Question

If the orthogonal trajectories of the family of ellipses \[ x^2 + 2y^2 = c_1, \quad c_1>0, \] are given by \[ y = c_2 x^\alpha, \quad c_2 \in \mathbb{R}, \] then \( \alpha = \) ................

Hint:

The orthogonal trajectory of a family of curves can be found by solving for the negative reciprocal of the slope of the original curves.

Answer 1

Correct Answer: 1.9 - 2.1

Step 1: Understanding the orthogonal trajectory.
For a family of curves, the orthogonal trajectory is a curve that intersects each curve of the family at a right angle. In this case, the family of ellipses is \(x^2 + 2y^2 = c_1\), and we are given that the orthogonal trajectories are given by \(y = c_2 x^\alpha\), where \(c_2\) is a constant.

Step 2: Calculating the slope of the ellipse.
To find the slope of the ellipse at any point, we implicitly differentiate the equation \(x^2 + 2y^2 = c_1\): \[ 2x + 4yy' = 0 \quad \Rightarrow \quad y' = -\frac{x}{2y} \] This gives the slope of the ellipse at any point.

Step 3: Calculating the slope of the orthogonal trajectory.
The slope of the orthogonal trajectory is the negative reciprocal of the slope of the ellipse. So, for the orthogonal trajectory \(y = c_2 x^\alpha\), we differentiate to get: \[ y' = \alpha c_2 x^{\alpha - 1} \] For orthogonality, the product of the slopes must be -1: \[ \left(-\frac{x}{2y}\right) \cdot \left(\alpha c_2 x^{\alpha - 1}\right) = -1 \] Substituting \(y = c_2 x^\alpha\) into the equation and solving for \( \alpha \), we get: \[ \alpha = 2 \]
Step 4: Conclusion.
Therefore, \( \alpha = 2 \).