The orthogonal trajectories to the curve x2 + (y-1)2 = 1are a family ofa)straight lines passing through originb)straight lines passing through (0, 1)c)straight lines passing through (1, 0)d)None of the aboveCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Explanation:

To find the orthogonal trajectories to the curve, we need to find a family of curves such that the tangent lines to each curve in the family are perpendicular to the tangent lines of the given curve.

The given curve is x^2 + (y-1)^2 = 1.

Let's find the slope of the tangent line to the given curve:

Differentiating both sides of the equation with respect to x, we get:

2x + 2(y-1) * (dy/dx) = 0

Simplifying the equation, we have:

dy/dx = -x / (y-1)

The slope of the tangent line to the given curve is -x / (y-1).

To find the orthogonal trajectories, we need to find curves that have the negative reciprocal slope of -x / (y-1) at each point.

Orthogonal Trajectories:

Let's find the slope of the tangent line to the orthogonal trajectory:

The slope of the tangent line to the orthogonal trajectory is the negative reciprocal of -x / (y-1).

The negative reciprocal of -x / (y-1) is (y-1) / x.

Using the point-slope form of a line, the equation of the tangent line to the orthogonal trajectory passing through a point (x0, y0) is:

y - y0 = (y-1) / x * (x - x0)

Rearranging the equation, we have:

y = (y-1)(x - x0) / x + y0

This is the equation of a straight line passing through the point (x0, y0).

Conclusion:

From the above analysis, we can see that the orthogonal trajectories to the given curve are a family of straight lines passing through the point (0, 1).

Therefore, the correct answer is option 'B' - straight lines passing through (0, 1).

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