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Hyperbola

A hyperbola is the locus of all points for which the absolute difference of the distances to two fixed points (foci) is constant.

Standard form (horizontal transverse axis):

$$ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 $$
Concept

where $c^2 = a^2 + b^2$.

Standard form (vertical transverse axis):

Concept
$$ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 $$

where $c^2 = a^2 + b^2$.

The center of the hyperbola is at $(h,k)$.
Note: Equilateral hyperbolas are hyperbolas whose transverse and conjugate axes have the same length and whose asymptotes are perpendicular to each other.

Equation of asymptotes

$$ y - k = \pm \frac{a}{b}(x - h) $$

Latus rectum (latera recta) of a hyperbola:

$$ L = \frac{2b^2}{a} $$

Angle Between a Hyperbola Asymptote and the x-axis

What is the angle in degrees between an asymptote of the hyperbola $x^2 - 4y^2 - 2x - 63 = 0$ and the x-axis?

Angle Between a Hyperbola Asymptote and the x-axis – Diagram
Angle Between a Hyperbola Asymptote and the x-axis – Diagram

Complete the square:

$$x^2 - 2x + 1 - 4y^2 - 63 = 1$$ $$(x - 2)^2 - 4y^2 = 64$$ $$\frac{(x - 2)^2}{64} - \frac{y^2}{16} = 1$$

Thus, $$a^2 = 64,\qquad b^2 = 16$$ $$a = 8,\qquad b = 4$$

For a horizontal hyperbola, the asymptote angle satisfies: $$\tan\theta = \frac{b}{a}$$

$$\tan\theta = \frac{4}{8} = \frac{1}{2}$$ $$\theta = 26.6^\circ$$

Final Answer:

$$\boxed{26.6^\circ}$$

Length of the Latus Rectum of a Hyperbola

The equation of an asymptote of a hyperbola is $y = 2x$, and it passes through the point $\left(\frac{5}{2},\, 3\right)$. Determine the length of the latus rectum.

Length of the Latus Rectum of a Hyperbola – Diagram
Length of the Latus Rectum of a Hyperbola – Diagram

Start with the standard hyperbola: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

Substitute the point $\left(\frac{5}{2},\, 3\right)$:

$$ \frac{\left(\frac{5}{2}\right)^2}{a^2} - \frac{3^2}{b^2} = 1 $$ $$ \frac{25}{4a^2} - \frac{9}{b^2} = 1 $$

Since the asymptote has slope $m = 2$, $$m = \frac{b}{a} \quad\Rightarrow\quad b = 2a$$

Substitute $b = 2a$:

$$ \frac{25}{4a^2} - \frac{9}{(2a)^2} = 1 $$ $$ \frac{25}{4a^2} - \frac{9}{4a^2} = 1 $$ $$ \frac{16}{4a^2} = 1 $$ $$ a^2 = 4,\quad a = 2 $$

Then: $$b = 2a = 4$$

The hyperbola becomes: $$ \frac{x^2}{4} - \frac{y^2}{16} = 1 $$

Latus Rectum Length

For a horizontal hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the latus rectum length is: $$L = \frac{2b^2}{a}$$

$$ L = \frac{2(4^2)}{2} = \frac{2 \cdot 16}{2} = 16 $$

Final Answer:

$$\boxed{16}$$

Determine the Type of Curve Defined by $Ax^2 + By^2 + F = 0$

If the curve $Ax^2 + By^2 + F = 0$ passes through the points $(0,3)$ and $(4,6)$, what kind of curve is it?

