Conic Sections

Equation of Hyperbola

A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant. ‘Difference’ means the distance to the ‘farther’ point minus the distance to the ‘closer’ point. The two fixed points are the foci and the mid-point of the line segment joining the foci is the centre of the hyperbola.

The line through the foci is called the transverse axis. Also, the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola.

hyperbola

  • The distance between the two foci is: 2c
  • The distance between two vertices is: 2a (this is also the length of the transverse axis)
  • The length of the conjugate axis is 2b … where b = √ (c2 – a2)

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Finding the Constant P1F2 – P1F1

Refer the diagram below:

hyperbola

We take a point P at A and B as shown above. Therefore, by the definition of a hyperbola, we have

BF1 – BF2 = AF2 – AF1
∴ BA + AF1 – BF2 = AB + BF2 – AF1

Solving the equation, we get, AF1 = BF2
Hence, BF1 – BF2 = BA + AF1 – BF2 = BA = 2a.

Browse more Topics Under Conic Sections

Eccentricity

Like in the ellipse, e = c/a is the eccentricity in a hyperbola. Also, ‘c’ is always greater than or equal to ‘a’. Hence, the eccentricity is never less than one.

Standard Equation of Hyperbola

When the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis, then the equation of the hyperbola is the simplest. Here are two such possible orientations:

hyperbola

Of these, let’s derive the equation for the hyperbola shown in Fig.3 (a) with the foci on the x-axis. Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2.

Also, let O be the origin and the line through O through F2 be the positive x-axis and that through F1 as the negative x-axis. The line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c,0) and F2 be (c,0) shown in Fig. 3 (a).

Derivation of the Equation

hyperbola

Now, we take a point P(x, y) on the hyperbola such that, PF1 – PF2 = 2a

By the distance formula, we have,
√ {(x + c)2 + y2} – √ {(x – c)2 + y2} = 2a
Or, √ {(x + c)2 + y2} = 2a + √ {(x – c)2 + y2}

Further, let’s square both the sides. Hence, we have
(x + c)2 + y2 = 4a2 + 4a√ {(x – c)2 + y2} + (x – c)2 + y2

On simplifying the equation, we get
√ {(x – c)2 + y2} = x(c/a) – a

We square both sides again and simplify it further to get,
x2/a2 – y2/(c2 – a2) = 1

We know that c2 – a2 = b2. Therefore, we have
x2/a2 – y2/b2 = 1

Therefore, we can say that any point on the hyperbola satisfies the equation:

x2/a2 – y2/b2 = 1 … (1)

Let’s look at the converse situation now. If P(x, y) satisfies equation (1) with 0 < a < c, then
y2 = b2{(x2 – a2)/a2}

Therefore, PF1 = √ {(x + c)2 + y2}
= √ {(x + c)2 + b2[(x2 – a2)/a2])}

On simplifying the equation, we get PF1 = a + x(c/a)
Using similar calculations for PF2, we get PF2 = a – x(c/a)

In hyperbola c > a; and since P is to the right of the line x = a, x > a, and (c/a)x > a. Therefore, a – (c/a)x becomes negative.
Thus, PF2 = (c/a)x – a
Therefore, PF1 – PF2 = {a + x(c/a)} – {x(c/a) – a} = a + x(c/a) – x(c/a) + a = 2a.

Also, note that if P is to the left of the line x = – a, then,
PF1 = – {a + (c/a)x} and PF2 = a – (c/a)x
In this case, PF1 – PF2 = 2a.
Hence, it is evident that any point that satisfies the equation x2/a2 – y2/b2 = 1, lies on the hyperbola.

Note

Solving the equation, we get

x2/a2 = 1 + y2/b2 ≥ 1

Therefore, no portion of the curve lies between the lines x = + a and x = – a. Similarly, we can derive the equation of the hyperbola in Fig. 3 (b) as

y2/a2 – x2/b2 = 1

These two equations are known as the Standard Equations of Hyperbolas.

Observations

  • A hyperbola is symmetric with respect to both the coordinate axes. In simple words, if (m, n) is a point on the hyperbola, then (- m, n), (m, – n) and (- m, – n) also fall on it.
  • The foci always lie on the transverse axis. The denominator of the positive term gives the transverse axis. For e.g. the transverse axis of x2/9 – y2/16 = 1 is along the x-axis and has length = 2a = 2 x √9 = 2 x 3 = 6.

Latus rectum

Latus rectum of a hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose endpoints lie on the hyperbola. The length of the latus rectum in hyperbola is 2b2/a.

Solved Examples for You

Question: Find the equation of the hyperbola where foci are (0, ±12) and the length of the latus rectum is 36.

Solution: The foci are (0, ±12). Hence, c = 12. Length of the latis rectum = 36 = 2b2/a

∴ b2 = 18a

Hence, from c2 = a2 + b2, we have 122 = a2 + 18a
Or, 144 = a2 + 18a
i.e. a2 + 18a – 144 = 0
Solving it, we get a = – 24, 6

Since ‘a’ cannot be negative, we take a = 6 and so b2 = 36a/2 = (36 x 6)/2 = 108. Therefore, the equation of the required hyperbola is y2/36 – x2/108 = 1.

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