To solve this problem, we need to determine the relationship between the tangents to the given curve at its points of intersection with the X-axis.
The curve equation provided is:
\(ax^2 + 2hxy + by^2 = 1\).
To find where the curve crosses the X-axis, we set \(y = 0\) since points on the X-axis have zero y-coordinates:
Substituting \(y = 0\) in the equation \(ax^2 + 2hxy + by^2 = 1\), we get:
\(ax^2 = 1\).
This implies:
\(x^2 = \frac{1}{a}\).
The solutions to this equation are:
\(x = \pm \frac{1}{\sqrt{a}}\).
So, the points of intersection with the X-axis are:
\(\left( \frac{1}{\sqrt{a}}, 0 \right)\) and \(\left( -\frac{1}{\sqrt{a}}, 0 \right)\).
Next, we need to calculate the slope of tangents to the curve at these points. The slope \(m\) of the tangent to the curve \(ax^2 + 2hxy + by^2 = 1\) at point \((x_1, y_1)\) is given by:
\(m = \frac{-\left(2hx_1 + 2by_1\right)}{2ax_1 + 2hy_1}\).
For the point \(\left( \frac{1}{\sqrt{a}}, 0 \right)\):
\(m_1 = \frac{-(2h \cdot \frac{1}{\sqrt{a}} + 2b \cdot 0)}{2a \cdot \frac{1}{\sqrt{a}} + 2h \cdot 0} = \frac{-2h/\sqrt{a}}{2a/\sqrt{a}} = \frac{-h}{a}\).
For the point \(\left( -\frac{1}{\sqrt{a}}, 0 \right)\):
\(m_2 = \frac{-(2h \cdot -\frac{1}{\sqrt{a}} + 2b \cdot 0)}{2a \cdot -\frac{1}{\sqrt{a}} + 2h \cdot 0} = \frac{2h/\sqrt{a}}{-2a/\sqrt{a}} = \frac{h}{a}\).
Comparing the two slopes, \(m_1 = -\frac{h}{a}\) and \(m_2 = \frac{h}{a}\), we see that these slopes are negatives of each other, indicating that the tangents are parallel.
Thus, the correct answer to the question is that the tangents are parallel.