8.E: Conic Sections (Exercises)
( \newcommand{\kernel}{\mathrm{null}\,}\)
Exercise 8.E.1
Calculate the distance and midpoint between the given two points.
- (0,2) and (−4,−1)
- (6,0) and (−2,−6)
- (−2,4) and (−6,−8)
- (12,−1) and (52,−12)
- (0,−3√2) and (√5,−4√2)
- (−5√3,√6) and (−3√3,√6)
- Answer
-
1. Distance: 5 units; midpoint: (−2,12)
3. Distance: 4√10 units; midpoint: (−4,−2)
5. Distance: √7 units; midpoint: (√52,−7√22)
Exercise 8.E.2
Determine the area of a circle whose diameter is defined by the given two points.
- (−3,3) and (3,−3)
- (−2,−9) and (−10,−15)
- (23,−12) and (−13,32)
- (2√5,−2√2) and (0,−4√2)
- Answer
-
1. 18π square units
3. 5π4 square units
Exercise 8.E.3
Rewrite in standard form and give the vertex.
- y=x2−10x+33
- y=2x2−4x−1
- y=x2−3x−1
- y=−x2−x−2
- x=y2+10y+10
- x=3y2+12y+7
- x=−y2+8y−3
- x=5y2−5y+2
- Answer
-
1. y=(x−5)2+8; vertex: (5,8)
3. y=(x−32)2−134; vertex: (32,−134)
5. x=(y+5)2−15; vertex: (−15,−5)
7. x=−(y−4)2+13; vertex: (13,4)
Exercise 8.E.4
Rewrite in standard form and graph. Be sure to find the vertex and all intercepts.
- y=x2−20x+75
- y=−x2−10x+75
- y=−2x2−12x−24
- y=4x2+4x+6
- x=y2−10y+16
- x=−y2+4y+12
- x=−4y2+12y
- x=9y2+18y+12
- x=−4y2+4y+2
- x=−y2−5y+2
- Answer
-
1. y=(x−10)2−25;
Figure 8.E.1 3. y=−2(x+3)2−6;
Figure 8.E.2 5. x=(y−5)2−9;
Figure 8.E.3 7. x=−4(y−32)2+9;
Figure 8.E.4 9. x=−4(y−12)2+3;
Figure 8.E.5
Exercise 8.E.5
Determine the center and radius given the equation of a circle in standard form.
- (x−6)2+y2=9
- (x+8)2+(y−10)2=1
- x2+y2=5
- (x−38)2+(y+52)2=12
- Answer
-
1. Center: (6,0); radius: r=3
3. Center: (0,0); radius: r=√5
Exercise 8.E.6
Determine standard form for the equation of the circle:
- Center (−7,2) with radius r=10
- Center (13,−1) with radius r=23
- Center (0,−5) with radius r=2√7
- Center (1,0) with radius r=5√32
- Circle whose diameter is defined by (−4,10) and (−2,8)
- Circle whose diameter is defined by (3,−6) and (0,−4)
- Answer
-
1. (x+7)2+(y−2)2=100
3. x2+(y+5)2=28
5. (x+3)2+(y−9)2=2
Exercise 8.E.7
Find the x- and y-intercepts.
- (x−3)2+(y+5)2=16
- (x+5)2+(y−1)2=4
- x2+(y−2)2=20
- (x−3)2+(y+3)2=8
- x2+y2−12y+27=0
- x2+y2−4x+2y+1=0
- Answer
-
1. x-intercepts: none; y-intercepts: (0,−5±√7)
3. x-intercepts: (±4,0); y-intercepts: (0,2±2√5)
5. x-intercepts: none; y-intercepts: (0,3),(0,9)
Exercise 8.E.8
Graph.
- (x+8)2+(y−6)2=4
- (x−20)2+(y+152)2=2254
- x2+y2=24
- (x−1)2+y2=14
- x2+(y−7)2=27
- (x+1)2+(y−1)2=2
- Answer
-
1.
Figure 8.E.6 3.
Figure 8.E.7 5.
Figure 8.E.8
Exercise 8.E.9
Rewrite in standard form and graph.
- x2+y2−6x+4y−3=0
- x2+y2+8x−10y+16=0
- 2x2+2y2−2x−6y−3=0
- 4x2+4y2+8y+1=0
- x2+y2−5x+y−12=0
- x2+y2+12x−8y=0
- Answer
-
1. (x−3)2+(y+2)2=16;
Figure 8.E.9 3. (x−12)2+(y−32)2=4;
Figure 8.E.10 5. (x−52)2+(y+12)2=7;
Figure 8.E.11
Exercise 8.E.10
Given the equation of an ellipse in standard form, determine its center, orientation, major radius, and minor radius.
- (x+12)216+(y−10)24=1
- (x+3)23+y225=1
- x2+(y−5)212=1
- (x−8)25+(y+8)18=1
- Answer
-
1. Center: (−12,10); orientation: horizontal; major radius: 4 units; minor radius: 2 units
3. Center: (0,5); orientation: vertical; major radius: 2√3 units; minor radius: 1 unit
Exercise 8.E.11
Determine the standard form for the equation of the ellipse given the following information.
- Center (0,−4) with a=3 and b=4
- Center (3,8) with a=1 and b=√7
- Center (0,0) with a=5 and b=√2
- Center (−10,−30) with a=10 and b=1
- Answer
