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Conic Sections

18 practice questions with full worked solutions. Free, no account needed.

Problems & worked solutions

Problem #0182 International

Problem 1 — Conic Sections

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

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

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

Problem #0181 International

Problem 2 — Conic Sections

The radius of the circle x2+y2=25x^2 + y^2 = 25 is:

Show answer & worked solution
  1. A. 5/25/2
  2. B. 5\sqrt{5}
  3. C. 55✓ correct
  4. D. 2525

The standard form x2+y2=r2x^2 + y^2 = r^2 gives r=25=5r = \sqrt{25} = 5.

Problem #0183 International

Problem 3 — Conic Sections

The vertex of the parabola y=(x−1)2+3y = (x - 1)^2 + 3 is:

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

y=(x−h)2+ky = (x - h)^2 + k has vertex (h,k)=(1,3)(h, k) = (1, 3).

Problem #0933 US SAT

Problem 4 — Circle

The equation of a circle with centre (2,−3)(2, -3) and radius 55 is:

Show answer & worked solution
  1. A. (x−2)2+(y+3)2=25(x - 2)^{2} + (y + 3)^{2} = 25✓ correct
  2. B. (x+2)2+(y−3)2=25(x + 2)^{2} + (y - 3)^{2} = 25
  3. C. (x−2)2+(y+3)2=5(x - 2)^{2} + (y + 3)^{2} = 5
  4. D. (x−2)2−(y+3)2=25(x - 2)^{2} - (y + 3)^{2} = 25

Centre (h,k)=(2,−3)(h, k) = (2, -3) and r=5r = 5 (so r2=25r^{2} = 25):

(x−2)2+(y−(−3))2=25  ⟹  (x−2)2+(y+3)2=25(x - 2)^{2} + (y - (-3))^{2} = 25 \;\Longrightarrow\; (x - 2)^{2} + (y + 3)^{2} = 25.

Problem #0185 International

Problem 5 — Conic Sections

Find the radius of the circle x2+y2−4x+6y+4=0x^2 + y^2 - 4x + 6y + 4 = 0.

Show answer & worked solution
  1. A. 11
  2. B. 22
  3. C. 33✓ correct
  4. D. 44

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

Problem #0187 International

Problem 6 — Conic Sections

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

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

r=9+16=5r = \sqrt{9 + 16} = 5, so x2+y2=25x^2 + y^2 = 25.

Problem #0184 International

Problem 7 — Conic Sections

For the ellipse x216+y29=1\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1, the lengths of the semi-axes are:

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

a2=16⇒a=4a^2 = 16 \Rightarrow a = 4; b2=9⇒b=3b^2 = 9 \Rightarrow b = 3.

Problem #0186 International

Problem 8 — Conic Sections

For the hyperbola x29−y216=1\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1, the asymptotes have equations:

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

a=3a = 3, b=4b = 4, so asymptotes: y=±43xy = \pm \dfrac{4}{3}x.

Problem #0188 International

Problem 9 — Conic Sections

The parabola y2=8xy^2 = 8x has focus at:

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

4p=8⇒p=24p = 8 \Rightarrow p = 2. Focus: (2,0)(2, 0).

Problem #0935 US SAT

Problem 10 — Ellipse

The ellipse x225+y29=1\dfrac{x^{2}}{25} + \dfrac{y^{2}}{9} = 1 has semi-major axis of length:

Show answer & worked solution
  1. A. 33
  2. B. 44
  3. C. 55✓ correct
  4. D. 99

Here a2=25a^{2} = 25 and b2=9b^{2} = 9, so a=5a = 5 and b=3b = 3.

The semi-major axis is max⁡(a,b)=5\max(a, b) = 5.

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