If lines `2x+3y=10 and 2x-3y=10` are tangents at the extremities of a latus rectum of an ellipse, whose centre is origin, then the length of the latus rectum is :

Options

A. `(110)/(27)`
B. `(98)/(27)`
C. `(100)/(27)`
D. `(120)/(27)`

Answer

To find the length of the latus rectum of the ellipse whose tangents at the extremities of the latus rectum are given by the equations \(2x + 3y = 10\) and \(2x - 3y = 10\), we can follow these steps:

### Step 1: Identify the equations of the tangents
The given equations of the tangents are:
1. \(2x + 3y = 10\)
2. \(2x - 3y = 10\)

### Step 2: Convert the equations to slope-intercept form
To find the slopes of the tangents, we can convert these equations to the slope-intercept form \(y = mx + b\).

1. For \(2x + 3y = 10\):
\[ 3y = -2x + 10 \implies y = -\frac{2}{3}x + \frac{10}{3} \]
The slope \(m_1 = -\frac{2}{3}\).

2. For \(2x - 3y = 10\):
\[ -3y = -2x + 10 \implies 3y = 2x - 10 \implies y = \frac{2}{3}x - \frac{10}{3} \]
The slope \(m_2 = \frac{2}{3}\).

### Step 3: Identify the eccentricity (E) and semi-major axis (A)
From the tangent equations, we can identify the values of \(E\) and \(A\).

1. The general form of the tangent to an ellipse at the extremities of the latus rectum is given by:
\[ Ex + y = A \]
Comparing with \(2x + 3y = 10\):
- From \(3y = -2x + 10\), we can see that \(E = \frac{2}{3}\) and \(A = \frac{10}{3}\).

### Step 4: Use the relationship between E, A, and B
The relationship between \(E\), \(A\), and \(B\) (semi-minor axis) is given by:
\[ E^2 = 1 - \frac{B^2}{A^2} \]
Substituting the values:
\[ \left(\frac{2}{3}\right)^2 = 1 - \frac{B^2}{\left(\frac{10}{3}\right)^2} \]
Calculating:
\[ \frac{4}{9} = 1 - \frac{B^2}{\frac{100}{9}} \]
\[ \frac{4}{9} = 1 - \frac{9B^2}{100} \]
Rearranging gives:
\[ \frac{9B^2}{100} = 1 - \frac{4}{9} = \frac{5}{9} \]
Thus:
\[ 9B^2 = \frac{500}{9} \]
\[ B^2 = \frac{500}{81} \]

### Step 5: Calculate the length of the latus rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2B^2}{A} \]
Substituting the values of \(B^2\) and \(A\):
\[ L = \frac{2 \times \frac{500}{81}}{\frac{10}{3}} = \frac{1000/81}{10/3} = \frac{1000 \times 3}{81 \times 10} = \frac{3000}{810} = \frac{100}{27} \]

### Final Answer
The length of the latus rectum is:
\[ \frac{100}{27} \]