-
1. x29+(y+4)216=1
3. x225+y22=1
Exercise 8.E.12
Find the x- and y-intercepts.
- (x+2)24+y29=1
- (x−1)22+(y+1)23=1
- 5x2+2y2=20
- 5(x−3)2+6y2=120
- Answer
-
1. x -intercepts: (−4,0),(0,0);y -intercepts: (0,0)
3. x -intercepts: (±2,0);y -intercepts: (0,±√10)
Exercise 8.E.13
Graph.
- (x−10)225+(y+5)24=1
- (x+6)29+(y−8)236=1
- (x−32)24+(y−72)2=1
- (x−23)2+y24=1
- x22+y25=1
- (x+2)28+(y−3)212=1
- Answer
-
1.
Figure 8.E.12 3.
Figure 8.E.13 5.
Figure 8.E.14
Exercise 8.E.14
Rewrite in standard form and graph.
- 4x2+9y2−8x+90y+193=0
- 9x2+4y2+108x−80y+580=0
- x2+9y2+6x+108y+324=0
- 25x2+y2−350x−8y+1,216=0
- 8x2+12y2−16x−36y−13=0
- 10x2+2y2−50x+14y+7=0
- Answer
-
1. (x−1)29+(y+5)24=1;
Figure 8.E.15 3. (x+3)29+(y+6)2=1;
Figure 8.E.16 5. (x−1)26+(y−32)24=1;
Figure 8.E.17
Exercise 8.E.15
Given the equation of a hyperbola in standard form, determine its center, which way the graph opens, and the vertices.
- (x−10)24−(y+5)216=1
- (x+7)22−(y−8)28=1
- (y−20)23−(x−15)2=1
- 3y2−12(x−1)2=36
- Answer
-
1. Center: (10,-5); opens left and right; vertices: (8,-5),(12,-5)
3. Center: (15,20); opens upward and downward; vertices: (15,20-\sqrt{3}),(15,20+\sqrt{3})
Exercise \PageIndex{16}
Determine the standard form for the equation of the hyperbola.
- Center (-25,10), a=3, b=\sqrt{5}, opens up and down.
- Center (9,-12), a=5 \sqrt{3}, b=7, opens left and right.
- Center (-4,0), a=1, b=6, opens left and right.
- Center (-2,-3), a=10 \sqrt{2}, b=2 \sqrt{3}, opens up and down.
- Answer
-
1. \frac{(y-10)^{2}}{5}-\frac{(x+25)^{2}}{9}=1
3. (x+4)^{2}-\frac{y^{2}}{36}=1
Exercise \PageIndex{17}
Find the x- and y-intercepts.
- \frac{(x-1)^{2}}{4}-\frac{(y+3)^{2}}{9}=1
- \frac{(x+4)^{2}}{8}-\frac{(y-2)^{2}}{12}=1
- 4(y-2)^{2}-x^{2}=16
- 6(y+1)^{2}-3(x-1)^{2}=18
- Answer
-
1. x -intercepts: (1 \pm 2 \sqrt{2}, 0) ; y -intercepts: none
3. x -intercepts: (0,0) ; y -intercepts: (0,0),(0,4)
Exercise \PageIndex{18}
Graph.
- \frac{(x-10)^{2}}{25}-\frac{(y+5)^{2}}{100}=1
- \frac{(x-4)^{2}}{4}-\frac{(y-8)^{2}}{16}=1
- \frac{(y-3)^{2}}{9}-\frac{(x-6)^{2}}{81}=1
- \frac{(y+1)^{2}}{4}-\frac{(x+1)^{2}}{25}=1
- \frac{y^{2}}{27}-\frac{(x-3)^{2}}{9}=1
- \frac{x^{2}}{2}-\frac{y^{2}}{3}=1
- Answer
-
1.
Figure 8.E.18 3.
Figure 8.E.19 5.
Figure 8.E.20
Exercise \PageIndex{19}
Rewrite in standard form and graph.
- 4 x^{2}-9 y^{2}-8 x-90 y-257=0
- 9 x^{2}-y^{2}-108 x+16 y+224=0
- 25 y^{2}-2 x^{2}-100 y+50=0
- 3 y^{2}-x^{2}-2 x-10=0
- 8 y^{2}-12 x^{2}+24 y-12 x-33=0
- 4 y^{2}-4 x^{2}-16 y-28 x-37=0
- Answer
-
1. \frac{(x-1)^{2}}{9}-\frac{(y+5)^{2}}{4}=1;
Figure 8.E.21 3. \frac{(y-2)^{2}}{2}-\frac{x^{2}}{25}=1;
Figure 8.E.22 5. \frac{\left(y+\frac{3}{2}\right)^{2}}{6}-\frac{\left(x+\frac{1}{2}\right)^{2}}{4}=1
Figure 8.E.23
Exercise \PageIndex{20}
Identify the conic sections and rewrite in standard form.
- x^{2}+y^{2}-2 x-8 y+16=0
- x^{2}+2 y^{2}+4 x-24 y+74=0
- x^{2}-y^{2}-6 x-4 y+3=0
- x^{2}+y-10 x+22=0
- x^{2}+12 y^{2}-12 x+24=0
- x^{2}+y^{2}+10 y+22=0
- 4 y^{2}-20 x^{2}+16 y+20 x-9=0
- 16 x-16 y^{2}+24 y-25=0
- 9 x^{2}-9 y^{2}-6 x-18 y-17=0
- 4 x^{2}+4 y^{2}+4 x-8 y+1=0
- Answer
-
1. Circle;(x-1)^{2}+(y-4)^{2}=1
3. Hyperbola; \frac{(x-3)^{2}}{2}-\frac{(y+2)^{2}}{2}=1
5. Ellipse; \frac{(x-6)^{2}}{12}+y^{2}=1
7. Hyperbola; \frac{(y+2)^{2}}{5}-\left(x-\frac{1}{2}\right)^{2}=1
9. Hyperbola; \left(x-\frac{1}{3}\right)^{2}-(y+1)^{2}=1
Exercise \PageIndex{21}
Given the graph, write the equation in general form.
1.

