The problem asks us to find a family of curves that intersect each member of the given family of parabolas $y - 1 = Cx^2$ at right angles. Let’s solve this step by step.

Step 1: Understand the family of curves

The given family of parabolas is described by the equation: $y - 1 = Cx^2,$ where $C$ is a parameter. This represents a family of parabolas with their vertex at $(0, 1)$, opening upwards for $C > 0$ and downwards for $C < 0$.

Step 2: Orthogonal trajectories

To find the desired family of curves that intersect these parabolas at right angles, we use the concept of orthogonal trajectories.

For two families of curves to intersect at right angles, the slopes of their tangents at the point of intersection must satisfy: $m_1 \cdot m_2 = -1,$ where $m_1$ is the slope of the tangent to one curve, and $m_2$ is the slope of the tangent to the other curve.

Step 3: Find the slope of the tangent to the parabola

Differentiating the equation $y - 1 = Cx^2$ with respect to $x$, we get: $\frac{dy}{dx} = 2Cx.$ So, the slope of the tangent to the parabola at any point is $2Cx$.

Step 4: Orthogonal condition

Let the equation of the orthogonal trajectory be $F(x, y) = 0$. Its slope $\frac{dy}{dx}$ must satisfy the orthogonality condition: $\frac{dy}{dx} \cdot 2Cx = -1.$ Thus, $\frac{dy}{dx} = -\frac{1}{2Cx}.$

Step 5: Eliminate $C$

From the equation of the parabola $y - 1 = Cx^2$, we solve for $C$: $C = \frac{y - 1}{x^2}.$ Substitute this into $\frac{dy}{dx} = -\frac{1}{2Cx}$: $\frac{dy}{dx} = -\frac{1}{2 \cdot \frac{y - 1}{x^2} \cdot x}.$ Simplify: $\frac{dy}{dx} = -\frac{x^2}{2x(y - 1)} = -\frac{x}{2(y - 1)}.$

Step 6: Solve the differential equation

The differential equation for the orthogonal trajectories is: $\frac{dy}{dx} = -\frac{x}{2(y - 1)}.$ Separate the variables: $2(y - 1) \, dy = -x \, dx.$ Integrate both sides: $\int 2(y - 1) \, dy = \int -x \, dx.$ The left-hand side becomes: $\int 2(y - 1) \, dy = \int 2y \, dy - \int 2 \, dy = y^2 - 2y + C_1.$ The right-hand side becomes: $\int -x \, dx = -\frac{x^2}{2} + C_2.$ Equating both sides: $y^2 - 2y = -\frac{x^2}{2} + C,$ where $C = C_2 - C_1$ is an arbitrary constant.

Step 7: Final form of the family of curves

Rearranging the equation: $y^2 - 2y + \frac{x^2}{2} = C,$ which represents the family of curves that intersects each parabola $y - 1 = Cx^2$ at right angles.

Summary

The family of orthogonal trajectories is given by: $y^2 - 2y + \frac{x^2}{2} = C,$ where $C$ is an arbitrary constant.

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Related Questions:

1. What is the geometric significance of orthogonal trajectories?
2. How do you find orthogonal trajectories for other families of curves, such as circles or ellipses?
3. Could we solve this problem using a parametric approach instead of eliminating $C$?
4. What happens if the original family of parabolas had a different vertex position?
5. How can we confirm that these trajectories truly intersect the given parabolas at right angles?

Tip:

When solving for orthogonal trajectories, always focus on finding and eliminating the parameter that defines the original family of curves!