2. Find a family of curves such that each of its members intersects every parabola in the family y-1=Cx^2 at right angles.

Here's how to find the orthogonal family of curves:<br /><br />1. **Find the differential equation representing the given family of parabolas:**<br /><br />The given family is `y - 1 = Cx²`. We need to eliminate the arbitrary constant 'C'. Differentiating both sides with respect to 'x', we get:<br /><br />dy/dx = 2Cx<br /><br />From the original equation, C = (y - 1)/x². Substituting this into the derivative:<br /><br />dy/dx = 2[(y - 1)/x²] * x<br />dy/dx = 2(y - 1)/x<br /><br />This is the differential equation representing the given family of parabolas.<br /><br />2. **Find the differential equation of the orthogonal family:**<br /><br />For two curves to be orthogonal (intersect at right angles), the product of their slopes at the point of intersection must be -1. Let the slope of the orthogonal family be dy'/dx. Then:<br /><br />(dy/dx) * (dy'/dx) = -1<br /><br />Substitute the previously found dy/dx:<br /><br />[2(y - 1)/x] * (dy'/dx) = -1<br />dy'/dx = -x / [2(y - 1)]<br /><br />This is the differential equation representing the orthogonal family of curves.<br /><br />3. **Solve the differential equation to find the orthogonal family:**<br /><br />This is a separable differential equation. Rearranging:<br /><br />2(y - 1) dy' = -x dx<br /><br />Integrating both sides:<br /><br />∫ 2(y - 1) dy = ∫ -x dx<br />(y - 1)² = -x²/2 + k (where 'k' is the constant of integration)<br /><br />Therefore, the family of curves orthogonal to the given parabolas is:<br /><br />(y - 1)² = -x²/2 + k<br /><br />or equivalently<br /><br />x² + 2(y - 1)² = 2k<br /><br />Since 'k' is an arbitrary constant, we can replace 2k with another constant, say 'c':<br /><br />x² + 2(y - 1)² = c<br /><br />This represents a family of ellipses centered at (0, 1).<br /><br /><br />Therefore, the family of ellipses x² + 2(y - 1)² = c intersects each parabola of the family y - 1 = Cx² at right angles.<br />