Question:

<p><strong>Step 1:</strong> Consider an ellipse with equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The normal at any point $(a\cos\theta, b\sin\theta)$ on the ellipse has slope $\frac{a\sin\theta}{b\cos\theta}$.</p><p><strong>Step 2:</strong> The equation of the normal at point $(a\cos\theta, b\sin\theta)$ is: $\frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2$</p><p><strong>Step 3:</strong> For this normal to pass through a given external point $P(h, k)$, we substitute: $\frac{ah}{\cos\theta} - \frac{bk}{\sin\theta} = a^2 - b^2$</p><p><strong>Step 4:</strong> Let $t = \tan(\theta/2)$. Using the Weierstrass substitution $\cos\theta = \frac{1-t^2}{1+t^2}$ and $\sin\theta = \frac{2t}{1+t^2}$, the normal condition becomes a polynomial equation in $t$.</p><p><strong>Step 5:</strong> After substitution and simplification, this yields a quartic (degree 4) polynomial equation in $t$: $At^4 + Bt^3 + Ct^2 + Dt + E = 0$</p><p><strong>Step 6:</strong> A polynomial of degree 4 can have at most 4 real roots, which corresponds to at most 4 distinct values of $\theta$, and hence at most 4 normals to the ellipse passing through point $P$.</p><p><strong>Step 7:</strong> Similarly, for a hyperbola, the normal equation also reduces to a degree 4 polynomial equation, yielding the same maximum.</p><p><strong>∴ Answer:</strong> 4</p>