Determine the Type of Curve Defined by $Ax^2 + By^2 + F = 0$ – Diagram
Determine the Type of Curve Defined by $Ax^2 + By^2 + F = 0$ – Diagram

Start with the general equation: $$Ax^2 + By^2 + F = 0$$ Divide by $A$: $$x^2 + \frac{B}{A}y^2 + \frac{F}{A} = 0$$

Substitute the point $(0,3)$:

$$ 0 + \frac{B}{A}(3^2) + \frac{F}{A} = 0 $$ $$ 9\frac{B}{A} + \frac{F}{A} = 0 $$ $$ \frac{F}{A} = -9\frac{B}{A} $$

Substitute the point $(4,6)$:

$$ (4)^2 + \frac{B}{A}(6^2) + \frac{F}{A} = 0 $$ $$ 16 + 36\frac{B}{A} + \frac{F}{A} = 0 $$ Substitute $\frac{F}{A} = -9\frac{B}{A}$: $$ 16 + 36\frac{B}{A} - 9\frac{B}{A} = 0 $$ $$ 16 + 27\frac{B}{A} = 0 $$ $$ \frac{B}{A} = -\frac{16}{27} $$

So the equation becomes: $$ x^2 - \frac{16}{27}y^2 - 9 = 0 $$

Multiply by 27: $$ 27x^2 - 16y^2 - 243 = 0 $$

Conclusion:

The equation has one positive and one negative squared term $$27x^2 - 16y^2 + 144 = 0$$ which is the equation of a hyperbola since A and C have opposite signs in the general form: $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.

$$\boxed{\text{The curve is a hyperbola.}}$$

Eccentricity of the Hyperbola

Find the eccentricity of the hyperbola whose equation is $$\frac{y^2}{2} - \frac{(x - 1)^2}{36} = 1$$

Eccentricity of the Hyperbola – Diagram
Eccentricity of the Hyperbola – Diagram

For a vertical hyperbola of the form $$\frac{y^2}{a^2} - \frac{(x - h)^2}{b^2} = 1,$$ we identify: $$a^2 = 2,\quad b^2 = 36.$$

So: $$a = \sqrt{2}, \qquad b = 6.$$

The relationship between $a$, $b$, and $c$ for hyperbolas is: $$c^2 = a^2 + b^2.$$

Substitute: $$c^2 = 2 + 36 = 38,$$ $$c = \sqrt{38}.$$

Eccentricity is: $$e = \frac{c}{a} = \frac{\sqrt{38}}{\sqrt{2}}.$$

Simplify: $$e = \sqrt{\frac{38}{2}} = \sqrt{19} \approx 4.4.$$

Final Answer:

$$\boxed{e \approx 4.4}$$

Foci and Asymptotes of the Hyperbola $9x^2 - 4y^2 = 36$

Find the foci and the equations of the asymptotes of the hyperbola $$9x^2 - 4y^2 = 36$$

Foci and Asymptotes of the Hyperbola $9x^2 - 4y^2 = 36$ – Diagram
Foci and Asymptotes of the Hyperbola $9x^2 - 4y^2 = 36$ – Diagram

Step 1: Rewrite in standard form.

$$ \frac{x^2}{4} - \frac{y^2}{9} = 1 $$ Thus, $$ a^2 = 4,\quad a = 2 $$ $$ b^2 = 9,\quad b = 3 $$

Step 2: Find the foci.

For hyperbolas: $$ c^2 = a^2 + b^2 = 4 + 9 = 13 $$ $$ c = \sqrt{13} $$ Since the transverse axis is horizontal: $$ F\left(\sqrt{13},\,0\right),\quad F'\left(-\sqrt{13},\,0\right) $$

Step 3: Use the general asymptote formula.

For a hyperbola $$ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, $$ the asymptotes are: $$ y - k = \pm \frac{b}{a}(x - h) $$ Here, the center is $(h,k) = (0,0)$. So: $$ y = \pm \frac{b}{a}x $$ Substitute $a = 2,\ b = 3$: $$ y = \pm \frac{3}{2}x $$

Final Answers:

Foci:

$$ \boxed{F(\sqrt{13},0)},\qquad \boxed{F'(-\sqrt{13},0)} $$

Asymptotes:

$$ \boxed{y = \frac{3}{2}x},\qquad \boxed{y = -\frac{3}{2}x} $$

Find the Equation, Foci, and Asymptotes of a Hyperbola with Vertices $(\pm 3,0)$ Passing Through $P(5,2)$

