JEE Main 2023 · 25 January · morning · Single correct · +4 / −1
Let z₁=2+3i and z₂=3+4i. The set S = { z ∈ ℂ:| z - z₁ |² - | z - z₂ |² = | z₁ - z₂ |² } represents a
- (a) hyperbola with the length of the transverse axis 7
- (b) hyperbola with eccentricity 2
- (c) straight line with the sum of its intercepts on the coordinate axes equals -18
- (d) straight line with the sum of its intercepts on the coordinate axes equals 14
Answer and solution
Answer (d) straight line with the sum of its intercepts on the coordinate axes equals 14
Solution
Maths sir · 4 steps
Squared distances cancel the x² and y² terms, so what is left is only a straight line.
- Use coordinates. Put z = x + iy. Then |z-z₁|² = (x-2)² + (y-3)² and |z-z₂|² = (x-3)² + (y-4)². Also |z₁ - z₂|² = 1² + 1² = 2.
- Subtract. The x² and y² terms cancel. Using a² - b² = (a-b)(a+b): (x-2)²-(x-3)² = 2x-5 and (y-3)²-(y-4)² = 2y-7.
(2x-5)+(2y-7) = 2 ⇒ 2x+2y = 14 ⇒ x + y = 7
- Read the intercepts. The line x + y = 7 meets the x-axis at x = 7 and the y-axis at y = 7. As a check, z₂ = (3, 4) lies on it. By Pythagoras the angle at z₂ is 90°, so this is the line through z₂ perpendicular to z₁ z₂.
- So. S is a straight line whose intercepts add up to 7 + 7 = 14, which is option (d).
The idea to keep. A difference of distances equal to a constant gives a hyperbola. A difference of squared distances equal to a constant always gives a straight line, because the square terms cancel.
Answer from the NTA final answer key. Solution by PrepWiser.