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Conic sections — ACT practice questions

Conic sections on the ACT Mathematics test cover the graphs and standard equations of circles, ellipses, parabolas, and hyperbolas in the coordinate plane. A student must read an equation in standard form and pull out features such as the center, vertices, foci, directrices, and the slopes of a hyperbola's asymptotes. Items usually give a completed-square equation and ask which choice lists those features, the coordinates of the foci, or an equation of a parabola's directrix. Common traps include swapping a and b, mixing horizontal and vertical openings, and treating a translated graph as if it were centered at the origin.

  • Section: Mathematics 45 questions on the paper
  • 35 practice questions in the app

Sample questions

Question 1

Which of the following gives the center, vertices, and asymptote slopes of (x+1)216−(y−3)29=1\dfrac{(x+1)^2}{16}-\dfrac{(y-3)^2}{9}=1?

  1. ACenter (−1,3)(-1,3); vertices (−5,3)(-5,3) and (3,3)(3,3); slopes ±43\pm\dfrac43
  2. BCenter (1,−3)(1,-3); vertices (−3,−3)(-3,-3) and (5,−3)(5,-3); slopes ±34\pm\dfrac34
  3. CCenter (−1,3)(-1,3); vertices (−4,3)(-4,3) and (2,3)(2,3); slopes ±43\pm\dfrac43
  4. DCenter (−1,3)(-1,3); vertices (−5,3)(-5,3) and (3,3)(3,3); slopes ±34\pm\dfrac34
Show answer

Answer: D — Center (−1,3)(-1,3); vertices (−5,3)(-5,3) and (3,3)(3,3); slopes ±34\pm\dfrac34

The center is (−1,3)(-1,3), and the positive xx term gives a horizontal transverse axis with a=4a=4. Thus, the vertices are (−5,3)(-5,3) and (3,3)(3,3) and the asymptote slopes are ±b/a=±34\pm b/a=\pm\dfrac34.

Question 2

The parabola y2=12xy^2=12x is graphed in the standard (x,y)(x,y) coordinate plane. Which of the following is an equation of its directrix?

  1. Ax=−3x=-3
  2. Bx=3x=3
  3. Cy=−3y=-3
  4. Dy=3y=3
Show answer

Answer: A — x=−3x=-3

For y2=4pxy^2=4px with 4p=124p=12, p=3p=3. The parabola has vertex (0,0)(0,0) and opens right, so its directrix is the vertical line x=−3x=-3.

Question 3

In the standard (x,y)(x,y) coordinate plane, what are the coordinates of the foci of the hyperbola x216−y29=1\dfrac{x^2}{16}-\dfrac{y^2}{9}=1?

  1. A(±5,0)(\pm5,0)
  2. B(±4,0)(\pm4,0)
  3. C(±3,0)(\pm3,0)
  4. D(0,±5)(0,\pm5)
Show answer

Answer: A — (±5,0)(\pm5,0)

For x216−y29=1\dfrac{x^2}{16}-\dfrac{y^2}{9}=1, c2=a2+b2=16+9=25c^2=a^2+b^2=16+9=25, so c=5c=5. The transverse axis is horizontal, so the foci are (±5,0)(\pm5,0).

Practice 35 Conic sections questions in the app

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