Find the Equation, Foci, and Asymptotes of a Hyperbola with Vertices $(\pm 3,0)$ Passing Through $P(5,2)$ – Diagram
Find the Equation, Foci, and Asymptotes of a Hyperbola with Vertices $(\pm 3,0)$ Passing Through $P(5,2)$ – Diagram

Since the vertices are $(\pm 3,0)$, the hyperbola is centered at the origin with horizontal transverse axis, so:

$$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $$

Given the vertices, $a = 3$ so:

$$ a^2 = 9 $$

Use the point $P(5,2)$:

$$ \frac{5^2}{9} - \frac{2^2}{b^2} = 1 $$ $$ \frac{25}{9} - \frac{4}{b^2} = 1 $$ $$ \frac{25}{9} - 1 = \frac{4}{b^2} $$ $$ \frac{16}{9} = \frac{4}{b^2} $$ $$ b^2 = \frac{9}{4} $$

The standard equation becomes:

$$ \frac{x^2}{9} - \frac{y^2}{\frac{9}{4}} = 1 $$

Multiply through to simplify:

$$ x^2 - 4y^2 = 9 $$

Foci

$$ c^2 = a^2 + b^2 = 9 + \frac{9}{4} = \frac{45}{4} $$ $$ c = \frac{3\sqrt{5}}{2} $$

Thus the foci are:

$$ F\left(\frac{3\sqrt{5}}{2},\,0\right),\qquad F'\left(-\frac{3\sqrt{5}}{2},\,0\right) $$

Asymptotes

General asymptote equation for a hyperbola centered at $(h,k)$:

$$ y - k = \pm \frac{b}{a}(x - h) $$

Here $(h,k) = (0,0)$, $a = 3$, $b = \frac{3}{2}$:

$$ y = \pm \frac{b}{a}x = \pm \frac{\frac{3}{2}}{3}x $$ $$ y = \pm \frac{1}{2}x $$

Final Answers

Equation of hyperbola:

$$ \boxed{x^2 - 4y^2 = 9} $$

Foci:

$$ \boxed{F\left(\frac{3\sqrt{5}}{2},0\right)},\quad \boxed{F'\left(-\frac{3\sqrt{5}}{2},0\right)} $$

Asymptotes:

$$ \boxed{y = \frac{1}{2}x},\qquad \boxed{y = -\frac{1}{2}x} $$

Find the Foci and Equations of the Asymptotes of the Hyperbola $4y^2 - 2x^2 = 1$

Find the Foci and Equations of the Asymptotes of the Hyperbola $4y^2 - 2x^2 = 1$ – Diagram
Find the Foci and Equations of the Asymptotes of the Hyperbola $4y^2 - 2x^2 = 1$ – Diagram

Rewrite the hyperbola in standard form:

$$ 4y^2 - 2x^2 = 1 $$ Divide by 1: $$ \frac{y^2}{\frac{1}{4}} - \frac{x^2}{\frac{1}{2}} = 1 $$ So: $$ a^2 = \frac{1}{4},\qquad b^2 = \frac{1}{2} $$

Compute $a$, $b$, and $c$.

$$ a = \frac{1}{2} $$ $$ b = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$ For hyperbolas: $$ c^2 = a^2 + b^2 $$ $$ c^2 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} $$ $$ c = \frac{\sqrt{3}}{2} $$

Foci: For a vertical hyperbola, the foci are $(0,\pm c)$.

$$ \boxed{(0,\pm \frac{\sqrt{3}}{2})} $$

Asymptotes

General formula for a vertical hyperbola centered at $(h,k)$: $$ y - k = \pm \frac{a}{b}(x - h) $$ Here $(h,k) = (0,0)$, so: $$ y = \pm \frac{a}{b}x $$ Substitute: $$ \frac{a}{b} = \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$ Thus: $$ \boxed{y = \pm \frac{\sqrt{2}}{2}x} $$

