The equation of one asymptote of the hyperbola `14x^(2)+38xy+20y^(2)+x-7y-91=0" is "7x+5y-3=0`. Then the other asymptote is

Options

A. 2x+4y=1
B. 2x+4y+1=0
C. 2x+4y+3=0
D. 2x+4y+4=0

Answer

To solve the problem of finding the other asymptote of the hyperbola given one asymptote, we will follow these steps:

### Step 1: Write down the given hyperbola equation and the known asymptote
The hyperbola is given by the equation:
\[ 14x^2 + 38xy + 20y^2 + x - 7y - 91 = 0 \]
One of the asymptotes is given as:
\[ 7x + 5y - 3 = 0 \]

### Step 2: Modify the hyperbola equation
We know that the equations of the asymptotes can be derived from the hyperbola equation by replacing the constant term (-91) with a variable constant (let's denote it as \(\lambda\)):
\[ 14x^2 + 38xy + 20y^2 + x - 7y + \lambda = 0 \]

### Step 3: Identify coefficients
We will compare this modified equation with the general form:
\[ Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \]
From our modified hyperbola equation, we have:
- \(A = 14\)
- \(B = 20\)
- \(C = \lambda\)
- \(H = 19\) (since \(2H = 38\))
- \(G = \frac{1}{2}\) (since \(2G = 1\))
- \(F = -\frac{7}{2}\) (since \(2F = -7\))

### Step 4: Use the condition for asymptotes
The condition for the asymptotes is given by:
\[ ABC + 2FGH - A F^2 - B G^2 - C H^2 = 0 \]
Substituting the values we have:
\[ 14 \cdot 20 \cdot \lambda + 2 \cdot \left(-\frac{7}{2}\right) \cdot \frac{1}{2} \cdot 19 - 14 \cdot \left(-\frac{7}{2}\right)^2 - 20 \cdot \left(\frac{1}{2}\right)^2 - \lambda \cdot 19^2 = 0 \]

### Step 5: Simplify the equation
Calculating each term:
1. \(ABC = 280\lambda\)
2. \(2FGH = -7 \cdot \frac{1}{2} \cdot 19 = -66.5\)
3. \(AF^2 = 14 \cdot \frac{49}{4} = 171.5\)
4. \(BG^2 = 20 \cdot \frac{1}{4} = 5\)
5. \(CH^2 = \lambda \cdot 361\)

Putting it all together:
\[ 280\lambda - 66.5 - 171.5 - 5 - 361\lambda = 0 \]
This simplifies to:
\[ -81\lambda - 243 = 0 \]
Thus:
\[ \lambda = -3 \]

### Step 6: Substitute \(\lambda\) back into the hyperbola equation
Now, we substitute \(\lambda\) back into the modified hyperbola equation:
\[ 14x^2 + 38xy + 20y^2 + x - 7y + 3 = 0 \]

### Step 7: Find the other asymptote
Let the other asymptote be of the form:
\[ Px + Qy + R = 0 \]
We know one asymptote is \(7x + 5y - 3 = 0\). The product of the asymptotes gives the hyperbola equation. Thus:
- Coefficient of \(x^2\): \(7P = 14 \Rightarrow P = 2\)
- Coefficient of \(y^2\): \(5Q = 20 \Rightarrow Q = 4\)
- Constant term: \(-3R = 3 \Rightarrow R = -1\)

Thus, the other asymptote is:
\[ 2x + 4y + 1 = 0 \]

### Final Answer
The equation of the other asymptote is:
\[ 2x + 4y + 1 = 0 \]