The equation of the chord of `(x^(2))/(36)+(y^(2))/(9)=1` which is bisected at (2,1) is

Options

A. `x+2y-4=0`
B. `x-2y+4=0`
C. `x+2y+4=0`
D. `-x+2y+4=0`

Answer

To find the equation of the chord of the ellipse \(\frac{x^2}{36} + \frac{y^2}{9} = 1\) that is bisected at the point (2, 1), we can follow these steps:

### Step 1: Identify the given values
The given point where the chord is bisected is \((x_1, y_1) = (2, 1)\).

### Step 2: Write the equation of the ellipse
The equation of the ellipse is:
\[ \frac{x^2}{36} + \frac{y^2}{9} = 1 \]

### Step 3: Use the formula for the chord bisected at a point
The formula for the equation of the chord of an ellipse that is bisected at the point \((x_1, y_1)\) is given by:
\[ T = S_1 \]
where \(T\) is the equation of the tangent at the point \((x_1, y_1)\) and \(S_1\) is the value of the ellipse equation at the point \((x_1, y_1)\).

### Step 4: Calculate \(S_1\)
Substituting the values of \(x_1\) and \(y_1\) into the equation of the ellipse:
\[ S_1 = \frac{x_1^2}{36} + \frac{y_1^2}{9} - 1 \]
Substituting \(x_1 = 2\) and \(y_1 = 1\):
\[ S_1 = \frac{2^2}{36} + \frac{1^2}{9} - 1 = \frac{4}{36} + \frac{1}{9} - 1 \]
Converting \(\frac{1}{9}\) to have a common denominator:
\[ \frac{1}{9} = \frac{4}{36} \]
Thus:
\[ S_1 = \frac{4}{36} + \frac{4}{36} - 1 = \frac{8}{36} - 1 = \frac{8}{36} - \frac{36}{36} = \frac{8 - 36}{36} = \frac{-28}{36} = -\frac{7}{9} \]

### Step 5: Write the equation of the tangent \(T\)
The equation of the tangent at the point \((x_1, y_1)\) is given by:
\[ T: \frac{xx_1}{36} + \frac{yy_1}{9} - 1 = 0 \]
Substituting \(x_1 = 2\) and \(y_1 = 1\):
\[ \frac{2x}{36} + \frac{1y}{9} - 1 = 0 \]
This simplifies to:
\[ \frac{x}{18} + \frac{y}{9} - 1 = 0 \]

### Step 6: Set \(T = S_1\)
Now, substituting \(S_1\) into the equation:
\[ \frac{x}{18} + \frac{y}{9} = -\frac{7}{9} \]
To eliminate the fractions, multiply through by 18:
\[ x + 2y = -14 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ x + 2y + 14 = 0 \]

### Final Answer
Thus, the equation of the chord of the ellipse that is bisected at the point (2, 1) is:
\[ x + 2y + 14 = 0 \]