Solutions — Hyperbolas

1. Key features from equations. Fluency

   • (a) y = 5/x: Asymptotes: x = 0, y = 0. k = 5 > 0 ⇒ branches in Q1 and Q3. Domain: x ≠ 0. Range: y ≠ 0.
   • (b) y = −3/x: Asymptotes: x = 0, y = 0. k = −3 < 0 ⇒ branches in Q2 and Q4. Domain: x ≠ 0. Range: y ≠ 0.
   • (c) xy = 8 ⇒ y = 8/x: Rewrite as y = 8/x. k = 8 > 0. Asymptotes: x = 0, y = 0. Branches in Q1 and Q3.
   • (d) xy = −6 ⇒ y = −6/x: Rewrite as y = −6/x. k = −6 < 0. Asymptotes: x = 0, y = 0. Branches in Q2 and Q4.

2. y = 12/x — evaluate and solve. Fluency

   • (a) Find y: y(3) = 12/3 = 4. y(−4) = 12/(−4) = −3. y(1.5) = 12/1.5 = 8.
   • (b) Find x: 6 = 12/x ⇒ x = 2. −2 = 12/x ⇒ x = −6. 24 = 12/x ⇒ x = 0.5. x = 2, −6, 0.5 respectively.
   • (c) Does (3,3) lie on y=12/x?: Test: 12/3 = 4 ≠ 3. No, (3,3) does not lie on y = 12/x. (The point (3,4) does.)
   • (d) Point with x = 5: y = 12/5 = 2.4. Point: (5, 2.4).

3. Find k from a point. Fluency

   • (a) Through (3, 4): y = k/x: 4 = k/3 ⇒ k = 12. y = 12/x
   • (b) Through (−2, 7): 7 = k/(−2) ⇒ k = −14. y = −14/x
   • (c) Through (5, −3): −3 = k/5 ⇒ k = −15. y = −15/x
   • (d) Through (−6, −2): −2 = k/(−6) ⇒ k = 12. y = 12/x

4. Asymptotes of shifted hyperbolas. Fluency

   • (a) y = 2/(x−3)+1: Vertical: denominator=0 when x=3. Horizontal: y approaches 1 as x → ±∞. x = 3, y = 1
   • (b) y = −5/(x+2)−4: x+2=0 ⇒ x=−2. y approaches −4. x = −2, y = −4
   • (c) y = 3/x+7: x = 0 (denominator = 0). y approaches 7 as x → ±∞. x = 0, y = 7
   • (d) y = 4/(x−1): x−1 = 0 ⇒ x = 1. No vertical shift, so y approaches 0. x = 1, y = 0

5. Comparing hyperbolas. Understanding

   • (a) y=10/x vs y=1/x: y = 10/x has branches further from the origin — the curve is stretched outward. For example, at x = 1, y = 10 (vs y = 1). Same shape and asymptotes; y = 10/x is further from the origin.
   • (b) y=−4/x vs y=4/x: y = −4/x is a reflection of y = 4/x in the x-axis (or equivalently, the y-axis). Branches move from Q1/Q3 to Q2/Q4.
   • (c) Why (6,2) must be on the same curve: For any point (x,y) on y = k/x, we have xy = k. If (2,6) is on the curve then k = 2×6 = 12. Check (6,2): 6×2 = 12 = k ✓. The product xy = k is the same for both points.
   • (d) k when curve passes through (−3,−3): k = xy = (−3)(−3) = 9. Equation: y = 9/x.

6. Speed and travel time. Understanding

   • (a) Travel times: T(60)=240/60=4 h. T(80)=240/80=3 h. T(120)=240/120=2 h.
   • (b) Speed for 3 hours: 3 = 240/v ⇒ v = 240/3 = 80 km/h.
   • (c) Effect of doubling speed: If v doubles, T = 240/(2v) = (1/2)(240/v). Travel time halves. This is the defining feature of inverse proportion.
   • (d) Speed giving T = 0: T = 240/v = 0 has no solution (240/v can never equal 0 for any finite v). No — as v → ∞, T approaches 0 but never reaches it. Zero travel time would require infinite speed.

7. Read from the graph. Understanding

   • (a) Quadrants and sign of k: The branches lie in Quadrant 1 (positive x, positive y) and Quadrant 3 (negative x, negative y). Since both x and y have the same sign, k > 0 (k is positive).
   • (b) Asymptotes: The dashed red lines are the coordinate axes. The asymptotes are x = 0 (vertical, the y-axis) and y = 0 (horizontal, the x-axis).
   • (c) Find k from the labelled point: The point (2, 3) lies on the curve ⇒ k = x × y = 2 × 3 = 6.
   • (d) Equation: y = 6/x.

8. y = 6/(x−2)+3 — behaviour analysis. Understanding

   • (a) Asymptotes: x = 2 and y = 3.
   • (b) Values: y(5) = 6/3+3 = 2+3 = 5. y(0) = 6/(−2)+3 = −3+3 = 0. y(−1) = 6/(−3)+3 = −2+3 = 1.
   • (c) As x approaches 2 from right: As x → 2+, (x−2) → 0+, so 6/(x−2) → +∞, and y = 6/(x−2)+3 → +∞. The graph shoots up toward infinity near x = 2.
   • (d) As x becomes very large: As x → +∞, 6/(x−2) → 0, so y → 0+3 = 3. The graph approaches the horizontal asymptote y = 3.

9. Finding the equation. Problem Solving

   • (a) Equation form: Vertical asymptote x=4 ⇒ denominator is (x−4). Horizontal asymptote y=−1 ⇒ c=−1. y = k/(x−4) − 1.
   • (b) Find k from (6,4): 4 = k/(6−4) − 1 = k/2 − 1. So k/2 = 5 ⇒ k = 10.
   • (c) Full equation, y when x=5: y = 10/(x−4) − 1. y(5) = 10/1 − 1 = 10 − 1 = 9.

10. Gas pressure and volume (Boyle’s Law). Problem Solving

    • (a) Find k and equation: PV = k: 200 × 3 = 600. k = 600. Equation: P = 600/V.
    • (b) Pressures: P(2) = 600/2 = 300 kPa. P(4) = 600/4 = 150 kPa. P(6) = 600/6 = 100 kPa.
    • (c) Volume at P=150: 150 = 600/V ⇒ V = 600/150 = 4 litres.
    • (d) Volume halved ⇒ pressure?: If V becomes V/2, then P = 600/(V/2) = 2×(600/V) = 2P. Pressure doubles. This is Boyle’s Law: halving the volume doubles the pressure at constant temperature.