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Standard Equations of Hyperbola

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Standard Equations of Hyperbola - Lesson Summary

A hyperbola is the set of all that points in a plane such that the difference of their distances from two fixed points in the plane is constant.

P1F2 - P1F1 = P2F2 - P2F1 = P3F1 - P3F2

Foci, vertices, centre, transverse axis and conjugate axis of a hyperbola are as shown in the figure.

In a hyperbola, b2 = c2 - a2,

Where

c = Distance of a focus from the centre

a = Length of semi-transverse axis

b = Length of semi-conjugate axis

The difference between the distances of any point on a hyperbola from its foci is equal to the length of the transverse axis of the hyperbola.

P1F2 - P1F1 = P2F2 - P2F1 = P3F1 - P3F2 = 2a

Eccentricity is a measure of the deviation of a conic section from being a circle.

The eccentricity of a hyperbola is denoted by E, and is equal to the ratio of the distance of a focus from its centre and the length of its semi-transverse axis.

Eccentricity of a hyperbola (e) = c/a


Þ c = ea

Thus, the distance of a focus from the centre of a hyperbola is equal to the product of its eccentricity and the length of its semi-transverse axis.


Standard Equations of a Hyperbola

Consider a hyperbola with its centre at the origin and its transverse axis along the X- or the Y-axis, and then there are two distinct possibilities.

In the first case, we can have a hyperbola with its centre lying on the origin and its transverse axis along the X-axis. In this case, the foci of the hyperbola lie on the X-axis.

In the second case, we can have a hyperbola with its centre lying on the origin and its transverse axis along the Y-axis. In this case, the foci of the hyperbola lie on the Y-axis.

Case (i)


Equation of the hyperbola:

Consider a point P on the hyperbola with the coordinates X, Y.

Then, by definition, the difference of the distances PF1 and PF2 is constant and equal to 2A.

PF1 - PF2 = 2a ……(1)

The distance between the points (x1, y1) and (x2, y2) = √(x2 - x1)2 + (y2 - y1)2

Using distance formula, PF1 = √{(x - (-c)}2 + (y - 0)2

⇒ PF1 = √(x + c)2 + y2 ……(2)

Using distance formula, PF2 = √(x - c)}2 + (y - 0)2

⇒ PF2 = √(x - c)2 + y2 ……(3)

From equations (1), (2) and (3), we get

√(x + c)2 + y2 - √(x - c)2 + y2 = 2a

⇒ √(x + c)2 + y2 = 2a + √(x - c)2 + y2

Squaring both sides,

(x + c)2 + y2 = 4a2 + 4a√(x - c)2 + y2 + (x - c)2 + y2

⇒ 4a√(x - c)2 + y2 = (x + c)2 - (x - c)2 - 4a2

⇒ 4a√(x - c)2 + y2 = x2 + c2 + 2xc - x2 - c2 + 2xc - 4a2

⇒ 4a√(x - c)2 + y2 = 4xc - 4a2

a√(x - c)2 + y2 = xc - a2

⇒ √(x - c)2 + y2 = c/a x - a

Squaring both sides,

(x - c)2 + y2 = a2 + c2/a2 x - 2cx

x2 + c2 - 2cx + y2 = a2 + c2/a2 x2 - 2cx

x2 + c2 + y2 = a2 + c2/a2 x2

x2 - c2/a2 x2 + y2 = a2 - c2

⇒(1 - c2/a2)x2 + y2 = a2 - c2

⇒((a2 - c2)/a2) x2 + y2 = a2 - c2 ……(4)

⇒( - b2/a2) x2 + y2 = -b2 (Since b2 = c2 - a2)

Multiplying both sides by -1/b2,

x2/a2 - y2/b2 = 1 ……(5)

Therefore, equation x2/a2 - y2/b2 = 1 represents the given hyperbola if the coordinates of point P as derived from this equation satisfy the geometrical condition (PF1 - PF2) = 2a.

