Examples on Propositions of a Hyperbola

Hyperbola is defined as a conic with an eccentricity greater than 1 [e > 1]. It is the locus of a point where the distance from a fixed point called focus and from the line which is fixed, called directrix, is a constant called eccentricity (denoted as “e”).

(x²/ a²) – (y²/ b²) = 1 where b² = a² (e² – 1)

is the standard equation of a hyperbola. The equation of the hyperbola with focus (h, k), directrix lx + my + n = 0 and eccentricity ‘e’ is

(x – h)² + (y – k)² = e² (lx + my + n)²/(l² + m²)

Propositions of a Hyperbola

The properties and propositions of the hyperbola are as follows:

• The diameter of a hyperbola is the locus of the centre point of a group of parallel chords. It is given by y = b² x / (a² m), and m is the slope of the group of parallel chords.
• The bisection of the chords, which are parallel to each other by the two diameters of the hyperbola, are called conjugate diameters. It is given by y = mx and y = m₁x of the hyperbola (x²/ a²) – (y²/ b²) = 1 are conjugate if mm₁ = b² / a².
• The latus rectum is a chord through one of the foci and at a right angle to the transverse axis of a hyperbola. The length of the latus rectum is given by 2b² /a.
• The locus of the point of intersection of a pair of perpendicular tangents to a hyperbola is the director circle. It is given by (x²/a²) – (y²/b²) = 1 is x² + y² = a² – b²
• A line which is tangent to the hyperbola at a point of infinity but not infinity by itself can be the asymptote of the curve.

Propositions of a Hyperbola Solved Examples for IIT JEE

Example 1: What is the equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13?

Solution:

The conjugate axis is 5, and the distance between foci = 13

2b = 5 and 2ae = 13

b² = a² (e² − 1)

25 / 4 = [(13)² / 4e²] * (e² − 1)

25 / 4 = [169 / 4] − [169 / 4e²] or e² = 169 / 144

e = 13 / 12

or a = 6, b = 5 / 2 or a hyperbola is

25 x² − 144 y² = 900

Example 2: A hyperbola passes through points (3, 2) and (17, 12) and has its centre at the origin and transverse axis along the x-axis. Find the length of its transverse axis.

Solution:

Let the equation of hyperbola is (x²/a²) – (y²/b²) = 1

But it passes through (3, 2)

(9/ a²) − (4/b²) = 1 ….. (i)

Also, it passes through (17, 12)

Solving these, we get a = 1 and b = √2

Hence, the length of the transverse axis = 2a = 2

Example 3: The distance between the foci of a hyperbola is double the distance between its vertices, and the length of its conjugate axis is 6. Find the equation of the hyperbola referred to its axes as axes of coordinates.

Solution:

According to given conditions, 2ae = 2 * 2a or

√[(x − 2)² + y²] = 4 − √[(x − 2)² + y²] and

A = (0, 0) ; B = (4a, 4a)

Hence, a = 3 / √3 = √3

Therefore, equation is x² / 3 − y² / 9 = 1

i.e., 3x² − y² = 9

Example 4: What is the equation of the directrices of the conic x² + 2x − y² + 5 = 0?

Solution:

(x + 1)² − y² −1 + 5 = 0

− [(x + 1)² / 4] + y² / 4 = 1

Equation of directrices of y² / b² − x² / a² = 1 are y = ± be

Here b = 2, e = √1 + 1 = √2

Hence, y = ± 2 / √2

y = ± √2

Example 5: The value of m for which y = mx + 6 is a tangent to the hyperbola (x²/100) − (y²/49) = 1, is ______________.

Solution:

If y = mx + c touches (x²/a²) – (y²/b²) = 1, then c² = a²m² − b².

Here c = 6, a² = 100, b² = 49

∴ 36 = 100m² − 49

⇒ 100m² = 85

⇒ m = √[17 / 20]

Example 6: Find the equation of the axis of the given hyperbola

A) y = x + 1                                   

B) y = x − 1

C) y = x + 2                                  

D) y = x − 2

Solution:

The equation of tangent is equally inclined to the axis, i.e.,  

Equation of tangent

Given equation is

Now, on comparing,

Example 7: If q is the acute angle of intersection at a real point of intersection of the circle x² + y² = 5 and the parabola y² = 4x, then find the value of tan θ.

Solution: 

Solving equations x² + y² = 5 and y² = 4x, we get

x² + 4x – 5 = 0 ⇒ x = 1, -5

For x = 1, y² = 4 ⇒ y = ± 2

For x = -5, y² = -20 (imaginary values) 

The points are (1, 2)(1, 2).

m₁ for x² + y² = 5 at (1, 2) 

Similarly, m₂ for y² = 4x at (1, 2) is 1.

Example 8: If the foci of the ellipse (x²/16) + (y²/b²) = 1 and the hyperbola (x²/144) – (y²/81) = 1/25 coincide, then find the value of b².

Solution:

Hyperbola is 

Therefore, foci

Therefore, the focus of ellipse = (4e, 0), i.e., (3, 0)

Hence,

Ellipse and Hyperbola – JEE Important and Previous Year Questions