Free Gaokao Analytic Geometry (Conic Sections) Questions and Answers — Questions and Answers
Question 1: An ellipse has foci at F1(-1, 0) and F2(1, 0). The perimeter of a triangle formed by a point P on the ellipse and the two foci is 10. What is the standard equation of the ellipse?
- x²/25 + y²/24 = 1
- x²/16 + y²/15 = 1 (Correct answer)
- x²/10 + y²/9 = 1
- x²/16 + y²/17 = 1
Correct answer: x²/16 + y²/15 = 1
The perimeter of ΔPF1F2 is PF1 + PF2 + |F1F2|. By the definition of an ellipse, PF1 + PF2 = 2a. The distance between the foci, |F1F2|, is 2c. Here, 2c = 2, so c=1. The perimeter is 2a + 2c = 10, so 2a + 2 = 10, which gives 2a = 8 and a = 4. For an ellipse, b² = a² - c² = 4² - 1² = 15. The standard equation is x²/a² + y²/b² = 1.
Question 2: The line y = kx + 1 intersects the parabola y² = 4x at exactly one point. What is the value of k (where k ≠ 0)?
- k = 2
- k = -1
- k = 1/2
- k = 1 (Correct answer)
Correct answer: k = 1
Substitute the line equation into the parabola equation: (kx + 1)² = 4x. This expands to k²x² + 2kx + 1 = 4x, or k²x² + (2k - 4)x + 1 = 0. For the line to be tangent (intersect at one point), the discriminant (Δ) of this quadratic equation must be zero. Δ = b² - 4ac = (2k - 4)² - 4(k²)(1) = 0. Expanding this gives 4k² - 16k + 16 - 4k² = 0, which simplifies to -16k + 16 = 0, so k = 1.
Question 3: What are the equations of the asymptotes for the hyperbola x²/9 - y²/16 = 1?
- y = ±(3/4)x
- y = ±(9/16)x
- y = ±(4/3)x (Correct answer)
- y = ±(16/9)x
Correct answer: y = ±(4/3)x
For a hyperbola of the form x²/a² - y²/b² = 1, the equations of the asymptotes are y = ±(b/a)x. In this case, a² = 9, so a = 3, and b² = 16, so b = 4. Therefore, the asymptotes are y = ±(4/3)x. This can be rewritten as 3y = ±4x, or 4x ± 3y = 0.
Question 4: The distance between the two foci of an ellipse is 8, and its eccentricity is 4/5. What is the length of the major axis?
- 5
- 8
- 10 (Correct answer)
- 12.5
Correct answer: 10
The distance between the foci is 2c, so 2c = 8, which means c = 4. The eccentricity e is defined as c/a. We are given e = 4/5. So, c/a = 4/a = 4/5. Solving for a gives a = 5. The length of the major axis is 2a, which is 2 * 5 = 10.
Question 5: A point M is on the parabola x² = 8y. If the distance from M to the x-axis is 5, what is the distance from M to the focus F of the parabola?
- 3
- 5
- 7 (Correct answer)
- 8
Correct answer: 7
The equation is of the form x² = 2py, so 2p = 8, and p = 4. The focus F is at (0, p/2) = (0, 2), and the directrix is the line y = -p/2 = -2. The distance from M to the x-axis is its y-coordinate, so M has coordinates (x, 5). By the definition of a parabola, the distance from any point on it to the focus is equal to its distance to the directrix. The distance from M(x, 5) to the directrix y = -2 is |5 - (-2)| = 7.
Question 6: A hyperbola has the equation x²/a² - y²/12 = 1. If one of its foci is at (4, 0), what is the value of a?
- 4
- 2 (Correct answer)
- √28
- 16
Correct answer: 2
For a hyperbola, the relationship between a, b, and c is c² = a² + b². We are given that a focus is at (4, 0), so c = 4. We are also given b² = 12. Plugging these values into the formula: 4² = a² + 12. This gives 16 = a² + 12, so a² = 4. Since a must be positive, a = 2.
Free Gaokao Analytic Geometry (Conic Sections) Questions and Answers
An ellipse has foci at F1(-1, 0) and F2(1, 0).
The perimeter of a triangle formed by a point P on the ellipse and the two foci is 10.
What is the standard equation of the ellipse?