Hyperbola
Geometry of Hyperbola
IIT-JEE 2008 Paper 2
Grade 11
Question:
Consider a branch of the hyperbola $x^2 - 2y^2 - 2\sqrt{2} x - 4\sqrt{2} y - 6 = 0$ with vertex at point $A$. Let $B$ be one endpoint of its latus rectum and $C$ be the focus nearest to $A$. The area of triangle $ABC$ is:
(1) $1 - \sqrt{\frac{2}{3}}$
(2) $\sqrt{\frac{3}{2}} - 1$
(3) $1 + \sqrt{\frac{2}{3}}$
(4) $\sqrt{\frac{3}{2}} + 1$
Step-by-Step Solution
Key Concept: Complete the square: $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 8 \Rightarrow \frac{(x - \sqrt{2})^2}{8} - \frac{(y + \sqrt{2})^2}{4} = 1$. So $a^2 = 8$, $b^2 = 4, e = \sqrt{3/2}$. Vertex $A = (\sqrt{2} - 2\sqrt{2}, -\sqrt{2})$; compute $B$ and $C$ and find area.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)