For the hyperbola x²/16 − y²/9 = 1, what is the distance from the center to each focus?
Ac = √(16 − 9) = √7 ≈ 2.65
Bc = √(16 + 9) = 5
Cc = √16 = 4
Dc = √9 = 3
For a hyperbola, c² = a² + b², so c = √(16 + 9) = √25 = 5. The most tempting wrong answer is √7, using the ellipse formula c² = a² − b². The distinction matters: for an ellipse, foci are inside the curve (c < a), so subtracting b² shrinks c; for a hyperbola, foci are outside the curve (c > a), so adding b² enlarges c. If you get c < a for a hyperbola, you have used the wrong formula.
Question 2 Multiple Choice
Consider the hyperbola (y − 3)²/4 − (x + 1)²/25 = 1. A student claims a = 5 because 'a is always the larger denominator.' What is wrong with this reasoning?
ANothing — a is always the larger denominator in all conic sections
BFor hyperbolas, a is the denominator under the positive term; here a² = 4 so a = 2, regardless of which is larger
CThe formula only applies to hyperbolas centered at the origin; this one is shifted
Da and b switch definitions when the transverse axis is vertical
For hyperbolas, 'a' is defined as the denominator under the *positive* term and corresponds to the semi-transverse axis. In this equation the positive term is (y−3)²/4, so a² = 4 and a = 2. The value b² = 25 is larger, but that does not make b into a. This contrasts with ellipses, where a conventionally denotes the larger semi-axis. For hyperbolas, there is no requirement that a > b.
Question 3 True / False
For any hyperbola, the foci are located farther from the center than the vertices.
TTrue
FFalse
Answer: True
Vertices are a units from center; foci are c units from center. Since c² = a² + b² and b² > 0, we have c² > a², so c > a. The foci always lie beyond the vertices along the transverse axis. This is the opposite of ellipses, where c < a and foci lie between the center and the vertices — a useful contrast to remember.
Question 4 True / False
The asymptotes of a hyperbola pass through the vertices of the curve.
TTrue
FFalse
Answer: False
Asymptotes pass through the *center* of the hyperbola, not the vertices. The branches curve away from the vertices and approach the asymptotes only as they extend toward infinity — they never touch or cross the asymptotes at any point. At the vertices (the closest points on the branches to the center), the branches are actually farthest from the asymptotes. The 'box method' makes this clear: asymptotes are diagonals through the box's corners at (±a, ±b), while vertices sit on the sides of the box at (±a, 0) or (0, ±a).
Question 5 Short Answer
Why does a hyperbola produce two separate branches while an ellipse is a single closed curve? Connect the defining distance condition to the geometric difference.
Think about your answer, then reveal below.
Model answer: An ellipse is defined by a constant *sum* of distances: d₁ + d₂ = 2a. Since both distances are positive, any satisfying point lies in a bounded region pulling toward the oval between the foci. A hyperbola uses a constant *absolute difference*: |d₁ − d₂| = 2a. This condition can be satisfied in two distinct ways — d₁ − d₂ = 2a (points substantially closer to one focus, forming one branch) and d₂ − d₁ = 2a (points substantially closer to the other focus, forming the second branch). The absolute difference allows solutions on both sides of the two foci, producing two open outward-facing branches rather than one closed curve.
The algebraic switch from sum to absolute difference is the single change that produces the dramatic visual difference. The sum condition bounds points to a closed oval; the difference condition permits points anywhere that maintain the correct distance differential, which happens on two separate sides of the focal pair.