Question

Consider the family of curves x² + y² = 2x + 4y + k with a real parameter k > −5. Then the orthogonal trajectory to this family of curves passing through (2, 3) also passes through

• A. (3, 4)
• B. (−1, 1)
• C. (1, 0)
• D. (3, 5)

Answer 1

The Correct Option is A

To solve this problem, we need to analyze the family of curves given by the equation \(x^2 + y^2 = 2x + 4y + k\) and find their orthogonal trajectories. Let's follow these steps:

1. Rewrite the given equation for the family of curves: 

Rearrange the terms of the equation \(x^2 + y^2 = 2x + 4y + k\):

\(x^2 - 2x + y^2 - 4y = k\)

1. Complete the square for both \(x\) and \(y\) terms:

For \(x\): \(x^2 - 2x = (x - 1)^2 - 1\)

For \(y\): \(y^2 - 4y = (y - 2)^2 - 4\)

Substitute these into the equation:

\((x - 1)^2 - 1 + (y - 2)^2 - 4 = k\)

Rearrange this to:

\((x - 1)^2 + (y - 2)^2 = k + 5\)

1. Differentiate implicitly to find the slope of the given curves:

Differentiate both sides with respect to \(x\):

\(2(x - 1) + 2(y - 2)\frac{dy}{dx} = 0\)

Simplifying gives:

\(\frac{dy}{dx} = -\frac{x - 1}{y - 2}\)

1. Find the slope of the orthogonal trajectory by using the negative reciprocal:

The slope of the orthogonal trajectories is the negative reciprocal, so:

\(\frac{dy}{dx} = \frac{y - 2}{x - 1}\)

1. Find the equation of the orthogonal trajectory passing through the point (2, 3):

Using the initial condition \((x_0, y_0) = (2, 3)\), let us integrate the differential equation obtained for orthogonal trajectories:

\(\frac{dy}{y - 2} = \frac{dx}{x - 1}\)

Integrating both sides:

\(\ln|y - 2| = \ln|x - 1| + C\)

Using the initial condition (2, 3),

\(\ln|3 - 2| = \ln|2 - 1| + C \Rightarrow C = 0\)

Then the orthogonal trajectory simplifies to:

\(|y - 2| = |x - 1|\)

1. Find the specific trajectory passing through given options:

Check the point (3, 4):

\(|4 - 2| = |3 - 1| \Rightarrow 2 = 2\)

1. Conclusion:

Thus, the point (3, 4) lies on the orthogonal trajectory of the family of curves. Therefore, the correct option is (3, 4).