2.

3.

4.

5.

6.

- Answer
-
1. x^{2}+y^{2}+18 x-6 y+9=0
3. 9 x^{2}-y^{2}+72 x-12 y+72=0
5. 9 x^{2}+64 y^{2}+54 x-495=0
Exercise \PageIndex{22}
Solve.
- \left\{\begin{array}{l}{x^{2}+y^{2}=8} \\ {x-y=4}\end{array}\right.
- \left\{\begin{array}{l}{x^{2}+y^{2}=1} \\ {x+2 y=1}\end{array}\right.
- \left\{\begin{array}{c}{x^{2}+3 y^{2}=4} \\ {2 x-y=1}\end{array}\right.
- \left\{\begin{array}{c}{2 x^{2}+y^{2}=5} \\ {x+y=3}\end{array}\right.
- \left\{\begin{array}{c}{3 x^{2}-2 y^{2}=1} \\ {x-y=2}\end{array}\right.
- \left\{\begin{array}{c}{x^{2}-3 y^{2}=10} \\ {x-2 y=1}\end{array}\right.
- \left\{\begin{array}{c}{2 x^{2}+y^{2}=11} \\ {4 x+y^{2}=5}\end{array}\right.
- \left\{\begin{array}{l}{x^{2}+4 y^{2}=1} \\ {2 x^{2}+4 y=5}\end{array}\right.
- \left\{\begin{array}{c}{5 x^{2}-y^{2}=10} \\ {x^{2}+y=2}\end{array}\right.
- \left\{\begin{array}{l}{2 x^{2}+y^{2}=1} \\ {2 x-4 y^{2}=-3}\end{array}\right.
- \left\{\begin{array}{c}{x^{2}+4 y^{2}=10} \\ {x y=2}\end{array}\right.
- \left\{\begin{array}{l}{y+x^{2}=0} \\ {x y-8=0}\end{array}\right.
- \left\{\begin{array}{l}{\frac{1}{x}+\frac{1}{y}=10} \\ {\frac{1}{x}-\frac{1}{y}=6}\end{array}\right.
- \left\{\begin{array}{l}{\frac{1}{x}+\frac{1}{y}=1} \\ {y-x=2}\end{array}\right.
- \left\{\begin{array}{l}{x-2 y^{2}=3} \\ {y=\sqrt{x-4}}\end{array}\right.
- \left\{\begin{array}{c}{(x-1)^{2}+y^{2}=1} \\ {y-\sqrt{x}=0}\end{array}\right.
- Answer
-
1. (2,-2)
3. \left(-\frac{1}{13},-\frac{15}{13}\right),(1,1)
5. (-9,-11),(1,-1)
7. (-1,-3),(-1,3)
9. (-\sqrt{2}, 0),(\sqrt{2}, 0),(-\sqrt{7},-5),(\sqrt{7},-5)
11. (\sqrt{2}, \sqrt{2}) \cdot(-\sqrt{2},-\sqrt{2}) \cdot\left(2 \sqrt{2}, \frac{\sqrt{2}}{2}\right) \cdot\left(-2 \sqrt{2},-\frac{\sqrt{2}}{2}\right)
13. \left(\frac{1}{8}, \frac{1}{2}\right)
15. (5,1)
Sample Exam
Exercise \PageIndex{23}
- Given two points (-4,-6) and (2,-8):
- Calculate the distance between them.
- Find the midpoint between them.
- Determine the area of a circle whose diameter is defined by the points (4, −3) and (−1, 2).
- Answer
-
1. (1) 2\sqrt{10} units; (2) (-1,-7)
Exercise \PageIndex{24}
Rewrite in standard form and graph. Find the vertex and all intercepts if any.
