If `a=[t^(2)-3t+4] and b=[3+5t]`, where [.] donates the greatest integer function, then the latusrectum of the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1` at `t=3/2` is

Options

A. 20
B. 10
C. `1/5`
D. `1/10`

Answer

To solve the problem step by step, we will follow the same approach as in the video transcript.

### Step 1: Calculate the value of \( a \)
We have:
\[ a = \lfloor t^2 - 3t + 4 \rfloor \]
Substituting \( t = \frac{3}{2} \):
\[ t^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]
\[ 3t = 3 \times \frac{3}{2} = \frac{9}{2} \]
Now, substituting these values into the equation for \( a \):
\[ a = \lfloor \frac{9}{4} - \frac{9}{2} + 4 \rfloor \]
Converting \( \frac{9}{2} \) to a fraction with a denominator of 4:
\[ \frac{9}{2} = \frac{18}{4} \]
Now substituting:
\[ a = \lfloor \frac{9}{4} - \frac{18}{4} + \frac{16}{4} \rfloor = \lfloor \frac{7}{4} \rfloor \]
Since \( \frac{7}{4} = 1.75 \), the greatest integer less than or equal to \( 1.75 \) is:
\[ a = 1 \]

### Step 2: Calculate the value of \( b \)
We have:
\[ b = \lfloor 3 + 5t \rfloor \]
Substituting \( t = \frac{3}{2} \):
\[ b = \lfloor 3 + 5 \times \frac{3}{2} \rfloor \]
Calculating \( 5 \times \frac{3}{2} \):
\[ 5 \times \frac{3}{2} = \frac{15}{2} = 7.5 \]
Now substituting:
\[ b = \lfloor 3 + 7.5 \rfloor = \lfloor 10.5 \rfloor \]
The greatest integer less than or equal to \( 10.5 \) is:
\[ b = 10 \]

### Step 3: Write the equation of the ellipse
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting the values of \( a \) and \( b \):
\[ \frac{x^2}{1^2} + \frac{y^2}{10^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{1} + \frac{y^2}{100} = 1 \]

### Step 4: Identify the latus rectum
For an ellipse, the length of the latus rectum (LR) is given by:
\[ \text{Length of LR} = \frac{2a^2}{b} \]
Substituting the values of \( a \) and \( b \):
\[ \text{Length of LR} = \frac{2 \times 1^2}{10} = \frac{2}{10} = \frac{1}{5} \]

### Final Answer
Thus, the length of the latus rectum of the ellipse at \( t = \frac{3}{2} \) is:
\[ \frac{1}{5} \]