JEE Main 1998 Mathematics question on Conic Sections
If P=(x,y), F1=(3,0), F2=(−3,0) and 16x2+25y2=400, then PF1+PF2 equals
A.
8
B.
6
C.
10
D.
12
Show answer and solution
Correct option: C
Solution
The equation 16x² + 25y² = 400 simplifies to x²/25 + y²/16 = 1, an ellipse with a² = 25, b² = 16, so a = 5. Here c² = 25 - 16 = 9, giving c = 3, confirming F₁ = (3,0) and F₂ = (-3,0) are the foci. By the fundamental definition of an ellipse, the sum of distances from any point on it to the two foci equals 2a = 2×5 = 10.
[1998]
Practise Conic Sections the way it is asked
A timed set of real JEE Main Conic Sections questions, graded as the exam grades. No sign-up.
