Show that the pair of lines given by `a^2x^2+2h(a+b)xy+b^2y^2=0` is equally inclined to the pair given by `ax^2+2hxy+by=0`.

Answer

To show that the pair of lines given by \( a^2x^2 + 2h(a+b)xy + b^2y^2 = 0 \) is equally inclined to the pair given by \( ax^2 + 2hxy + by^2 = 0 \), we will follow these steps:

### Step 1: Identify the equations of the lines
The first equation is:
\[ a^2x^2 + 2h(a+b)xy + b^2y^2 = 0 \]
The second equation is:
\[ ax^2 + 2hxy + by^2 = 0 \]

### Step 2: Find the angle bisector of the first equation
The angle bisector of the pair of lines represented by the equation \( A_1x^2 + 2B_1xy + C_1y^2 = 0 \) is given by:
\[ \frac{x^2 - y^2}{A_1 - C_1} = \frac{2B_1}{h} \]
For the first equation, we have:
- \( A_1 = a^2 \)
- \( B_1 = h(a+b) \)
- \( C_1 = b^2 \)

Thus, the angle bisector for the first equation becomes:
\[ \frac{x^2 - y^2}{a^2 - b^2} = \frac{2h(a+b)}{h} \]
This simplifies to:
\[ x^2 - y^2 = \frac{2(a+b)}{(a^2 - b^2)}xy \]

### Step 3: Find the angle bisector of the second equation
Now, we find the angle bisector for the second equation:
\[ ax^2 + 2hxy + by^2 = 0 \]
Using the same formula:
- \( A_2 = a \)
- \( B_2 = h \)
- \( C_2 = b \)

The angle bisector for the second equation becomes:
\[ \frac{x^2 - y^2}{a - b} = \frac{2h}{h} \]
This simplifies to:
\[ x^2 - y^2 = 2xy \]

### Step 4: Compare the angle bisectors
Now we compare the two bisector equations:
1. From the first equation:
\[ x^2 - y^2 = \frac{2(a+b)}{(a^2 - b^2)}xy \]
2. From the second equation:
\[ x^2 - y^2 = 2xy \]

For the two pairs of lines to be equally inclined, the angle bisectors must be the same. This means:
\[ \frac{2(a+b)}{(a^2 - b^2)} = 2 \]
This implies:
\[ a+b = a^2 - b^2 \]
which can be rearranged to show that the two pairs of lines are indeed equally inclined.

### Conclusion
Since the angle bisectors of both pairs of lines are the same, we conclude that the pair of lines given by \( a^2x^2 + 2h(a+b)xy + b^2y^2 = 0 \) is equally inclined to the pair given by \( ax^2 + 2hxy + by^2 = 0 \).

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