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Conic Sections practice questions — SAT Math

18 free multiple-choice problems on conic sections, ordered to match SAT Math difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

Try 3 problems from this chapter

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Problem #0186 International
Mediumanalytic-geometry
For the hyperbola , the asymptotes have equations:

Problems & worked solutions

Problem #0186 International

Problem 1 — Conic Sections

For the hyperbola x29−y216=1, the asymptotes have equations:

Show answer & worked solution
  1. A. y=±x
  2. B. y=±3x/4
  3. C. y=±4x/3✓ correct
  4. D. y=±5x/3

a=3, b=4, so asymptotes: y=±43x.

Problem #0188 International

Problem 2 — Conic Sections

The parabola y2=8x has focus at:

Show answer & worked solution
  1. A. (0,2)
  2. B. (2,2)
  3. C. (2,0)✓ correct
  4. D. (8,0)

4p=8⇒p=2. Focus: (2,0).

Problem #0187 International

Problem 3 — Conic Sections

A circle centered at the origin passes through (3,4). Its equation is:

Show answer & worked solution
  1. A. x2+y2=5
  2. B. x2+y2=7
  3. C. x2+y2=25✓ correct
  4. D. (x−3)2+(y−4)2=25

r=9+16=5, so x2+y2=25.

Problem #0184 International

Problem 4 — Conic Sections

For the ellipse x216+y29=1, the lengths of the semi-axes are:

Show answer & worked solution
  1. A. a=16, b=9
  2. B. a=8, b=6
  3. C. a=4, b=3✓ correct
  4. D. a=2, b=3

a2=16⇒a=4; b2=9⇒b=3.

Problem #0185 International

Problem 5 — Conic Sections

Find the radius of the circle x2+y2−4x+6y+4=0.

Show answer & worked solution
  1. A. 1
  2. B. 2
  3. C. 3✓ correct
  4. D. 4

(x2−4x)+(y2+6y)=−4⇒(x−2)2−4+(y+3)2−9=−4⇒(x−2)2+(y+3)2=9. Radius =3.

Problem #0935 US SAT

Problem 6 — Conic Sections

The ellipse x225+y29=1 has semi-major axis of length:

Show answer & worked solution
  1. A. 3
  2. B. 4
  3. C. 5✓ correct
  4. D. 9

Here a2=25 and b2=9, so a=5 and b=3.

The semi-major axis is max⁡(a,b)=5.

Problem #0936 US SAT

Problem 7 — Conic Sections

The hyperbola x216−y29=1 has vertices at:

Show answer & worked solution
  1. A. (±4, 0)✓ correct
  2. B. (0, ±4)
  3. C. (±3, 0)
  4. D. (±5, 0)

a2=16, so a=4. The vertices are (±a,0)=(±4,0).

Problem #0934 US SAT

Problem 8 — Conic Sections

The vertex of the parabola y=x2−4x+7 is at:

Show answer & worked solution
  1. A. (4,7)
  2. B. (2,3)✓ correct
  3. C. (−2,3)
  4. D. (2,−3)

For y=ax2+bx+c, the vertex is at x=−b2a.

x=−−42(1)=2.

y=(2)2−4(2)+7=4−8+7=3.

Vertex: (2,3).

Problem #0181 International

Problem 9 — Conic Sections

The radius of the circle x2+y2=25 is:

Show answer & worked solution
  1. A. 5/2
  2. B. 5
  3. C. 5✓ correct
  4. D. 25

The standard form x2+y2=r2 gives r=25=5.

Problem #0182 International

Problem 10 — Conic Sections

The center of the circle (x−2)2+(y+3)2=16 is:

Show answer & worked solution
  1. A. (−2,3)
  2. B. (2,3)
  3. C. (2,−3)✓ correct
  4. D. (−2,−3)

Reading off: h=2, k=−3. Center: (2,−3).

8 more Conic Sections questions in the app

Also covered in Conic Sections practice across every exam.

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