8.E: Conic Sections (Exercises)

Last updated

Oct 6, 2021
Save as PDF
- 8.5: Solving Nonlinear Systems
- 9: Sequences, Series, and the Binomial Theorem

Anonymous
LibreTexts

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$ $\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

Exercise $\PageIndex{1}$

Calculate the distance and midpoint between the given two points.

$(0,2)$ and $(-4,-1)$
$(6,0)$ and $(-2,-6)$
$(-2,4)$ and $(-6,-8)$
$\left(\frac{1}{2},-1\right)$ and $\left(\frac{5}{2},-\frac{1}{2}\right)$
$(0,-3 \sqrt{2})$ and $(\sqrt{5},-4 \sqrt{2})$
$(-5 \sqrt{3}, \sqrt{6})$ and $(-3 \sqrt{3}, \sqrt{6})$

Answer

1. Distance: $5$ units; midpoint: $\left(-2, \frac{1}{2}\right)$

3. Distance: $4\sqrt{10}$ units; midpoint: $(-4,-2)$

5. Distance: $\sqrt{7}$ units; midpoint: $\left(\frac{\sqrt{5}}{2},-\frac{7 \sqrt{2}}{2}\right)$

Exercise $\PageIndex{2}$

Determine the area of a circle whose diameter is defined by the given two points.

$(-3,3)$ and $(3,-3)$
$(-2,-9)$ and $(-10,-15)$
$\left(\frac{2}{3},-\frac{1}{2}\right)$ and $\left(-\frac{1}{3}, \frac{3}{2}\right)$
$(2 \sqrt{5},-2 \sqrt{2})$ and $(0,-4 \sqrt{2})$

Answer

1. $18\pi$ square units

3. $\frac{5 \pi}{4}$ square units

Exercise $\PageIndex{3}$

Rewrite in standard form and give the vertex.

$y=x^{2}-10 x+33$
$y=2 x^{2}-4 x-1$
$y=x^{2}-3 x-1$
$y=-x^{2}-x-2$
$x=y^{2}+10 y+10$
$x=3 y^{2}+12 y+7$
$x=-y^{2}+8 y-3$
$x=5 y^{2}-5 y+2$

Answer

1. $y=(x-5)^{2}+8 ;$ vertex: $(5,8)$

3. $y=\left(x-\frac{3}{2}\right)^{2}-\frac{13}{4} ;$ vertex: $\left(\frac{3}{2},-\frac{13}{4}\right)$

5. $x=(y+5)^{2}-15 ;$ vertex: $(-15,-5)$

7. $x=-(y-4)^{2}+13 ;$ vertex: $(13,4)$

Exercise $\PageIndex{4}$

Rewrite in standard form and graph. Be sure to find the vertex and all intercepts.

$y=x^{2}-20 x+75$
$y=-x^{2}-10 x+75$
$y=-2 x^{2}-12 x-24$
$y=4 x^{2}+4 x+6$
$x=y^{2}-10 y+16$
$x=-y^{2}+4 y+12$
$x=-4 y^{2}+12 y$
$x=9 y^{2}+18 y+12$
$x=-4 y^{2}+4 y+2$
$x=-y^{2}-5 y+2$

Answer

1. $y=(x-10)^{2}-25$ ;

3. $y=-2(x+3)^{2}-6$ ;

5. $x=(y-5)^{2}-9$ ;

7. $x=-4\left(y-\frac{3}{2}\right)^{2}+9$ ;

9. $x=-4\left(y-\frac{1}{2}\right)^{2}+3$ ;

Exercise $\PageIndex{5}$

Determine the center and radius given the equation of a circle in standard form.

$(x-6)^{2}+y^{2}=9$
$(x+8)^{2}+(y-10)^{2}=1$
$x^{2}+y^{2}=5$
$\left(x-\frac{3}{8}\right)^{2}+\left(y+\frac{5}{2}\right)^{2}=\frac{1}{2}$

Answer

1. Center: $(6,0) ;$ radius: $r=3$

3. Center: $(0,0) ;$ radius: $r=\sqrt{5}$

Exercise $\PageIndex{6}$

Determine standard form for the equation of the circle:

Center $(-7,2)$ with radius $r=10$
Center $\left(\frac{1}{3},-1\right)$ with radius $r=\frac{2}{3}$
Center $(0,-5)$ with radius $r=2 \sqrt{7}$
Center $(1,0)$ with radius $r=\frac{5 \sqrt{3}}{2}$
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. $x^{2}+(y+5)^{2}=28$

5. $(x+3)^{2}+(y-9)^{2}=2$

Exercise $\PageIndex{7}$

Find the $x$ - and $y$ -intercepts.

$(x-3)^{2}+(y+5)^{2}=16$
$(x+5)^{2}+(y-1)^{2}=4$
$x^{2}+(y-2)^{2}=20$
$(x-3)^{2}+(y+3)^{2}=8$
$x^{2}+y^{2}-12 y+27=0$
$x^{2}+y^{2}-4 x+2 y+1=0$

Answer

1. $x$ -intercepts: none; $y$ -intercepts: $(0,-5 \pm \sqrt{7})$

3. $x$ -intercepts: $(\pm 4,0)$ ; $y$ -intercepts: $(0,2 \pm 2 \sqrt{5})$

5. $x$ -intercepts: none; $y$ -intercepts: $(0,3),(0,9)$

Exercise $\PageIndex{8}$

Graph.

