Ellipse
Sum of focal distances
Grade 11
Question:
<p>A coplanar beam of light emerging from a point source have the equation \(lx - y + 2(1 + l) = 0\), \(l \in \mathbb{R}\); the rays of the beam strike an elliptical surface and get reflected inside the ellipse. The reflected rays form another convergent beam having the equation \(mx - y + 2(1 - m) = 0\), \(m \in \mathbb{R}\). Further it is found that the foot of the perpendicular from the point (2, 2) upon any tangent to the ellipse lies on the circle \(x^2 + y^2 - 4y - 5 = 0\). The least value of total distance travelled by an incident ray and the corresponding reflected ray is equal to:</p>
<p>(a) 6</p>
<p>(b) 3</p>
<p>(c) \(\sqrt{5}\)</p>
<p>(d) \(2\sqrt{5}\)</p>
Step-by-Step Solution
Key Concept: The total distance from one focus to any point on an ellipse to the other focus is always equal to \(2a\), the length of the major axis.
<p>By the reflection property of an ellipse, if a ray from one focus reflects off the ellipse, it passes through the other focus. The total distance travelled by the ray from one focus to a point on the ellipse and then to the other focus is constant and equals \(2a\). For the given ellipse with semi-major axis \(a = 3\), the total distance is \(2a = 6\). This is the same for all incident rays and their corresponding reflections.</p>
Correct Answer: A