Equation of hyperbola passing through the point `(1,-1)` and having asymptotes `x+2y+3=0` and `3x+4y+5=0` is ………….

Answer

To find the equation of the hyperbola that passes through the point (1, -1) and has the given asymptotes, we can follow these steps:

### Step 1: Write the equations of the asymptotes
The asymptotes are given as:
1. \( x + 2y + 3 = 0 \)
2. \( 3x + 4y + 5 = 0 \)

### Step 2: Form the equation of the hyperbola
The equation of the hyperbola can be expressed as the product of the equations of its asymptotes plus a constant \( \lambda \):
\[ (x + 2y + 3)(3x + 4y + 5) + \lambda = 0 \]

### Step 3: Expand the product
Now, we will expand the product:
\[ (x + 2y + 3)(3x + 4y + 5) = x(3x + 4y + 5) + 2y(3x + 4y + 5) + 3(3x + 4y + 5) \]
Calculating each term:
- \( x(3x + 4y + 5) = 3x^2 + 4xy + 5x \)
- \( 2y(3x + 4y + 5) = 6xy + 8y^2 + 10y \)
- \( 3(3x + 4y + 5) = 9x + 12y + 15 \)

Combining these, we get:
\[ 3x^2 + (4xy + 6xy) + 8y^2 + (5x + 9x) + (10y + 12y) + 15 \]
This simplifies to:
\[ 3x^2 + 10xy + 8y^2 + 14x + 22y + 15 \]

Thus, the equation of the hyperbola becomes:
\[ 3x^2 + 10xy + 8y^2 + 14x + 22y + 15 + \lambda = 0 \]

### Step 4: Substitute the point (1, -1)
Now we substitute the point (1, -1) into the equation to find \( \lambda \):
\[ 3(1)^2 + 10(1)(-1) + 8(-1)^2 + 14(1) + 22(-1) + 15 + \lambda = 0 \]
Calculating this gives:
\[ 3 - 10 + 8 + 14 - 22 + 15 + \lambda = 0 \]
This simplifies to:
\[ 3 - 10 + 8 + 14 - 22 + 15 = 8 + \lambda = 0 \]
Thus, we find:
\[ \lambda = -8 \]

### Step 5: Substitute \( \lambda \) back into the hyperbola equation
Now we substitute \( \lambda \) back into the hyperbola equation:
\[ 3x^2 + 10xy + 8y^2 + 14x + 22y + 15 - 8 = 0 \]
This simplifies to:
\[ 3x^2 + 10xy + 8y^2 + 14x + 22y + 7 = 0 \]

### Final Answer
The equation of the hyperbola is:
\[ 3x^2 + 10xy + 8y^2 + 14x + 22y + 7 = 0 \]

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