$(x+8)^{2}+(y-6)^{2}=4$
$(x-20)^{2}+\left(y+\frac{15}{2}\right)^{2}=\frac{225}{4}$
$x^{2}+y^{2}=24$
$(x-1)^{2}+y^{2}=\frac{1}{4}$
$x^{2}+(y-7)^{2}=27$
$(x+1)^{2}+(y-1)^{2}=2$

Answer

Exercise $\PageIndex{9}$

Rewrite in standard form and graph.

$x^{2}+y^{2}-6 x+4 y-3=0$
$x^{2}+y^{2}+8 x-10 y+16=0$
$2 x^{2}+2 y^{2}-2 x-6 y-3=0$
$4 x^{2}+4 y^{2}+8 y+1=0$
$x^{2}+y^{2}-5 x+y-\frac{1}{2}=0$
$x^{2}+y^{2}+12 x-8 y=0$

Answer

1. $(x-3)^{2}+(y+2)^{2}=16$ ;

3. $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=4$ ;

5. $\left(x-\frac{5}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=7$ ;

Exercise $\PageIndex{10}$

Given the equation of an ellipse in standard form, determine its center, orientation, major radius, and minor radius.

$\frac{(x+12)^{2}}{16}+\frac{(y-10)^{2}}{4}=1$
$\frac{(x+3)^{2}}{3}+\frac{y^{2}}{25}=1$
$x^{2}+\frac{(y-5)^{2}}{12}=1$
$\frac{(x-8)^{2}}{5}+\frac{(y+8)}{18}=1$

Answer

1. Center: $(−12, 10)$ ; orientation: horizontal; major radius: $4$ units; minor radius: $2$ units

3. Center: $(0, 5)$ ; orientation: vertical; major radius: $2\sqrt{3}$ units; minor radius: $1$ unit

Exercise $\PageIndex{11}$

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 $b=\sqrt{7}$
Center $(0,0)$ with $a=5$ and $b=\sqrt{2}$
Center $(-10,-30)$ with $a=10$ and $b=1$

Answer

1. $\frac{x^{2}}{9}+\frac{(y+4)^{2}}{16}=1$

3. $\frac{x^{2}}{25}+\frac{y^{2}}{2}=1$

Exercise $\PageIndex{12}$

Find the $x$ - and $y$ -intercepts.

$\frac{(x+2)^{2}}{4}+\frac{y^{2}}{9}=1$
$\frac{(x-1)^{2}}{2}+\frac{(y+1)^{2}}{3}=1$
$5 x^{2}+2 y^{2}=20$
$5(x-3)^{2}+6 y^{2}=120$

Answer

1. $x$ -intercepts: $(-4,0),(0,0) ; y$ -intercepts: $(0,0)$

3. $x$ -intercepts: $(\pm 2,0) ; y$ -intercepts: $(0, \pm \sqrt{10})$

Exercise $\PageIndex{13}$

Graph.

$\frac{(x-10)^{2}}{25}+\frac{(y+5)^{2}}{4}=1$
$\frac{(x+6)^{2}}{9}+\frac{(y-8)^{2}}{36}=1$
$\frac{\left(x-\frac{3}{2}\right)^{2}}{4}+\left(y-\frac{7}{2}\right)^{2}=1$
$\left(x-\frac{2}{3}\right)^{2}+\frac{y^{2}}{4}=1$
$\frac{x^{2}}{2}+\frac{y^{2}}{5}=1$
$\frac{(x+2)^{2}}{8}+\frac{(y-3)^{2}}{12}=1$

Answer

Exercise $\PageIndex{14}$

Rewrite in standard form and graph.

$4 x^{2}+9 y^{2}-8 x+90 y+193=0$
$9 x^{2}+4 y^{2}+108 x-80 y+580=0$
$x^{2}+9 y^{2}+6 x+108 y+324=0$
$25 x^{2}+y^{2}-350 x-8 y+1,216=0$
$8 x^{2}+12 y^{2}-16 x-36 y-13=0$
$10 x^{2}+2 y^{2}-50 x+14 y+7=0$

Answer

1. $\frac{(x-1)^{2}}{9}+\frac{(y+5)^{2}}{4}=1$ ;

3. $\frac{(x+3)^{2}}{9}+(y+6)^{2}=1$ ;

5. $\frac{(x-1)^{2}}{6}+\frac{\left(y-\frac{3}{2}\right)^{2}}{4}=1$ ;

Exercise $\PageIndex{15}$

Given the equation of a hyperbola in standard form, determine its center, which way the graph opens, and the vertices.

$\frac{(x-10)^{2}}{4}-\frac{(y+5)^{2}}{16}=1$
$\frac{(x+7)^{2}}{2}-\frac{(y-8)^{2}}{8}=1$
$\frac{(y-20)^{2}}{3}-(x-15)^{2}=1$
$3 y^{2}-12(x-1)^{2}=36$

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

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. $\frac{(x-1)^{2}}{9}-\frac{(y+5)^{2}}{4}=1$ ;

3. $\frac{(y-2)^{2}}{2}-\frac{x^{2}}{25}=1$ ;

5. $\frac{\left(y+\frac{3}{2}\right)^{2}}{6}-\frac{\left(x+\frac{1}{2}\right)^{2}}{4}=1$

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.

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):
1. Calculate the distance between them.
2. 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. $y=-(x-3)^{2}+4$ ;

3. $x=-3\left(y-\frac{1}{2}\right)^{2}+\frac{7}{4}$ ;

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

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. $\frac{(x-8)^{2}}{4}+\frac{(y+2)^{2}}{9}=1$ ;

3. $(x+10)^{2}+(y-10)^{2}=100$ ;

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

Search

Text Color

Text Size

Margin Size

Font Type

Sample Exam

Support Center

How can we help?