AP PreCalculus Practice Quiz: Conic Sections
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 10 questions to check your progress.
Question 1 of 10
All Questions (10)
A) A circle is a type of hyperbola with equal radii.
B) An ellipse is a special case of a circle.
C) A circle is a special case of an ellipse where the horizontal and vertical radii are equal.
D) There is no analytical relationship between a circle and an ellipse.
Correct Answer: C
The provided content states that the equation for an ellipse is $\\frac{(x-h)^2}{a^2} + \\frac{(y-k)^2}{b^2} = 1$, where 'a' is the horizontal radius and 'b' is the vertical radius. It explicitly mentions that 'A circle is a special case where a=b.'
A) (3, 5)
B) (-3, 5)
C) (5, -3)
D) (2, 5)
Correct Answer: B
The standard form for a parabola with a vertical axis of symmetry is $y-k = a(x-h)^2$, where the vertex is $(h, k)$. In the given equation, $y-5 = 2(x-(-3))^2$, so $h = -3$ and $k = 5$. The vertex is (-3, 5).
A) (1, -8)
B) (16, 9)
C) (-1, 8)
D) (4, 3)
Correct Answer: C
The standard form for an ellipse is $\\frac{(x-h)^2}{a^2} + \\frac{(y-k)^2}{b^2} = 1$, with the center at $(h, k)$. In this equation, $(x-(-1))^2$ means $h = -1$, and $(y-8)^2$ means $k = 8$. Therefore, the center is (-1, 8).
A) $(x-2)^2 + (y+2)^2 = 4$
B) $y+4 = 3(x-1)^2$
C) $\\frac{(x-3)^2}{4} + \\frac{(y+1)^2}{9} = 1$
D) $\\frac{(y-5)^2}{16} - \\frac{(x-2)^2}{25} = 1$
Correct Answer: D
A hyperbola is characterized by the subtraction of the squared terms for x and y. Option A is a circle (a special ellipse), B is a parabola, and C is an ellipse (addition of terms). Option D has the form $\\frac{(y-k)^2}{a^2} - \\frac{(x-h)^2}{b^2} = 1$, which is a hyperbola.
A) Horizontal radius = 25, Vertical radius = 49
B) Horizontal radius = 5, Vertical radius = 7
C) Horizontal radius = 7, Vertical radius = 5
D) Horizontal radius = 2, Vertical radius = 3
Correct Answer: B
In the standard equation $\\frac{(x-h)^2}{a^2} + \\frac{(y-k)^2}{b^2} = 1$, 'a' is the horizontal radius and 'b' is the vertical radius. Here, $a^2 = 25$, so the horizontal radius $a=\\sqrt{25}=5$. And $b^2 = 49$, so the vertical radius $b=\\sqrt{49}=7$.
A) $\\frac{(y-1)^2}{9} - \\frac{(x+2)^2}{4} = 1$
B) $\\frac{(x+2)^2}{4} - \\frac{(y-1)^2}{9} = 1$
C) $\\frac{(x+2)^2}{4} + \\frac{(y-1)^2}{9} = 1$
D) $x+2 = 4(y-1)^2$
Correct Answer: B
A hyperbola opens horizontally when the $(x-h)^2$ term is positive. The form is $\\frac{(x-h)^2}{a^2} - \\frac{(y-k)^2}{b^2} = 1$. Option B matches this form. Option A opens vertically because the $(y-k)^2$ term is positive. Option C is an ellipse. Option D is a parabola that opens horizontally.
A) $y+1 = a(x-4)^2$
B) $y-4 = a(x+1)^2$
C) $x+1 = a(y-4)^2$
D) $x-4 = a(y+1)^2$
Correct Answer: D
A parabola with a horizontal axis of symmetry has the form $(x-h) = a(y-k)^2$. The vertex is $(h, k)$. Substituting the vertex $(4, -1)$ gives $h=4$ and $k=-1$, resulting in the equation form $x-4 = a(y-(-1))^2$, which simplifies to $x-4 = a(y+1)^2$.
A) $\\frac{(x+3)^2}{4} + \\frac{y^2}{36} = 1$
B) $\\frac{(x-3)^2}{4} + \\frac{y^2}{36} = 1$
C) $\\frac{(x+3)^2}{2} + \\frac{y^2}{6} = 1$
D) $\\frac{(x+3)^2}{36} + \\frac{y^2}{4} = 1$
Correct Answer: A
The standard form is $\\frac{(x-h)^2}{a^2} + \\frac{(y-k)^2}{b^2} = 1$. The center $(h, k)$ is $(-3, 0)$. The horizontal radius $a=2$, so $a^2=4$. The vertical radius $b=6$, so $b^2=36$. Substituting these values gives $\\frac{(x-(-3))^2}{4} + \\frac{(y-0)^2}{36} = 1$, which is $\\frac{(x+3)^2}{4} + \\frac{y^2}{36} = 1$.
A) $\\frac{y^2}{a^2} - \\frac{x^2}{b^2} = 1$
B) $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$
C) $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$
D) $y-k = a(x-h)^2$
Correct Answer: C
A hyperbola with a vertical line of symmetry opens horizontally (left and right). Its equation is of the form $\\frac{(x-h)^2}{a^2} - \\frac{(y-k)^2}{b^2} = 1$. Since the center is $(0,0)$, this becomes $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$.
A) An ellipse centered at (1, -2).
B) A hyperbola centered at (1, -2).
C) An ellipse centered at (-1, 2).
D) A hyperbola centered at (-1, 2).
Correct Answer: B
To identify the conic, we must put the equation in standard form by making the right side equal to 1. Divide the entire equation by 36: $\\frac{4(x-1)^2}{36} - \\frac{9(y+2)^2}{36} = \\frac{36}{36}$. This simplifies to $\\frac{(x-1)^2}{9} - \\frac{(y+2)^2}{4} = 1$. This is the standard form of a hyperbola, $\\frac{(x-h)^2}{a^2} - \\frac{(y-k)^2}{b^2} = 1$, with a center at $(h, k) = (1, -2)$.