The two tangents to the curve \(ax^2 + 2hxy + by^2 = 1\), \(a>0\) at the points, where it crosses X-axis, are
Show Hint
- parallel
- perpendicular
- inclined at an angle \(\frac{\pi}{4}\)
- None of these
The Correct Option is A
Solution and Explanation
To solve this problem, we need to determine the nature of the tangents to the given conic section equation at the points where it crosses the X-axis.
The curve is given by the equation \(ax^2 + 2hxy + by^2 = 1\). This equation represents a general conic section.
The condition for the curve to cross the X-axis is \(y = 0\). Substituting \(y=0\) into the equation gives us:
\(ax^2 = 1 \Rightarrow x^2 = \frac{1}{a}\)
Thus, the points where the curve crosses the X-axis are \((x_1, 0)\) and \((x_2, 0)\) where \(x_1 = \frac{1}{\sqrt{a}}\) and \(x_2 = -\frac{1}{\sqrt{a}}\).
Now, we find the equation of the tangents at these points. The general formula for the tangent to a conic \(ax^2 + 2hxy + by^2 = 1\) at a point \((x_1, y_1)\) is:
\(axx_1 + h(xy_1 + yx_1) + byy_1 = 1\).
For the point \((x_1, 0)\): Substitute \(y_1 = 0\), the tangent equation becomes:
\(axx_1 = 1\)
For the point \((x_2, 0)\): Substitute \(y_1 = 0\), the tangent equation becomes:
\(axx_2 = 1\)
Both equations are of the form \(axx_1 = 1\) and \(ax(-x_1) = 1\) respectively.
These equations show that both tangents are vertical lines, which means they are parallel to each other. Therefore, the two tangents at the points where the curve crosses the X-axis are parallel.
Hence, the correct answer is: Parallel.
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