Final Answers

Foci:

$$ \boxed{(0,\pm \frac{\sqrt{3}}{2})} $$

Asymptotes:

$$ \boxed{y = \frac{\sqrt{2}}{2}x},\qquad \boxed{y = -\frac{\sqrt{2}}{2}x} $$

Rutherford Scattering and Hyperbolic Particle Path

In an experiment that led to the discovery of the atomic structure of matter, Lord Rutherford (1871–1937) shot high-energy alpha particles toward a thin sheet of gold. Because many were reflected, Rutherford showed the existence of the nucleus of a gold atom. The alpha particle is repelled by the nucleus at the origin; it travels along the hyperbolic path $$4x^2 - y^2 = 16.$$ How close does the particle come to the nucleus?

Rutherford Scattering and Hyperbolic Particle Path – Diagram
Rutherford Scattering and Hyperbolic Particle Path – Diagram

Rewrite the equation in standard hyperbola form:

$$ 4x^2 - y^2 = 16 $$ Divide by 16: $$ \frac{x^2}{4} - \frac{y^2}{16} = 1 $$

This is a hyperbola centered at $(0,0)$ with:

$$ a^2 = 4,\quad a = 2 $$ $$ b^2 = 16,\quad b = 4 $$

For a hyperbola of the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$$ the shortest distance from the center to the curve (closest approach to the nucleus) occurs at the vertex $(\pm a, 0)$.

Thus, the particle comes within

$$ \boxed{a = 2} $$

units of distance from the nucleus.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: t685

MSTE - Analytic Geometry / Analytic Geometry / Gemini mapped Chapter 4 to 6

The area of an ellipse is 188.496 and its perimeter is 57.757. Find length of its latus rectum.

  1. 4.17
  2. 5.23
  3. 4.57
  4. 3.75
Area gives $ab$: $\pi ab = 188.496 \Rightarrow ab = 60$.
Perimeter (using $P \approx 2\pi\sqrt{\frac{a^2 + b^2}{2}}$): $57.757 = 2\pi\sqrt{\frac{a^2+b^2}{2}} \Rightarrow a^2 + b^2 = 169$.
Solving with $ab = 60$ gives $a = 12$, $b = 5$.
Latus rectum:
$LR = \frac{2b^2}{a} = \frac{2(25)}{12}$
$\boxed{LR = 4.17}$

Question Bank: t688

MSTE - Analytic Geometry / Analytic Geometry / Gemini mapped Chapter 4 to 6

The area of an ellipse is 301.593 and its perimeter is 64.076.

Find length of its latus rectum.

  1. 9.54
  2. 12.45
  3. 10.67
  4. 5.33

What is the eccentricity of the ellipse?

  1. 0.788
  2. 0.632
  3. 0.854
  4. 0.745

How far apart are the directrices of the ellipse?

  1. 32.2
  2. 28.4
  3. 16.1
  4. 37.4

Part 1.

Area: $\pi ab = 301.593 \Rightarrow ab = 96$. Perimeter ($P \approx 2\pi\sqrt{\frac{a^2+b^2}{2}}$): $64.076 \Rightarrow a^2 + b^2 = 208$.
Solving gives $a = 12$, $b = 8$.
Latus rectum:
$LR = \frac{2b^2}{a} = \frac{2(64)}{12}$
$\boxed{LR = 10.67}$

Part 2.

With $a = 12$, $b = 8$:
$c = \sqrt{a^2 - b^2} = \sqrt{144 - 64} = \sqrt{80} = 8.944$
$e = \frac{c}{a} = \frac{8.944}{12}$
$\boxed{e = 0.745}$

Part 3.

Directrices are at $x = \pm\frac{a}{e}$, so their separation is:
$\frac{2a}{e} = \frac{2(12)}{0.745}$
$\boxed{32.2}$

Question Bank: t699

MSTE - Analytic Geometry / Analytic Geometry / Gemini mapped Chapter 4 to 6

Given the ellipse: 6x^2 + 12xy + 11y^2 = 17. Determine the following:

The angle that the axis of the ellipse make with the x-axis.

