<p>The number of normals that can be drawn to an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) from an exterior point is ___ in general.</p>
Step-by-Step Solution
Key Concept: A normal to an ellipse at point (a cos θ, b sin θ) has slope b²tan(θ)/a² perpendicular to the tangent. From an exterior point, we can draw multiple normals by solving the normal equation, which is a polynomial of degree 4 in the parameter, yielding at most 4 real solutions.
<p><strong>Step 1:</strong> The equation of normal to ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a\cos\theta, b\sin\theta)$ is:</p><p>$$\frac{a^2 x}{a\cos\theta} - \frac{b^2 y}{b\sin\theta} = a^2 - b^2$$</p><p><strong>Step 2:</strong> For a normal to pass through exterior point $(h, k)$, we substitute and get:</p><p>$$\frac{a^2 h}{\cos\theta} - \frac{b^2 k}{\sin\theta} = a^2 - b^2$$</p><p><strong>Step 3:</strong> This equation, when manipulated (letting $t = \tan(\theta/2)$), reduces to a polynomial equation of degree 4 in the parameter.</p><p><strong>Step 4:</strong> A quartic equation can have at most 4 real roots, and for a general exterior point, all 4 roots are real and distinct, giving 4 distinct normal lines.</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4