Question

Let \( S \) be the family of orthogonal trajectories of the family of curves \[ 2x^2 + y^2 = k, \text{ for } k \in \mathbb{R} \text{ and } k>0. \] If \( \ell \in S \) and \( C \) passes through the point \( (1, 2) \), then \( C \) also passes through

• A. \( (4, -\sqrt{2}) \)
• B. \( (2, -4) \)
• C. \( (2, 2\sqrt{2}) \)
• D. \( (4, 2\sqrt{2}) \)

Hint:

To find the orthogonal trajectories, differentiate the given equation implicitly, then solve for the new family by taking the negative reciprocal of the derivative.

Answer 1

The Correct Option is C

Step 1: Equation of the family of curves.
We are given the family of curves \( 2x^2 + y^2 = k \), where \( k \) is a constant. To find the orthogonal trajectories, we differentiate implicitly with respect to \( x \) to get the slope of the tangent line. This will give the equation of the orthogonal trajectories. \[ 4x + 2yy' = 0 \implies y' = -\frac{2x}{y}. \] The family of orthogonal trajectories will have slopes that are the negative reciprocal of this slope.
Step 2: Use the point \( (1, 2) \).
Substitute the point \( (1, 2) \) into the equation of the curve \( 2x^2 + y^2 = k \) to find the value of \( k \): \[ 2(1)^2 + (2)^2 = k \implies k = 8. \] So, the equation of the curve is \( 2x^2 + y^2 = 8 \). Now, we find the equation of the orthogonal trajectory. The orthogonal trajectory will pass through \( (1, 2) \) and the new point will be \( (2, 2\sqrt{2}) \).
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(C)} \).