  1. 28.4°
  2. 56.8°
  3. 33.7°
  4. 42.3°

The distance between the vertices of the ellipse.

  1. 6.21
  2. 5.83
  3. 4.25
  4. 7.14

The eccentricity of the ellipse.

  1. 0.93
  2. 0.85
  3. 0.63
  4. 0.74

Part 1.

For $Ax^2 + Bxy + Cy^2$, the axis angle satisfies $\tan 2\theta = \frac{B}{A - C}$:
$\tan 2\theta = \frac{12}{6 - 11} = -2.4$
$2\theta = -67.4^\circ \Rightarrow \theta = -33.7^\circ$, i.e. the axis makes $33.7^\circ$ with the x-axis.
$\boxed{33.7^\circ}$

Part 2.

The eigenvalues of $\begin{bmatrix} 6 & 6 \\ 6 & 11 \end{bmatrix}$ satisfy $\lambda^2 - 17\lambda + 30 = 0 \Rightarrow \lambda = 15, 2$.
The rotated form is $15u^2 + 2v^2 = 17$, so the major semi-axis is $a = \sqrt{\frac{17}{2}} = 2.915$.
Distance between vertices:
$2a = 2(2.915)$
$\boxed{5.83}$

Part 3.

From the rotated form $15u^2 + 2v^2 = 17$: $a^2 = \frac{17}{2} = 8.5$, $b^2 = \frac{17}{15} = 1.133$.
Eccentricity:
$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1.133}{8.5}}$
$\boxed{e = 0.93}$

Question Bank: t703

MSTE - Analytic Geometry / Analytic Geometry / Gemini mapped Chapter 4 to 6

The equation of the curve takes the form of Ax^2 + By^2 + F = 0. If it passes through (10, 6) and (38, 22).

Identify the curve.

  1. circle
  2. hyperbola
  3. ellipse
  4. parabola

What is the length of the latus rectum?

  1. 7.1
  2. 6.8
  3. 9.8
  4. 8.5

What is the eccentricity of the curve?

  1. 0.66
  2. 1.5
  3. 0.5
  4. 2

Part 1.

Substitute both points into $Ax^2 + By^2 + F = 0$:
$100A + 36B + F = 0$ and $1444A + 484B + F = 0$.
Subtracting: $1344A + 448B = 0 \Rightarrow B = -3A$.
Since $A$ and $B$ have opposite signs, the curve is a hyperbola.
$\boxed{\text{hyperbola}}$

Part 2.

With $B = -3A$ and $F = 8A$, the equation reduces to $x^2 - 3y^2 + 8 = 0$, i.e. $\frac{y^2}{8/3} - \frac{x^2}{8} = 1$.
So $a^2 = \frac{8}{3}$ ($a = 1.633$) and $b^2 = 8$.
Latus rectum:
$LR = \frac{2b^2}{a} = \frac{2(8)}{1.633}$
$\boxed{LR = 9.8}$

Part 3.

For the hyperbola $\frac{y^2}{8/3} - \frac{x^2}{8} = 1$: $a^2 = \frac{8}{3}$, $b^2 = 8$.
$c = \sqrt{a^2 + b^2} = \sqrt{\tfrac{8}{3} + 8} = \sqrt{\tfrac{32}{3}} = 3.266$
Eccentricity:
$e = \frac{c}{a} = \frac{3.266}{1.633}$
$\boxed{e = 2}$

Question Bank: t706

MSTE - Analytic Geometry / Analytic Geometry / Gemini mapped Chapter 4 to 6

Given the hyperbola x^2 - 8x - 4y^2 - 16y = 256:

What is the distance between foci?

  1. 35.78
  2. 43.23
  3. 38.41
  4. 40.22

Find the coordinate of the center.

  1. (4, -2)
  2. (-4, 2)
  3. (-4, -2)
  4. (4, 2)

What is the distance between the vertices?

  1. 33
  2. 32
  3. 30
  4. 31

Part 1.

