Ellipse
Geometry Optimization
JEE Advanced 2013 Paper 1
Grade 11
Question:
A vertical line through $h$ intersects the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$ at $P, Q$ and the tangents to the ellipse at $P, Q$ meet at $R$. Let $\Delta(h) = \text{area of } \triangle PQR$, $\Delta_1 = \max_{1/2 \leq h \leq 1} \Delta(h)$, $\Delta_2 = \min_{1/2 \leq h \leq 1} \Delta(h)$. Find $\frac{8}{\sqrt{5}} \Delta_1 - 8 \Delta_2$.
Step-by-Step Solution
Key Concept: $P = (h, 3(1 - h^2/4)^{1/2}/\sqrt{3})$; chord $PQ$ is vertical. Tangent at $P$: $xh/4 + yy_P/3 = 1$ meets tangent at $Q$ at $R = (4/h, 0)$. $|PQ| = 2\sqrt{3(1 - h^2/4)}$; height from $R = \text{horizontal distance} = 4/h - h$. $\Delta(h) = 3(1 - h^2/4)(4/h - h)$. Evaluate at endpoints.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: $8\sqrt{5}$