Table of Contents
Last modified on September 5th, 2024
Let us imagine a person holding two strings, each tied to a different peg in the ground. Now, if we stretch these strings to their maximum length while keeping them taut and then move a pencil around in such a way that the difference in distance from the pencil to each peg is always the same, the shape that forms is called a hyperbola.
Mathematically, a hyperbola is a type of conic section that results when a plane intersects both halves of a double right circular cone at an angle. This intersection of the plane and cone generates two unbounded curves that are mirror images. These curves, known as branches, together form the hyperbola.
A hyperbola can also be defined as a curved plane formed by a point moving in such a way that the difference in its distances to two fixed points (called foci) remains constant.
Nuclear plant cooling towers, certain buildings, and the orbits of some celestial objects are a few real-life examples of hyperbola.
It is the ratio of the distance from the center to the focus (c units) to the distance from the center to the vertex (a units). This value is always greater than 1 (e > 1). Eccentricity can also be defined as the ratio of the distance from any point on the hyperbola to the focus, compared to its distance from the directrix – a line perpendicular to the hyperbola’s axis of symmetry and parallel to its latus rectum.
The formula to calculate the eccentricity of a hyperbola is:
Based on its center and the orientation of its branches, the equation of a hyperbola can be written in two forms: standard and parametric.
The standard form of the equation of a horizontal hyperbola centered at (0, 0) is given by:
${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$
Here,
Thus, a horizontal hyperbola has its transverse axis aligned along the x-axis.
The standard form of the equation of a vertical hyperbola centered at (0, 0) is given by:
${\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1}$
Here,
Thus, a vertical hyperbola has its transverse axis aligned along the y-axis.
Let us consider a point P on the hyperbola, and the foci are F and F’
As we know, the difference in distances from any point on the hyperbola to its foci is always constant.
Thus, the difference in the distance from P to the foci F and F’ is 2a
PF’ – PF = 2a …..(i)
If the coordinates of P is (x, y), F is (c, 0), and F’ is (-c, 0)
Thus, PF’ = ${\sqrt{\left( x+c\right) ^{2}+y^{2}}}$ and PF = ${\sqrt{\left( x-c\right) ^{2}+y^{2}}}$
Now, substituting the values of PF’ and PF in (i), we get
${\sqrt{\left( x+c\right) ^{2}+y^{2}}-\sqrt{\left( x-c\right) ^{2}+y^{2}}=2a}$
⇒ ${\sqrt{\left( x+c\right) ^{2}+y^{2}}=2a+\sqrt{\left( x-c\right) ^{2}+y^{2}}}$
Squaring both sides and simplifying, we get
⇒ ${\left( x+c\right) ^{2}+y^{2}=4a^{2}+\left( x-c\right) ^{2}+y^{2}+4a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$
⇒ ${x^{2}+c^{2}+2cx+y^{2}=4a^{2}+x^{2}+c^{2}-2cx+y^{2}+4a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$
⇒ ${2cx=4a^{2}-2cx+4a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$
⇒ ${4cx-4a^{2}=4a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$
⇒ ${cx-a^{2}=a\sqrt{\left( x-c\right) ^{2}+y^{2}}}$
Again, squaring both sides and simplifying further, we get
⇒ ${\left( cx-a^{2}\right) ^{2}=a^{2}\left\{ \left( x-c\right) ^{2}+y^{2}\right\}}$
⇒ c2x2 + a4 – 2a2cx = a2(x2 + c2 – 2cx + y2)
⇒ c2x2 + a4 – 2a2cx = a2x2 + a2c2 – 2a2cx + a2y2
⇒ c2x2 + a4 = a2x2 + a2c2 + a2y2
⇒ c2x2 – a2x2 – a2y2 = a2c2 – a4
⇒ x2(c2 – a2) – a2y2 = a2(c2 – a2)
Dividing both sides by a2(c2 – a2),
⇒ ${\dfrac{x^{2}\left( c^{2}-a^{2}\right) }{a^{2}\left( c^{2}-a^{2}\right) }-\dfrac{a^{2}y^{2}}{a^{2}\left( c^{2}-a^{2}\right) }=\dfrac{a^{2}\left( c^{2}-a^{2}\right) }{a^{2}\left( c^{2}-a^{2}\right) }}$
⇒ ${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{c^{2}-a^{2}}=1}$ …..(ii)
Since c2 = a2 + b2, so from (ii), we get
${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$, which is the standard equation of the horizontal hyperbola centered at (0, 0).
