Conic Sections
Conic Section
Allen Star Batch
Grade 11
Question:
The straight line $\frac{x}{4} + \frac{y}{3} = 1$ intersects the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ at two points $A$ and $B$, there is a point $P$ on this ellipse such that the area of $\triangle PAB$ is equal to $6\left(\sqrt{2} - 1\right)$. Then the number of such points $P$ is ______.
Step-by-Step Solution
Key Concept: Perpendicular distance from a point to a line equals the product condition on the ellipse parameters.
The perpendicular distance from point $P$ to line $h$ is found using $\frac{1}{2} \times h = 6(\sqrt{2}-1)$, where $AP \cdot AB = 6(\sqrt{2}-1)$. Solving for $h$ gives $h = \frac{12(\sqrt{2}-1)}{5}$. The distance from line $AB$ (equation $4y + 3x = 12\sqrt{2}$) is $\frac{12(\sqrt{2}-1)}{5}$, and there are exactly three such points satisfying the given conditions.
Correct Answer: 3