Ellipse
Focal Distance Property
IIT-JEE 1998
Grade 11
Question:
If $P = (x, y)$, $F_1 = (3, 0)$, $F_2 = (-3, 0)$ and $16x^2 + 25y^2 = 400$, then $PF_1 + PF_2$ equals:
(1) $8$
(2) $10$
(3) $6$
(4) $12$
Step-by-Step Solution
Key Concept: Rewrite as $x^2/25 + y^2/16 = 1$; $a = 5$. The foci are $(\pm 3, 0)$ confirming $c = 3$. By the focal-distance property, $PF_1 + PF_2 = 2a = 10$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)