Similarly, we can also derive the equation of a vertical hyperbola.
Graph the hyperbola ${\dfrac{x^{2}}{25}-\dfrac{y^{2}}{144}=1}$ and mark its vertices, co-vertices, foci, and asymptotes.
Given, ${\dfrac{x^{2}}{25}-\dfrac{y^{2}}{144}=1}$
Here,
a = 5, b = 12, and the transverse axis is along the x-axis.
As we know, the standard equation of a hyperbola with the center (0, 0) is
${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$
Here,
The coordinates of the vertices are (±a, 0) = (±5, 0)
The coordinates of the co-vertices are (0, ±b) = (0, ±12)
Since c2 = a2 + b2
⇒ c = ${\sqrt{a^{2}+b^{2}}}$
⇒ c = ${\sqrt{5^{2}+12^{2}}}$ = ${\sqrt{25+144}}$ = ${\sqrt{169}}$ = 13
The coordinates of the foci are (±c, 0) = (±13, 0)
The equations of the asymptotes are y = ${\pm \dfrac{b}{a}x}$ = ${\pm \dfrac{12}{5}x}$
By plotting the vertices, co-vertices, foci, and asymptotes on the curves, we get the hyperbola alongside.
The standard form of the equation of a horizontal hyperbola centered at (h, k) is given by:
${\dfrac{\left( x-h\right) ^{2}}{a^{2}}-\dfrac{\left( y-k\right) ^{2}}{b^{2}}=1}$
Here,
The standard form of the equation of a vertical hyperbola centered at (h, k) is given by:
${\dfrac{\left( y-k\right) ^{2}}{a^{2}}-\dfrac{\left( x-h\right) ^{2}}{b^{2}}=1}$
Here,
To graph a hyperbola, if its equation is not in the standard form, we convert the equation to standard form by completing the square.
Graph the hyperbola given by the equation 25x2 – 16y2 – 150x + 160y – 175 = 400. Find its center, vertices, co-vertices, foci, and asymptotes.
Given, 25x2 – 16y2 – 150x + 160y – 175 = 400
Converting the equation in the standard form,
25x2 – 150x – 16y2 + 160y – 175 = 400
⇒ (5x)2 – 2 ⋅ 5x ⋅ 15 + (15)2 – {(4y)2 – 2 ⋅ 4y ⋅ 20 + (20)2} = 400
⇒ (5x – 15)2 – (4y – 20)2 = 400
⇒ 25(x – 3)2 – 16(y – 5)2 = 400
⇒ ${\dfrac{25\left( x-3\right) ^{2}}{400}-\dfrac{16\left( y-5\right) ^{2}}{400}=\dfrac{400}{400}}$
⇒ ${\dfrac{\left( x-3\right) ^{2}}{16}-\dfrac{\left( y-5\right) ^{2}}{25}=1}$, which is the standard equation of a hyperbola in the form of ${\dfrac{\left( x-h\right) ^{2}}{a^{2}}-\dfrac{\left( y-k\right) ^{2}}{b^{2}}=1}$
Here, a = 4, b = 5, and the transverse axis is parallel to the x-axis.
