -0.1t = 1 - \frac{1}{2} \]
\[ -0.1t = \frac{1}{2} \]
Now, multiplying both sides by -1:
\[ 0.1t = -\frac{1}{2} \]
Dividing both sides by 0.1:
\[ t = -\frac{1/2}{0.1} = -5 \]

### Step 8: Conclusion
Since time cannot be negative, we interpret that the eccentricity will reach the value of 1 at \( t = 5 \) seconds.

Thus, the time at which the eccentricity of the ellipse coincides with that of the auxiliary circle is:
\[ \boxed{5 \text{ seconds}} \]