The equation of the chord of the ellipse `2x^2+ 5y^2 =20` which is bisected at the point `(2, 1)` is

Options

A. `4x+5y+13=0`
B. `4x+5y=13`
C. `5x+4y+13=0`
D. none of these

Answer

To find the equation of the chord of the ellipse \(2x^2 + 5y^2 = 20\) that is bisected at the point \((2, 1)\), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form
The given ellipse equation is:
\[ 2x^2 + 5y^2 = 20 \]
We can divide the entire equation by 20 to rewrite it in standard form:
\[ \frac{x^2}{10} + \frac{y^2}{4} = 1 \]

### Step 2: Identify the parameters \(a\) and \(b\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 10\) (thus \(a = \sqrt{10}\))
- \(b^2 = 4\) (thus \(b = 2\))

### Step 3: Use the chord midpoint formula
For an ellipse of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the equation of the chord bisected at the point \((h, k)\) is given by:
\[ a h x + b k y = a h^2 + b k^2 \]
In our case, \(h = 2\) and \(k = 1\).

### Step 4: Substitute \(a\), \(b\), \(h\), and \(k\) into the formula
Substituting the values:
- \(a = \sqrt{10}\)
- \(b = 2\)
- \(h = 2\)
- \(k = 1\)

The equation becomes:
\[ \sqrt{10} \cdot 2 \cdot x + 2 \cdot 1 \cdot y = \sqrt{10} \cdot 2^2 + 2 \cdot 1^2 \]

### Step 5: Simplify the equation
Calculating the left side:
\[ 2\sqrt{10} x + 2y \]
Calculating the right side:
\[ \sqrt{10} \cdot 4 + 2 \cdot 1 = 4\sqrt{10} + 2 \]
Thus, we have:
\[ 2\sqrt{10} x + 2y = 4\sqrt{10} + 2 \]

### Step 6: Rearranging the equation
To express this in a standard linear form, we can rearrange it:
\[ 2\sqrt{10} x + 2y - 4\sqrt{10} - 2 = 0 \]
Dividing the entire equation by 2 for simplicity:
\[ \sqrt{10} x + y - 2\sqrt{10} - 1 = 0 \]

### Final Equation
Thus, the equation of the chord of the ellipse that is bisected at the point \((2, 1)\) is:
\[ \sqrt{10} x + y - 2\sqrt{10} - 1 = 0 \]