Gaokao Exam Analytic Geometry (Conic Sections) 2 — Questions and Answers

Question 1: The equation of an ellipse is x²/25 + y²/16 = 1. What is the distance between its two foci?

6 (Correct answer)
8
10
3

Correct answer: 6

For x²/a² + y²/b² = 1 with a > b, c² = a² - b² = 25 - 16 = 9, so c = 3. The distance between foci is 2c = 6.

In the standard ellipse equation x²/a² + y²/b² = 1 where a > b > 0, the foci are located at (±c, 0) where c² = a² - b². Here a² = 25 and b² = 16, so c² = 9 and c = 3. The total distance between the two foci is 2c = 6.

Question 2: A parabola has the equation y² = 8x. What are the coordinates of its focus?

(2, 0) (Correct answer)
(0, 2)
(4, 0)
(-2, 0)

Correct answer: (2, 0)

The standard form y² = 4px gives focus at (p, 0). Here 4p = 8 so p = 2, giving focus (2, 0).

For a parabola in the form y² = 4px, the focus is at (p, 0) and the directrix is x = -p. Comparing y² = 8x with y² = 4px gives 4p = 8, so p = 2. Therefore the focus is at (2, 0).

Question 3: Which of the following is the equation of a hyperbola with real axis along the y-axis?

y²/9 - x²/4 = 1 (Correct answer)
x²/9 - y²/4 = 1
x²/4 + y²/9 = 1
y²/9 + x²/4 = 1

Correct answer: y²/9 - x²/4 = 1

A hyperbola with real axis along the y-axis has the form y²/a² - x²/b² = 1. Option A matches this form.

A hyperbola's standard forms are: x²/a² - y²/b² = 1 (transverse axis along x-axis) and y²/a² - x²/b² = 1 (transverse axis along y-axis). Option A, y²/9 - x²/4 = 1, has the y² term positive, so the real (transverse) axis lies along the y-axis.

Question 4: If the eccentricity of an ellipse is 1/2, and one semi-axis is a = 4, what is the length of the other semi-axis b?

2√3 (Correct answer)
2
√15
3

Correct answer: 2√3

Eccentricity e = c/a = 1/2, so c = 2. Then b² = a² - c² = 16 - 4 = 12, giving b = 2√3.

The eccentricity of an ellipse is e = c/a. With e = 1/2 and a = 4, we get c = 2. Using the relationship b² = a² - c², we calculate b² = 16 - 4 = 12, so b = 2√3 ≈ 3.46.

Question 5: A circle passes through the origin and has center at (3, 4). What is the equation of this circle?

(x-3)² + (y-4)² = 25 (Correct answer)
(x-3)² + (y-4)² = 5
x² + y² = 25
(x+3)² + (y+4)² = 25

Correct answer: (x-3)² + (y-4)² = 25

The radius is the distance from center (3,4) to origin (0,0): r = √(9+16) = 5. Equation: (x-3)² + (y-4)² = 25.

The radius r is found by the distance formula between center (3, 4) and point (0, 0): r = √((3-0)² + (4-0)²) = √(9 + 16) = √25 = 5. The circle's equation in standard form is (x-3)² + (y-4)² = r² = 25.

Question 6: The asymptotes of the hyperbola x²/4 - y²/9 = 1 are:

y = ±(3/2)x (Correct answer)
y = ±(2/3)x
y = ±(4/9)x
y = ±2x

Correct answer: y = ±(3/2)x

For x²/a² - y²/b² = 1, asymptotes are y = ±(b/a)x. Here a = 2, b = 3, so y = ±(3/2)x.

For the standard hyperbola x²/a² - y²/b² = 1, the asymptotes are the lines y = ±(b/a)x. With a² = 4 (a = 2) and b² = 9 (b = 3), the asymptotes are y = ±(3/2)x. These are the lines the hyperbola approaches but never crosses.

Gaokao Exam Analytic Geometry (Conic Sections) 2

The equation of an ellipse is x²/25 + y²/16 = 1.
What is the distance between its two foci?