The mid point of the chord 4x-3y=5 of the hyperbola `2x^(2)-3y^(2)=12` is

Options

A. (2,1)
B. (5,5)
C. (1,-1/3)
D. (1/2,-1)

Answer

To find the midpoint of the chord \(4x - 3y = 5\) of the hyperbola \(2x^2 - 3y^2 = 12\), we can follow these steps:

### Step 1: Express \(y\) in terms of \(x\) from the chord equation
The equation of the chord is given as:
\[ 4x - 3y = 5 \]
Rearranging this equation to express \(y\) in terms of \(x\):
\[ 3y = 4x - 5 \implies y = \frac{4x - 5}{3} \]

**Hint:** Rearranging the equation of the chord helps isolate one variable.

### Step 2: Substitute \(y\) into the hyperbola equation
Now we substitute \(y\) into the hyperbola equation \(2x^2 - 3y^2 = 12\):
\[ 2x^2 - 3\left(\frac{4x - 5}{3}\right)^2 = 12 \]
Calculating \(y^2\):
\[ y^2 = \left(\frac{4x - 5}{3}\right)^2 = \frac{(4x - 5)^2}{9} \]
Substituting this back into the hyperbola equation:
\[ 2x^2 - 3\left(\frac{(4x - 5)^2}{9}\right) = 12 \]
Multiplying through by 9 to eliminate the fraction:
\[ 18x^2 - 3(4x - 5)^2 = 108 \]

**Hint:** Substituting \(y\) into the hyperbola equation allows us to work with a single variable.

### Step 3: Expand and simplify the equation
Now, we expand \((4x - 5)^2\):
\[ (4x - 5)^2 = 16x^2 - 40x + 25 \]
Substituting this back gives:
\[ 18x^2 - 3(16x^2 - 40x + 25) = 108 \]
Distributing \(-3\):
\[ 18x^2 - 48x^2 + 120x - 75 = 108 \]
Combining like terms:
\[ -30x^2 + 120x - 75 - 108 = 0 \implies -30x^2 + 120x - 183 = 0 \]
Dividing the entire equation by -3:
\[ 10x^2 - 40x + 61 = 0 \]

**Hint:** Simplifying the equation helps in finding the roots more easily.

### Step 4: Find the sum of the roots
Using the quadratic formula, the sum of the roots \(x_1 + x_2\) can be found using:
\[ x_1 + x_2 = -\frac{b}{a} = -\frac{-40}{10} = 4 \]

**Hint:** The sum of the roots can be derived directly from the coefficients of the quadratic equation.

### Step 5: Calculate the x-coordinate of the midpoint
The x-coordinate of the midpoint is:
\[ \frac{x_1 + x_2}{2} = \frac{4}{2} = 2 \]

**Hint:** The midpoint is the average of the x-coordinates of the endpoints of the chord.

### Step 6: Find the corresponding y-coordinate
Now we substitute \(x = 2\) back into the equation for \(y\):
\[ y = \frac{4(2) - 5}{3} = \frac{8 - 5}{3} = \frac{3}{3} = 1 \]

**Hint:** Substituting back into the equation for \(y\) gives us the corresponding y-coordinate.

### Final Result
Thus, the midpoint of the chord is:
\[ \boxed{(2, 1)} \]