This precalculus conic sections worksheet answer key solves all 26 problems of the conic sections: parabolas, ellipses and hyperbolas worksheet step by step.
Each solution shows the definition used, the values of \(a\), \(b\), \(c\) or \(p\), and the final equation or feature list. Graphs confirm the key points. Check your standard forms by substituting a known point.
Precalculus Conic Sections Worksheet Answer Key: problems 1–5
Problem 1: Write the equation of a circle
Use \((x-h)^{2}+(y-k)^{2}=r^{2}\).
- \(x^{2}+y^{2}=49\)
- \((x-3)^{2}+(y+2)^{2}=25\)
- \((x+4)^{2}+(y-1)^{2}=7\)
(a) \(x^{2}+y^{2}=49\) (b) \((x-3)^{2}+(y+2)^{2}=25\) (c) \((x+4)^{2}+(y-1)^{2}=7\)
Problem 2: Center and radius from an equation
- Center \((-4,\ 1)\), \(r=\sqrt{49}=7\).
- Center \((0,\ 5)\), \(r=6\).
- Center \((2,\ -3)\), \(r=\sqrt{20}=2\sqrt{5}\approx 4.47\).
(a) \((-4,\ 1)\), 7 (b) \((0,\ 5)\), 6 (c) \((2,\ -3)\), \(2\sqrt{5}\)
Problem 3: Circle through a point
\(r^{2}=(4-1)^{2}+(6-2)^{2}=9+16=25\), so \(r=5\).
\((x-1)^{2}+(y-2)^{2}=25\)
Problem 4: Circle from a diameter
The center is the midpoint: \((\frac{-2+6}{2},\ \frac{3+7}{2})=(2,\ 5)\). The radius squared is half the diameter squared over two: \(AB^{2}=8^{2}+4^{2}=80\), so \(r^{2}=\frac{80}{4}=20\).
\((x-2)^{2}+(y-5)^{2}=20\)
Problem 5: Complete the square: circles
- \((x^{2}-8x+16)+(y^{2}+6y+9)=-9+16+9\), so \((x-4)^{2}+(y+3)^{2}=16\): center \((4,\ -3)\), \(r=4\).
- \((x^{2}+10x+25)+(y^{2}-4y+4)=-13+25+4\), so \((x+5)^{2}+(y-2)^{2}=16\): center \((-5,\ 2)\), \(r=4\).
- \((x^{2}+2x+1)+(y^{2}-6y+9)=15+1+9\), so \((x+1)^{2}+(y-3)^{2}=25\): center \((-1,\ 3)\), \(r=5\).
(a) \((4,\ -3)\), 4 (b) \((-5,\ 2)\), 4 (c) \((-1,\ 3)\), 5
Precalculus Conic Sections Worksheet Answer Key: problems 6–10
Problem 6: Parabola from focus and directrix
The vertex is halfway between focus and directrix: \((0,\ 0)\). The focus is \(p=3\) above the vertex, so \(x^{2}=4py=12y\).
\(x^{2}=12y\)
Problem 7: Focus and directrix from an equation
- \(4p=16\), so \(p=4\). The parabola opens right: focus \((4,\ 0)\), directrix \(x=-4\).
- \(4p=-20\), so \(p=-5\). It opens down: focus \((0,\ -5)\), directrix \(y=5\).
(a) \(p=4\), \((4,\ 0)\), \(x=-4\) (b) \(p=-5\), \((0,\ -5)\), \(y=5\)
Problem 8: Equal distances
Distance to the focus: \(\sqrt{(4-0)^{2}+(2-2)^{2}}=4\). Distance to the directrix: \(2-(-2)=4\). The distances are equal.
Check with the equation \(x^{2}=8y\): \(4^{2}=16=8(2)\). Both distances equal \(4\), so \(P\) is on the parabola.
Problem 9: Parabola with a shifted vertex
The focus is \(4-1=3\) above the vertex, so \(p=3\) and the parabola opens up. The form is \((x-h)^{2}=4p(y-k)\).