Now,
The center is (h, k) = (3, 5)
The coordinates of the vertices are (h ± a, k) = (3 ± 4, 5) = (-1, 5) or (7, 5)
The coordinates of the co-vertices are (h, k ± b) = (3, 5 ± 5) = (3, 0) or (3, 10)
Since c2 = a2 + b2 ⇒ c = ${\sqrt{a^{2}+b^{2}}}$ = ${\sqrt{4^{2}+5^{2}}}$ = ${\sqrt{41}}$
The coordinates of the foci are (h ± c, k)= (3 ± ${\sqrt{41}}$, 5) = (3 + ${\sqrt{41}}$, 5) or (3 – ${\sqrt{41}}$, 5)
The equations of the asymptotes are y = ${\pm \dfrac{b}{a}\left( x-h\right) +k}$ = ${\pm \dfrac{5}{4}\left( x-3\right) +5}$
By plotting the center, vertices, co-vertices, foci, and asymptotes on the curves, we get the hyperbola alongside.
The parametric form of a hyperbola is a way to express the coordinates of any point on the hyperbola using a single parameter, typically denoted by 𝑡.
The parametric form of the hyperbola is written as:
x = a sec(t)
y = b tan(t), here 0° ≤ t ≤ 360°
This form is particularly useful for simplifying the process of graphing the hyperbola and solving problems.
If (x0, y0) are the coordinates of the center of the hyperbola, here is the summary of the formulas describing it:
| Property | Formula |
|---|---|
| Equation of Hyperbola | ${\dfrac{\left( x-x_{0}\right) ^{2}}{a^{2}}-\dfrac{\left( y-y_{0}\right) ^{2}}{b^{2}}=1}$ |
| Major-axis | y = y0, and its length is 2a |
| Minor-axis | x = x0, and its length is 2b |
| Eccentricity | ${e=\dfrac{\sqrt{a^{2}+b^{2}}}{a}}$ |
| Asymptotes | ${y=y_{0}-\left( \dfrac{b}{a}\right) x+\left( \dfrac{b}{a}\right) x_{0}}$ and ${y=y_{0}+\left( \dfrac{b}{a}\right) x-\left( \dfrac{b}{a}\right) x_{0}}$ |
| Vertex | (a, y0) and (-a, y0) |
| Focus (foci) | ${\left( x_{0}+\sqrt{a^{2}+b^{2}},y_{0}\right)}$ and ${\left( x_{0}-\sqrt{a^{2}+b^{2}},y_{0}\right)}$ |
| Latus rectum | ${\dfrac{2b^{2}}{a}}$ |
If the equation of the hyperbola is ${\dfrac{\left( x-4\right) ^{2}}{9}-\dfrac{\left( y+2\right) ^{2}}{16}=1}$, find its asymptote.
Given, the equation of the hyperbola is ${\dfrac{\left( x-4\right) ^{2}}{9}-\dfrac{\left( y+2\right) ^{2}}{16}=1}$
As we know, the formulas to find asymptotes are:
${y=y_{0}-\left( \dfrac{b}{a}\right) x+\left( \dfrac{b}{a}\right) x_{0}}$ and ${y=y_{0}+\left( \dfrac{b}{a}\right) x-\left( \dfrac{b}{a}\right) x_{0}}$
Here,
a = 3, b = 4, and (x0, y0) = (4, -2)
Now, the asymptotes are:
${y=-2-\left( \dfrac{4}{3}\right) x+\left( \dfrac{4}{3}\right) 4}$ = ${-\left( \dfrac{4}{3}\right) x+\dfrac{10}{3}}$
${y=-2+\left( \dfrac{4}{3}\right) x-\left( \dfrac{4}{3}\right) 4}$ = ${\left( \dfrac{4}{3}\right) x-\dfrac{22}{3}}$
Thus, the asymptotes of the hyperbola are ${-\left( \dfrac{4}{3}\right) x+\dfrac{10}{3}}$ and ${\left( \dfrac{4}{3}\right) x-\dfrac{22}{3}}$
Last modified on September 5th, 2024