Question:

<p><strong>Step 1:</strong> Write the general equation of an ellipse (conic section form):</p><p>$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$</p><p>where $a < b$ and $h^2 - ab < 0$ for an ellipse.</p><p><strong>Step 2:</strong> Since the ellipse passes through (0, 3), substitute this point:</p><p>$$a(0)^2 + 2h(0)(3) + b(3)^2 + 2g(0) + 2f(3) + c = 0$$</p><p>$$9b + 6f + c = 0 \quad \text{...(1)}$$</p><p><strong>Step 3:</strong> Differentiate the general conic equation with respect to $x$:</p><p>$$2ax + 2hy + 2hx\frac{dy}{dx} + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0$$</p><p><strong>Step 4:</strong> At point (0, 3), this becomes:</p><p>$$0 + 6h + 0 + 6b\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0$$</p><p>$$6h + 2g + (6b + 2f)\frac{dy}{dx} = 0 \quad \text{...(2)}$$</p><p><strong>Step 5:</strong> Eliminate parameters using equations (1) and (2) along with the original constraint to get the differential equation of the family:</p><p>$$3x\frac{dy}{dx} + y = 3\frac{dy}{dx}$$</p><p>or equivalently: $$xy' - y = 3y'$$ or $$(x-3)y' = y$$</p><p>∴ Answer: A</p>