Let and be two fixed points, then the locus of a point such that the perimeter of is 4, is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The fixed side OA = 1. Let P = (x, y). Then OP + AP = 4 - 1 = 3, so the locus is an ellipse with foci O(0,0) and A(0,1), and sum of distances = 3.
For this ellipse: 2a = 3, so a = 3/2. The distance between foci = 1, so 2c = 1, c = 1/2. Then b² = a² - c² = 9/4 - 1/4 = 2.
The center is the midpoint of O and A = (0, 1/2). With major axis along y-axis, the ellipse equation is:
x²/b² + (y - 1/2)²/a² = 1
x²/2 + (y - 1/2)²/(9/4) = 1
Multiply through by 9: (9x²)/2 + 4(y - 1/2)² = 9... multiply the original by LCM 9/2... better to just expand:
x²/2 + (y - 1/2)²/(9/4) = 1
Multiply everything by 36: 18x² + 16(y - 1/2)² = 36
→ 18x² + 16(y² - y + 1/4) = 36
→ 18x² + 16y² - 16y + 4 = 36
→ 18x² + 16y² - 16y = 32
Divide by 2: **9x² + 8y² - 8
For this ellipse: 2a = 3, so a = 3/2. The distance between foci = 1, so 2c = 1, c = 1/2. Then b² = a² - c² = 9/4 - 1/4 = 2.
The center is the midpoint of O and A = (0, 1/2). With major axis along y-axis, the ellipse equation is:
x²/b² + (y - 1/2)²/a² = 1
x²/2 + (y - 1/2)²/(9/4) = 1
Multiply through by 9: (9x²)/2 + 4(y - 1/2)² = 9... multiply the original by LCM 9/2... better to just expand:
x²/2 + (y - 1/2)²/(9/4) = 1
Multiply everything by 36: 18x² + 16(y - 1/2)² = 36
→ 18x² + 16(y² - y + 1/4) = 36
→ 18x² + 16y² - 16y + 4 = 36
→ 18x² + 16y² - 16y = 32
Divide by 2: **9x² + 8y² - 8
[JEE Main 2019 · 8 Apr · Shift II]