Precalculus Conic Sections Worksheet Solutions (PDF)

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This conics practice solutions for Precalculus covers conic sections (Common Core G-GPE.A.1, G-GPE.A.2, G-GPE.A.3), with a free PDF to print.

Use this key to check each conic. Solutions show completing the square or the relation among a, b and c, with final answers in bold.

Precalculus · Common Core State Standards: G-GPE.A.1, G-GPE.A.2, G-GPE.A.3

Conics practice solutions: exercises 1 to 10

Exercise 1 – Equation of a circle

-4-2246810-8-6-4-2246xy(3, -2)r = 5
Center (3, -2) and radius 5.

\((x-3)^{2}+(y-(-2))^{2}=5^{2}\).

\((x-3)^{2}+(y+2)^{2}=25\)

Exercise 2 – Center and radius

Complete the squares: \((x^{2}-6x+9)+(y^{2}+4y+4)=12+9+4\), so \((x-3)^{2}+(y+2)^{2}=25\).

Center \((3,-2)\); radius \(5\)

Exercise 3 – Circle through a point

\(r^{2}=(1-(-2))^{2}+(2-6)^{2}=9+16=25\).

\((x+2)^{2}+(y-6)^{2}=25\)

Exercise 4 – Parts of a parabola

-6-5-4-3-2-1123456-3-2-112345xyF (0, 2)directrix y = -2V
Focus (0, 2), directrix y = -2.

\(4p=8\), so \(p=2\). The vertex is \((0,0)\), the focus is \((0,2)\), the directrix is \(y=-2\).

Vertex \((0,0)\); focus \((0,2)\); directrix \(y=-2\)

Exercise 5 – Parabola opening left

\(4p=-12\), so \(p=-3\). The parabola opens left: focus \((-3,0)\), directrix \(x=3\).

Focus \((-3,0)\); directrix \(x=3\)

Exercise 6 – Equation from focus and directrix

The vertex is the midpoint \((0,0)\), and the parabola opens down with \(p=-5\). \(x^{2}=4py=-20y\).

\(x^{2}=-20y\)

Exercise 7 – Shifted parabola

\(h=2\), \(k=-1\), \(4p=4\), so \(p=1\). The focus is \(1\) above the vertex, and the directrix is \(1\) below.

Vertex \((2,-1)\); focus \((2,0)\); directrix \(y=-2\)

Exercise 8 – Complete the square for a parabola

\(y-7+4=x^{2}-4x+4\), so \((x-2)^{2}=y-3\). \(4p=1\), \(p=\frac14\). The vertex is \((2,3)\).

\((x-2)^{2}=y-3\); focus \(\left(2,\frac{13}{4}\right)\)

Exercise 9 – Parts of an ellipse

\(a=10\), \(b=6\), \(c=\sqrt{100-36}=8\). The major axis is horizontal.

\(a=10,b=6,c=8\); vertices \((\pm10,0)\); foci \((\pm8,0)\)

Exercise 10 – Vertical ellipse

The larger denominator is under \(y^{2}\), so the major axis is vertical: \(a^{2}=36\), \(b^{2}=20\), \(c=\sqrt{36-20}=4\).

\((0,\pm4)\)

Exercises 11 and up

Exercise 11 – Ellipse from vertices and foci

\(a=6\), \(c=4\), so \(b^{2}=a^{2}-c^{2}=36-16=20\).

\(\frac{x^{2}}{36}+\frac{y^{2}}{20}=1\)

Exercise 12 – Shifted ellipse

Center \((1,-2)\). \(a=3\), \(b=2\), \(c=\sqrt{9-4}=\sqrt5\). The major axis is horizontal.

Center \((1,-2)\); foci \((1\pm\sqrt5,\,-2)\)

Exercise 13 – Standard form of an ellipse

Divide by 36.

\(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)

Exercise 14 – Parts of a hyperbola

\(a=7\), \(b=24\), \(c=\sqrt{49+576}=25\). The asymptotes are \(y=\pm\frac{b}{a}x\).

Vertices \((\pm7,0)\); foci \((\pm25,0)\); \(y=\pm\frac{24}{7}x\)

Exercise 15 – Vertical hyperbola

\(a^{2}=25\), \(b^{2}=144\), \(c=\sqrt{169}=13\). With the \(y^{2}\) term first, the branches open up and down and the asymptotes are \(y=\pm\frac{5}{12}x\).

Foci \((0,\pm13)\); \(y=\pm\frac{5}{12}x\)

Exercise 16 – Hyperbola from vertices and foci

\(a=4\), \(c=5\), so \(b^{2}=c^{2}-a^{2}=9\).

\(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

Exercise 17 – Identify the conic

(a) equal coefficients, same sign: circle. (b) one squared variable: parabola. (c) opposite signs: hyperbola. (d) same sign, different coefficients: ellipse.

(a) circle; (b) parabola; (c) hyperbola; (d) ellipse

Exercise 18 – Satellite dish

-3-2-112312xyfocus
The focus is 1 foot above the vertex.

The rim point is \((2,1)\): \(4=4p(1)\), so \(p=1\).

\(x^{2}=4y\); focus \(1\) ft above the vertex

Exercise 19 – Whispering gallery

-60-40-20204060-20-101020xyF₂F₁100 ft
Foci about 45.8 feet from the center.

\(a=50\), \(b=20\), \(c=\sqrt{2500-400}\approx45.83\). The foci are \(2c\) apart.

\(2c\approx 91.7\text{ ft}\)

Exercise 20 – Elliptical arch

-30-20-101020301020xy15 ft
Height 17.32 feet at 15 feet from the center.

\(a=30\), \(b=20\): \(\frac{x^{2}}{900}+\frac{y^{2}}{400}=1\). At \(x=15\): \(y^{2}=400\left(1-\frac{225}{900}\right)=300\), so \(y=\sqrt{300}\approx17.32\).

\(\frac{x^{2}}{900}+\frac{y^{2}}{400}=1\); \(17.32\text{ ft}\)

Key facts behind this conics practice solutions

Key facts
  • Circle: (x – h)^2 + (y – k)^2 = r^2.
  • Parabola: (x – h)^2 = 4p(y – k); focus is p from the vertex and the directrix is on the opposite side.
  • Ellipse: x^2/a^2 + y^2/b^2 = 1 with c^2 = a^2 – b^2; the foci lie on the major axis.
  • Hyperbola: x^2/a^2 – y^2/b^2 = 1 with c^2 = a^2 + b^2 and asymptotes y = plus or minus (b/a) x.
  • Complete the square to find the center or vertex.

How to use this conics practice solutions

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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Download the conics practice solutions as a PDF with the button below and keep it with your notes.

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Precalculus Conic Sections Worksheet Solutions

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