Conic Sections
Exercise 10.4 • Hyperbolas
Hyperbola Parameters:
Relation: $c^2 = a^2 + b^2$, Eccentricity $e = c/a$, Latus Rectum = $2b^2/a$
Horizontal ($\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$): Foci $(\pm c, 0)$, Vertices $(\pm a, 0)$
Vertical ($\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$): Foci $(0, \pm c)$, Vertices $(0, \pm a)$
Note: For Hyperbolas, $a$ is always the denominator of the positive term.
Relation: $c^2 = a^2 + b^2$, Eccentricity $e = c/a$, Latus Rectum = $2b^2/a$
Horizontal ($\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$): Foci $(\pm c, 0)$, Vertices $(\pm a, 0)$
Vertical ($\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$): Foci $(0, \pm c)$, Vertices $(0, \pm a)$
Note: For Hyperbolas, $a$ is always the denominator of the positive term.
Q1
1. $\frac{x^2}{16} – \frac{y^2}{9} = 1$
[Image of a horizontal hyperbola]
Identify
Form $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$ (Horizontal).
Values
$a^2 = 16 \Rightarrow a = 4$. $b^2 = 9 \Rightarrow b = 3$.
Find c
$c^2 = a^2 + b^2 = 16 + 9 = 25 \Rightarrow c = 5$.
Results
Foci: $(\pm 5, 0)$, Vertices: $(\pm 4, 0)$
$e = \frac{c}{a} = \frac{5}{4}$, L.R. = $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{9}{2}$
$e = \frac{c}{a} = \frac{5}{4}$, L.R. = $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{9}{2}$
Foci $(\pm 5, 0)$, Vertices $(\pm 4, 0)$, $e=5/4$, LR=9/2
Q2
2. $\frac{y^2}{9} – \frac{x^2}{27} = 1$
[Image of a vertical hyperbola]
Identify
Form $\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$ (Vertical).
Values
$a^2 = 9 \Rightarrow a = 3$. $b^2 = 27 \Rightarrow b = 3\sqrt{3}$.
Find c
$c^2 = 9 + 27 = 36 \Rightarrow c = 6$.
Foci $(0, \pm 6)$, Vertices $(0, \pm 3)$, $e=2$, LR=18
Q3
3. $9y^2 – 4x^2 = 36$
Rewrite
Divide by 36: $\frac{y^2}{4} – \frac{x^2}{9} = 1$ (Vertical).
Values
$a^2 = 4 \Rightarrow a = 2$. $b^2 = 9 \Rightarrow b = 3$.
c
$c = \sqrt{4 + 9} = \sqrt{13}$.
Foci $(0, \pm \sqrt{13})$, Vertices $(0, \pm 2)$, $e=\frac{\sqrt{13}}{2}$, LR=9
Q4
4. $16x^2 – 9y^2 = 576$
Rewrite
Divide by 576: $\frac{x^2}{36} – \frac{y^2}{64} = 1$ (Horizontal).
Values
$a^2 = 36 \Rightarrow a = 6$. $b^2 = 64 \Rightarrow b = 8$.
c
$c = \sqrt{36 + 64} = \sqrt{100} = 10$.
Foci $(\pm 10, 0)$, Vertices $(\pm 6, 0)$, $e=5/3$, LR=$\frac{64}{3}$
Q5
5. $5y^2 – 9x^2 = 36$
Rewrite
$\frac{y^2}{36/5} – \frac{x^2}{4} = 1$ (Vertical).
Values
$a^2 = \frac{36}{5}$. $b^2 = 4$.
c
$c = \sqrt{\frac{36}{5} + 4} = \sqrt{\frac{56}{5}} = 2\sqrt{\frac{14}{5}}$.
e
$e = \frac{c}{a} = \frac{\sqrt{56}/\sqrt{5}}{6/\sqrt{5}} = \frac{\sqrt{56}}{6} = \frac{2\sqrt{14}}{6} = \frac{\sqrt{14}}{3}$.
LR
$\frac{2b^2}{a} = \frac{2(4)}{6/\sqrt{5}} = \frac{8\sqrt{5}}{6} = \frac{4\sqrt{5}}{3}$.
Foci $(0, \pm \sqrt{56/5})$, Vertices $(0, \pm \frac{6}{\sqrt{5}})$
Q6
6. $49y^2 – 16x^2 = 784$
Rewrite
$\frac{y^2}{16} – \frac{x^2}{49} = 1$ (Vertical).
