Find the middle point of the chord intercepted on the line 2x-y+3 =0 by the ellipse `x^2/10 + y^2/6 = 1`

Options

A. `((30)/(23),-(9)/(23))`
B. `(-(30)/(23),(9)/(23))`
C. `(-(30)/(23),-(9)/(23))`
D. `((30)/(23),(9)/(23))`

Answer

To find the middle point of the chord intercepted on the line \(2x - y + 3 = 0\) by the ellipse \(\frac{x^2}{10} + \frac{y^2}{6} = 1\), we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse is in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, we have:
- \(a^2 = 10\) (thus \(a = \sqrt{10}\))
- \(b^2 = 6\) (thus \(b = \sqrt{6}\))

### Step 2: Use the chord midpoint formula
For a chord with a midpoint \((h, k)\) in the ellipse, the equation is given by:
\[ \frac{hx}{a^2} + \frac{ky}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} \]
Substituting \(a^2\) and \(b^2\):
\[ \frac{hx}{10} + \frac{ky}{6} = \frac{h^2}{10} + \frac{k^2}{6} \]

### Step 3: Rewrite the line equation
The line equation is given as \(2x - y + 3 = 0\). Rearranging gives:
\[ y = 2x + 3 \]

### Step 4: Substitute \(y\) in the chord equation
Substituting \(y = 2x + 3\) into the chord equation:
\[ \frac{hx}{10} + \frac{k(2x + 3)}{6} = \frac{h^2}{10} + \frac{k^2}{6} \]
This simplifies to:
\[ \frac{hx}{10} + \frac{2kx}{6} + \frac{3k}{6} = \frac{h^2}{10} + \frac{k^2}{6} \]

### Step 5: Combine like terms
Combining the \(x\) terms:
\[ \left(\frac{h}{10} + \frac{2k}{6}\right)x + \frac{3k}{6} = \frac{h^2}{10} + \frac{k^2}{6} \]
Let’s denote \(A = \frac{h}{10} + \frac{2k}{6}\) and \(B = \frac{h^2}{10} + \frac{k^2}{6} - \frac{3k}{6}\).

### Step 6: Set up the ratios
From the line equation and the chord equation, we can set up the ratios:
\[ \frac{h}{10} : \frac{2k}{6} : 1 \]
This gives:
\[ \frac{h}{10} = \frac{2k}{6} \implies h = \frac{10k}{3} \]

### Step 7: Substitute back into the equation
Now substitute \(h = \frac{10k}{3}\) into the chord equation:
\[ \frac{\frac{10k}{3}}{10} + \frac{2k}{6} = \frac{\left(\frac{10k}{3}\right)^2}{10} + \frac{k^2}{6} \]
This simplifies to:
\[ \frac{k}{3} + \frac{k}{3} = \frac{100k^2}{90} + \frac{k^2}{6} \]
Combining the left side gives:
\[ \frac{2k}{3} = \frac{100k^2}{90} + \frac{15k^2}{90} = \frac{115k^2}{90} \]

### Step 8: Solve for \(k\)
Cross-multiplying gives:
\[ 2k \cdot 90 = 3 \cdot 115k^2 \implies 180k = 345k^2 \implies k(345k - 180) = 0 \]
Thus, \(k = 0\) or \(k = \frac{180}{345} = \frac{12}{23}\).

### Step 9: Find \(h\)
Using \(k = \frac{12}{23}\):
\[ h = \frac{10 \cdot \frac{12}{23}}{3} = \frac{120}{69} = \frac{40}{23} \]

### Step 10: Conclusion
Thus, the midpoint of the chord is:
\[ \left(h, k\right) = \left(-\frac{30}{23}, \frac{12}{23}\right) \]

### Final Answer
The middle point of the chord is \(\left(-\frac{30}{23}, \frac{12}{23}\right)\).