\((x-2)^{2}=12(y-1)\). The directrix is \(3\) below the vertex: \(y=1-3=-2\).
\((x-2)^{2}=12(y-1)\); directrix \(y=-2\)
Problem 10: Parabola opening left
The vertex is \((0,\ 0)\). The focus is \(3\) units left, so \(p=-3\) and \(y^{2}=4px=-12x\).
Check \((-3,\ 6)\): \(36=-12(-3)\). Distance to focus \(6\), to directrix \(3-(-3)=6\). \(y^{2}=-12x\)
Precalculus Conic Sections Worksheet Answer Key: problems 11–15
Problem 11: Ellipse features
\(a^{2}=25\) and \(b^{2}=9\), so \(a=5\), \(b=3\). Then \(c^{2}=a^{2}-b^{2}=16\) and \(c=4\).
\(a=5\), \(b=3\), \(c=4\); vertices \((\pm 5,\ 0)\); co-vertices \((0,\ \pm 3)\); foci \((\pm 4,\ 0)\)
Problem 12: A vertical ellipse
The larger denominator is under \(y\): \(a^{2}=25\), \(b^{2}=16\), so \(a=5\), \(b=4\) and the major axis is vertical. \(c^{2}=25-16=9\), \(c=3\).
Vertices \((0,\ \pm 5)\), foci \((0,\ \pm 3)\); the major axis is the \(y\)-axis
Problem 13: Ellipse from its foci
\(2a=10\), so \(a=5\). The foci give \(c=3\). Then \(b^{2}=a^{2}-c^{2}=25-9=16\).
\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Problem 14: A shifted ellipse
Center \((2,\ -1)\). \(a^{2}=36\), \(b^{2}=20\), so \(a=6\) and \(c^{2}=36-20=16\), \(c=4\). The major axis is horizontal.
Vertices: \((2\pm 6,\ -1)\), that is \((-4,\ -1)\) and \((8,\ -1)\). Foci: \((2\pm 4,\ -1)\), that is \((-2,\ -1)\) and \((6,\ -1)\). The sum is \(2a=12\).
Center \((2,\ -1)\); vertices \((-4,\ -1)\), \((8,\ -1)\); foci \((-2,\ -1)\), \((6,\ -1)\); sum \(12\)
Problem 15: Hyperbola features
\(a=4\), \(b=3\), \(c^{2}=16+9=25\), \(c=5\). The branches open left and right.
Vertices \((\pm 4,\ 0)\); foci \((\pm 5,\ 0)\); asymptotes \(y=\pm\frac{3}{4}x\)
Precalculus Conic Sections Worksheet Answer Key: problems 16–20
Problem 16: A vertical hyperbola
The positive term is \(y\), so the branches open up and down. \(a^{2}=9\), \(b^{2}=16\), \(a=3\), \(b=4\), \(c^{2}=9+16=25\), \(c=5\).
The asymptotes have slope \(\pm\frac{a}{b}=\pm\frac{3}{4}\).
Vertices \((0,\ \pm 3)\); foci \((0,\ \pm 5)\); asymptotes \(y=\pm\frac{3}{4}x\)
Problem 17: Hyperbola from its foci
\(2a=8\), so \(a=4\). The foci give \(c=5\), and \(b^{2}=c^{2}-a^{2}=25-16=9\).
Check \((5,\ 2.25)\): distances \(2.25\) and \(10.25\) differ by \(8\). \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)
Problem 18: Identify an ellipse
\(4(x^{2}-4x+4)+9(y^{2}+2y+1)=11+16+9=36\), so \(4(x-2)^{2}+9(y+1)^{2}=36\).
Divide by 36: \(\frac{(x-2)^{2}}{9}+\frac{(y+1)^{2}}{4}=1\). It is an ellipse with center \((2,\ -1)\), \(a=3\), \(b=2\), \(c=\sqrt{9-4}=\sqrt{5}\).