Values
$a^2 = 16 \Rightarrow a = 4$. $b^2 = 49 \Rightarrow b = 7$.
c
$c = \sqrt{16 + 49} = \sqrt{65}$.
Foci $(0, \pm \sqrt{65})$, Vertices $(0, \pm 4)$, $e=\frac{\sqrt{65}}{4}$, LR=$\frac{49}{2}$
Q7
7. Vertices $(\pm 2, 0)$, Foci $(\pm 3, 0)$
Type
Horizontal (x-axis). $a=2, c=3$.
Find b
$b^2 = c^2 – a^2 = 9 – 4 = 5$.
$\frac{x^2}{4} – \frac{y^2}{5} = 1$
Q8
8. Vertices $(0, \pm 5)$, Foci $(0, \pm 8)$
Type
Vertical (y-axis). $a=5, c=8$.
Find b
$b^2 = 64 – 25 = 39$.
$\frac{y^2}{25} – \frac{x^2}{39} = 1$
Q9
9. Vertices $(0, \pm 3)$, Foci $(0, \pm 5)$
Type
Vertical. $a=3, c=5$.
Find b
$b^2 = 25 – 9 = 16$.
$\frac{y^2}{9} – \frac{x^2}{16} = 1$
Q10
10. Foci $(\pm 5, 0)$, Transverse Axis Length = 8
Values
Horizontal. $c=5$. $2a=8 \Rightarrow a=4$.
Find b
$b^2 = 25 – 16 = 9$.
$\frac{x^2}{16} – \frac{y^2}{9} = 1$
Q11
11. Foci $(0, \pm 13)$, Conjugate Axis Length = 24
Values
Vertical. $c=13$. $2b=24 \Rightarrow b=12$.
Find a
$c^2 = a^2 + b^2 \Rightarrow 169 = a^2 + 144 \Rightarrow a^2 = 25$.
$\frac{y^2}{25} – \frac{x^2}{144} = 1$
Q12
12. Foci $(\pm 3\sqrt{5}, 0)$, Latus Rectum = 8
Values
Horizontal. $c = 3\sqrt{5} \Rightarrow c^2 = 45$. LR $\frac{2b^2}{a} = 8 \Rightarrow b^2 = 4a$.
Equation
$c^2 = a^2 + b^2 \Rightarrow 45 = a^2 + 4a \Rightarrow a^2 + 4a – 45 = 0$.
Solve a
$(a+9)(a-5) = 0 \Rightarrow a=5$ (since $a>0$).
Find b
$b^2 = 4(5) = 20$.
$\frac{x^2}{25} – \frac{y^2}{20} = 1$
Q13
13. Foci $(\pm 4, 0)$, Latus Rectum = 12
Values
Horizontal. $c=4 \Rightarrow c^2=16$. LR $\frac{2b^2}{a} = 12 \Rightarrow b^2 = 6a$.
Equation
$16 = a^2 + 6a \Rightarrow a^2 + 6a – 16 = 0$.
Solve a
$(a+8)(a-2) = 0 \Rightarrow a=2$.
Find b
$b^2 = 6(2) = 12$.
$\frac{x^2}{4} – \frac{y^2}{12} = 1$
Q14
14. Vertices $(\pm 7, 0)$, $e = 4/3$
Values
Horizontal. $a=7$. $c/a = 4/3 \Rightarrow c = 7(4/3) = 28/3$.
Find b
$b^2 = c^2 – a^2 = (\frac{28}{3})^2 – 7^2 = \frac{784}{9} – 49 = \frac{784 – 441}{9} = \frac{343}{9}$.
$\frac{x^2}{49} – \frac{9y^2}{343} = 1$
Q15
15. Foci $(0, \pm \sqrt{10})$, passing through $(2, 3)$
Type
Vertical. $\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$. $c^2 = a^2+b^2 = 10 \Rightarrow b^2 = 10-a^2$.
Substitute
Point $(2,3) \Rightarrow \frac{9}{a^2} – \frac{4}{10-a^2} = 1$.
Solve
$9(10-a^2) – 4a^2 = a^2(10-a^2) \Rightarrow 90 – 9a^2 – 4a^2 = 10a^2 – a^4$.
Quadratic
$a^4 – 23a^2 + 90 = 0 \Rightarrow (a^2-18)(a^2-5) = 0$.
Check
If $a^2=18$, $b^2 = 10-18 = -8$ (Invalid). So $a^2=5, b^2=5$.
$\frac{y^2}{5} – \frac{x^2}{5} = 1$ ($y^2 – x^2 = 5$)