From equation (5),

y2 = b2 (x2/a2 - 1)

From equation (2),

PF1 = √(x + c)2 + y2

⇒ PF1 = √(x + c)2 + b2 (x2/a2 - 1)

= √ x2 + c2 + 2cx + b2x2/a2 - b2

= √ x2(1+ b2/a2 + 2cx + c2 - b2

= √ x2((a2+b2)/a2)+ 2cx + c2 - b2

= √c2/a2 .x2+ 2cx + a2 (Since b2 = c2 - a2)

= √(a + cx/a)2

⇒ PF1 = a + cx/a

From equation (3),

PF2 = √(x - c)2 + y2

⇒ PF1 = √(x - c)2 + b2 (x2/a2 - 1)

= √ x2 + c2 - 2cx + b2x2/a2 - b2

= √ x2(1+ b2/a2) - 2cx + c2 - b2

= √ x2( (a2+ b2)/a2) - 2cx + c2 - b2

= √c2/a2 . x2 - 2cx + a2 (Since b2 = c2 - a2)

= √( c/a x - a)2

⇒ PF2 = c/a x - a

⇒ PF1 - PF2 = a + c/a . x - c/a . x + a = 2a

Case I: The equation, x2/a2 - y2/b2 = 1, is the standard equation for a hyperbola with its centre at the origin and the transverse axis lying along the X-axis.

Case II: Similarly, the standard equation for a hyperbola with its centre at the origin and the transverse axis along the Y-axis is y2/a2 - x2/b2 = 1.

These two equations are called the standard equations of a hyperbola having its centre at the origin and the transverse axis along the X- or Y-axis.

If the coefficient of x2 is positive, then the transverse axis lies along the X-axis.

If the coefficient of y2 is positive, then the transverse axis lies along the Y-axis.

Note that both the standard equations of a hyperbola contain even powers of x and y.


A hyperbola with its centre lying on the origin and the transverse axis along either the X- or the Y-axis is symmetrical about both the coordinate axes.

⇒ If (x, y) lies on the hyperbola, then (-x, y), (x, -y) and (-x, -y) also lie on it.

If a = b, then the hyperbola is called an equilateral hyperbola.

Case I: Centre (0, 0), transverse axis along the X-axis.

x2/a2 - y2/b2 = 1

⇒ x2/a2  = 1 + y2/b2

⇒ |x/a | ≥ 1

⇒x ≤ -a or x ≥ a

⇒ No point on the curve of the hyperbola lies between the lines x = -a and x = a.

Case II: Centre (0, 0), transverse axis along the Y-axis.

    y2/a2 - x2/b2 = 1

⇒ y2/a2 = 1 + x2/b2

⇒ |y/a |≥ 1

⇒ y ≤ -a or y ≥ a

⇒ No point on the curve of the hyperbola lies between the lines y = -a and y = a.

Expression for Length of Latus Rectum

A line segment passing through a focus and perpendicular to the transverse axis, and with its end points lying on the curve of a hyperbola, is called its latus rectum.

Since a hyperbola has two foci, it has two latus recta.

Consider a hyperbola with its centre at the origin and the transverse axis along the X-axis and latus recta AB and CD.

Let AF2 = l

⇒ Coordinates of A are (c, l), where

c = Distance of focus F2 from centre O

Equation of the given hyperbola: x2/a2 - y2/b2 = 1

Sine A (c, l) lies on x2/a2 - y2/b2 = 1

⇒ c2/a2 - l2/b2 = 1

⇒ l2 = b2(c2/a2 - 1)

⇒ l2 = b2((c2-a2) / a2)

⇒ l2 = b2( b2/ a2) (Since b2 = c2 - a2)

⇒ l2 = b4/ a2

⇒ l = b2/a

AF2 = F2B = b2/a

⇒ AB = AF2 + F2B = 2b2/a

Length of latus rectum of a hyperbola = 2b2/a,

Where,

a = Length of semi-transverse axis

b = Length of semi-conjugate axis.

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