Ellipse; center \((2,\ -1)\); foci \((2\pm\sqrt{5},\ -1)\)
Problem 19: Identify a hyperbola
\((x^{2}+6x+9)-4(y^{2}-4y+4)=23+9-16=16\), so \((x+3)^{2}-4(y-2)^{2}=16\).
Divide by 16: \(\frac{(x+3)^{2}}{16}-\frac{(y-2)^{2}}{4}=1\). It is a hyperbola with center \((-3,\ 2)\), \(a=4\), \(b=2\), so the asymptotes have slope \(\pm\frac{1}{2}\).
Hyperbola; center \((-3,\ 2)\); asymptotes \(y-2=\pm\frac{1}{2}(x+3)\)
Problem 20: Identify a parabola
\(y^{2}-6y+9=8x-25+9\), so \((y-3)^{2}=8x-16=8(x-2)\).
Vertex \((2,\ 3)\); \(4p=8\), so \(p=2\) and the parabola opens right. Focus \((2+2,\ 3)=(4,\ 3)\); directrix \(x=2-2=0\).
\((y-3)^{2}=8(x-2)\); vertex \((2,\ 3)\); focus \((4,\ 3)\); directrix \(x=0\)
Precalculus Conic Sections Worksheet Answer Key: problems 21–25
Problem 21: Satellite dish
The rim point \((4,\ 2)\) satisfies \(x^{2}=4py\): \(16=4p(2)=8p\), so \(p=2\). The equation is \(x^{2}=8y\).
\(x^{2}=8y\); the receiver is \(2\) feet above the vertex
Problem 22: Whispering gallery
\(2a=100\) so \(a=50\); \(2b=60\) so \(b=30\). The equation is \(\frac{x^{2}}{2500}+\frac{y^{2}}{900}=1\).
\(c^{2}=2500-900=1600\), so \(c=40\). The foci are \((\pm 40,\ 0)\).
\(\frac{x^{2}}{2500}+\frac{y^{2}}{900}=1\); the foci are \(80\) feet apart
Problem 23: Sprinkler coverage
Boundary: \((x-20)^{2}+(y-30)^{2}=225\).
Flower: \((24-20)^{2}+(38-30)^{2}=16+64=80 < 225\), so it is inside. Bench: \((35-20)^{2}+(45-30)^{2}=225+225=450 > 225\), so it is outside.
\((x-20)^{2}+(y-30)^{2}=225\); the flower is watered, the bench is not
Problem 24: Navigation stations
\(2a=6\) so \(a=3\); \(c=5\) so \(b^{2}=25-9=16\). The equation is \(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\).
For \(x=6\): \(\frac{36}{9}-\frac{y^{2}}{16}=1\), so \(\frac{y^{2}}{16}=3\) and \(y^{2}=48\), \(y=4\sqrt{3}\approx 6.93\).
\(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\); \(y=4\sqrt{3}\approx 6.93\) miles
Problem 25: Elliptical garden
\(a=12\), \(b=9\), so \(c^{2}=144-81=63\) and \(c=3\sqrt{7}\approx 7.94\). The stakes are \(2c\approx 15.87\) feet apart.
A pencil on the taut string has distances to the stakes adding to \(2a=24\) feet. The loop = that sum plus the stake gap: \(24+15.87\approx 39.87\) feet.
Stakes about \(15.87\) feet apart; loop about \(39.87\) feet
Precalculus Conic Sections Worksheet Answer Key: problems 26–26
Problem 26: Circular fountain
\((x^{2}-12x+36)+(y^{2}-16y+64)=-75+36+64=25\), so \((x-6)^{2}+(y-8)^{2}=25\): center \((6,\ 8)\), radius \(5\).
The center is \(\sqrt{6^{2}+8^{2}}=10\) feet from the entrance, so the nearest edge point is \(10-5=5\) feet away.
Center \((6,\ 8)\), radius \(5\) ft; shortest distance \(5\) ft
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Precalculus Conic Sections: Parabolas, Ellipses and Hyperbolas worksheet answer key
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