- y=-x^{2}+6 x-5
- x=2 y^{2}+4 y-6
- x=-3 y^{2}+3 y+1
- Find the equation of a circle in standard form with center (−6, 3) and radius 2 \sqrt{5} units.
- Answer
-
1. y=-(x-3)^{2}+4;
Figure 8.E.30 3. x=-3\left(y-\frac{1}{2}\right)^{2}+\frac{7}{4};
Figure 8.E.31
Exercise \PageIndex{25}
Sketch the graph of the conic section given its equation in standard form.
- (x-4)^{2}+(y+1)^{2}=45
- \frac{(x+3)^{2}}{4}+\frac{y^{2}}{9}=1
- \frac{y^{2}}{3}-\frac{x^{2}}{9}=1
- \frac{x^{2}}{16}-(y-2)^{2}=1
- Answer
-
1.
Figure 8.E.32 3.
Figure 8.E.33
Exercise \PageIndex{26}
Rewrite in standard form and graph.
- 9 x^{2}+4 y^{2}-144 x+16 y+556=0
- x-y^{2}+6 y+7=0
- x^{2}+y^{2}+20 x-20 y+100=0
- 4 y^{2}-x^{2}+40 y-30 x-225=0
- Answer
-
1. \frac{(x-8)^{2}}{4}+\frac{(y+2)^{2}}{9}=1;
Figure 8.E.34 3. (x+10)^{2}+(y-10)^{2}=100;
Figure 8.E.35
Exercise \PageIndex{27}
Find the x- and y-intercepts.
- x=-2(y-4)^{2}+9
- \frac{(y-1)^{2}}{12}-(x+1)^{2}=1
- Answer
-
1. x -intercept: (-23,0) ; y -intercepts: \left(0, \frac{8 \pm 3 \sqrt{2}}{2}\right)
Exercise \PageIndex{28}
Solve.
- \left\{\begin{array}{l}{x+y=2} \\ {y=-x^{2}+4}\end{array}\right.
- \left\{\begin{array}{l}{y-x^{2}=-3} \\ {x^{2}+y^{2}=9}\end{array}\right.
- \left\{\begin{array}{c}{2 x-y=1} \\ {(x+1)^{2}+2 y^{2}=1}\end{array}\right.
- \left\{\begin{array}{c}{x^{2}+y^{2}=6} \\ {x y=3}\end{array}\right.
- Answer
-
1. (-1,3),(2,0)
3. \emptyset
Exercise \PageIndex{29}
- Find the equation of an ellipse in standard form with vertices (−3, −5) and (5, −5) and a minor radius 2 units in length.
- Find the equation of a hyperbola in standard form opening left and right with vertices (\pm \sqrt{5}, 0) and a conjugate axis that measures 10 units.
- Given the graph of the ellipse, determine its equation in general form.

4. A rectangular deck has an area of 80 square feet and a perimeter that measures 36 feet. Find the dimensions of the deck.
5. The diagonal of a rectangle measures 2\sqrt{13} centimeters and the perimeter measures 20 centimeters. Find the dimensions of the rectangle.
- Answer
-
1. \frac{(x-1)^{2}}{16}+\frac{(y+5)^{2}}{4}=1
3. 4 x^{2}+25 y^{2}-24 x-100 y+36=0
5. 6 centimeters by 4 centimeters