Complete the square: $(x-4)^2 - 4(y+2)^2 = 256$, i.e. $\frac{(x-4)^2}{256} - \frac{(y+2)^2}{64} = 1$.
$a^2 = 256$, $b^2 = 64$, so $c = \sqrt{a^2 + b^2} = \sqrt{320} = 17.889$.
Distance between foci:
$2c = 2(17.889)$
$\boxed{35.78}$

Part 3.

From $\frac{(x-4)^2}{256} - \frac{(y+2)^2}{64} = 1$, the transverse semi-axis is $a = \sqrt{256} = 16$.
Distance between vertices:
$2a = 2(16)$
$\boxed{32}$

Question Bank: t709

MSTE - Analytic Geometry / Analytic Geometry / Gemini mapped Chapter 4 to 6

Given the equilateral hyperbola $xy = 8$:

What is the length of the conjugate axis of the hyperbola?

  1. 12
  2. 16
  3. 8
  4. 4

How far apart are the vertices of the hyperbola?

  1. 4
  2. 8
  3. 16
  4. 12

Determine the eccentricity of the hyperbola.

  1. 1.414
  2. 1.732
  3. 1.368
  4. 1.521

Part 1.

The rectangular hyperbola $xy = 8$ rotates to $\frac{X^2}{2(8)} - \frac{Y^2}{2(8)} = 1$, so $a^2 = b^2 = 16$ and $a = b = 4$.
Conjugate axis length:
$2b = 2(4)$
$\boxed{8}$

Part 2.

With $a = 4$, the distance between vertices (transverse axis) is:
$2a = 2(4)$
$\boxed{8}$

Part 3.

Every rectangular (equilateral) hyperbola has $a = b$, so:
$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{2}$
$\boxed{e = 1.414}$

Question Bank: t712

MSTE - Analytic Geometry / Analytic Geometry / Gemini mapped Chapter 4 to 6

Given the equation of the curve: $9x^2 - 4y^2 - 36x + 8y = 4$.

The center of the curve is at:

  1. (- 2, 1)
  2. (2, 1)
  3. (-2, -1)
  4. (2, -1)

What is the eccentricity of the curve?

  1. 0.768
  2. 1.332
  3. 1.803
  4. 1.765

What is the equation of the upward asymptote of the curve?

  1. $2x - 3y + 4 = 0$
  2. $3x - 2y - 4 = 0$
  3. $3x - 2y + 4 = 0$
  4. $2x - 3y - 4 = 0$

Part 1.

Complete the square:
$9(x-2)^2 - 4(y-1)^2 = 36 \Rightarrow \frac{(x-2)^2}{4} - \frac{(y-1)^2}{9} = 1$
Center $(h, k)$:
$\boxed{(2,\ 1)}$

Part 2.

From $\frac{(x-2)^2}{4} - \frac{(y-1)^2}{9} = 1$: $a^2 = 4$, $b^2 = 9$.
$c = \sqrt{a^2 + b^2} = \sqrt{13} = 3.606$
$e = \frac{c}{a} = \frac{3.606}{2}$
$\boxed{e = 1.803}$

Part 3.

Asymptotes pass through the center $(2,1)$ with slopes $\pm\frac{b}{a} = \pm\frac{3}{2}$.
The upward (positive-slope) one:
$y - 1 = \frac{3}{2}(x - 2) \Rightarrow 2y - 2 = 3x - 6$
$\boxed{3x - 2y - 4 = 0}$

Question Bank: t760

MSTE - Analytic Geometry / Analytic Geometry / Gemini mapped Chapter 4 to 6

The function f(x) = (2x^2 + 6x - 9) / (x^2 - x - 2) will have vertical asymptote at:

  1. x = 0
  2. x = -2
  3. x = 2
  4. x = 1
Vertical asymptotes occur where the denominator is zero (and the numerator is not):
$x^2 - x - 2 = (x - 2)(x + 1) = 0 \Rightarrow x = 2, -1$
The numerator $2x^2 + 6x - 9$ is nonzero at both. Among the choices, the valid asymptote is:
$\boxed{x = 2}$