Hyperbola – Study Notes
Manipal MET – Mathematics
1. Simple Explanation
A hyperbola is one of the four conic sections (the others being ellipse, parabola, and circle).
It is the set of all points in the plane for which the difference of the distances to two fixed points (the foci) is constant.
Geometric definition
For any point (P) on a hyperbola,
[ |PF_1 - PF_2| = 2a ]
where (F_1) and (F_2) are the foci and (2a) is a positive constant.
The hyperbola consists of two symmetric branches opening away from the foci.
2. Key Points
| Topic | Details |
|---|---|
| Standard Forms | 1. Horizontal transverse axis: (\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1) 2. Vertical transverse axis: (\displaystyle \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2}=1) |
| Vertices | Points where the hyperbola intersects the transverse axis: Horizontal: ((h\pm a, k)) Vertical: ((h, k\pm a)) |
| Foci | ((h\pm c, k)) or ((h, k\pm c)) where (c = \sqrt{a^2 + b^2}) |
| Asymptotes | Straight lines that the hyperbola approaches as (x\to\pm\infty): Horizontal: (y-k = \pm \frac{b}{a} (x-h)) Vertical: (x-h = \pm \frac{b}{a} (y-k)) |
| Eccentricity | (e = \dfrac{c}{a} > 1) |
| Area of a branch | Infinite (unbounded). |
| Equation from geometric data | 1. Find centre (C) as midpoint of foci. 2. Compute (a = \frac{1}{2}) of the constant difference. 3. Compute (b^2 = c^2 - a^2). 4. Write standard form. |
| Special cases | - Rectangular hyperbola: (a = b) → asymptotes are perpendicular. - Degenerate hyperbola: line pair when (b^2 = 0). |
| Transformation | Rotation, translation, scaling preserve hyperbola type. |
3. Worked Example
Problem
Find the equation of the hyperbola whose foci are at (F_1(0,-4)) and (F_2(6,2)) and which passes through the point (P(2,0)).
Step 1: Find the centre
The centre is the midpoint of the foci: [ C!\left(\frac{0+6}{2},,\frac{-4+2}{2}\right) = C(3,-1) ]
Step 2: Compute (c) – distance from centre to a focus
[ c = \sqrt{(6-3)^2 + (2-(-1))^2} = \sqrt{3^2 + 3^2} = \sqrt{18}=3\sqrt{2} ]
Step 3: Determine the constant difference (2a)
For any point (P) on the hyperbola:
[
|PF_1 - PF_2| = 2a
]
Compute distances:
[
PF_1 = \sqrt{(2-0)^2 + (0+4)^2} = \sqrt{4+16}= \sqrt{20}=2\sqrt5
]
[
PF_2 = \sqrt{(2-6)^2 + (0-2)^2} = \sqrt{(-4)^2+(-2)^2} = \sqrt{16+4}=\sqrt{20}=2\sqrt5
]
So (PF_1 = PF_2), thus (|PF_1-PF_2| = 0).
This cannot be a hyperbola (difference must be non‑zero).
Conclusion: The point (P(2,0)) does not lie on any hyperbola with those foci.
(If the problem intended the sum of distances to be constant, it would be an ellipse. Be careful to read “difference” vs “sum”.)
Corrected Example – Let’s use a proper hyperbola:
New Problem
Find the equation of the hyperbola with foci (F_1(-2,0)) and (F_2(2,0)) and passing through (P(3,4)).
- Centre (C = (0,0)).
- c = distance from centre to a focus = 2.
- Constant difference (2a = |PF_1 - PF_2|).
[ PF_1 = \sqrt{(3+2)^2 + 4^2} = \sqrt{25+16}= \sqrt{41} ] [ PF_2 = \sqrt{(3-2)^2 + 4^2} = \sqrt{1+16}= \sqrt{17} ] [ 2a = |\sqrt{41}-\sqrt{17}| = \sqrt{41}-\sqrt{17} ] So (a = \frac{\sqrt{41}-\sqrt{17}}{2}) (≈ 1.12). - Compute (b^2 = c^2 - a^2)
[ c^2 = 4,\quad a^2 \approx 1.25 ; \Rightarrow; b^2 \approx 4-1.25