Question

Tangents are drawn to a hyperbola from a point on one of the branches of its conjugate hyperbola

Answer 1

The hyperbola, whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola, is called the conjugate hyperbola of the given hyperbola, and the two hyperbolas are conjugate to one another. Thus, the hyperbolas x2/a – y2/b2 = 1 and y2/b2 – x2/a2 = 1 are conjugate hyperbolas.

Solved Example 1: 

Tangents are drawn to a hyperbola from any point on one of the branches of the conjugate hyperbola. Show that their chord of contact will touch the other branch of the conjugate hyperbola.

Solution:            

Let the hyperbola be x2/a2 –y2/b2 = 1. So its conjugate hyperbola is x2/a2 – y2/b2 = –1. Let any point on it be (a tan θ, b sec θ). Now equation of chord of contact will be x/a tan θ – y/b sec θ = 1 => x/a (–tan θ) – y/b (–sec θ) = –1. So, it is a tangent to the conjugate hyperbola at point (–a tan θ, – b sec θ) which will obviously on the other branch.

Solved Example 2: 

If e1 and e2 are the eccentricities of the hyperbola x2/a2– y2/b2 = 1 and its conjugate hyperbola, prove that e1-2+ e2-2 = 1.

Solution:             

The eccentricity e1 of the given hyperbola is obtained from

    b2 = a2(e12 - 1)                                                  ….(1)

 

The eccentricity e2 of the conjugate hyperbola is given by

 

                              a2 = b2 (e22 -1)                                                             ….(2)

 

                              Multiply (1) and (2).

 

                              We get 1 = (e12 - 1) (e22 - 1)

 

                              => 0 = e12 e22 - e12 - e22 => e1-2 + e2-2 = 1 

Answer 2

Answer:

Step-by-step explanation:

Hi,

Let the hyperbola be x²/a² –y²/b² = 1.

So its equation of conjugate hyperbola will be x²/a² –y²/b² = –1.

Let any point on the conjugate hyperbola be P (a tan θ, b sec θ).

Now equation of chord of contact drawn to the hyperbola from P will be

x/a tan θ – y/b sec θ = 1

which can be rewritten in the form

x/a (–tan θ) – y/b (–sec θ) = –1.

which is clearly a tangent to the conjugate hyperbola at point

(–a tan θ, – b sec θ) which is obviously on the other branch.

Hence, the chord of contact touches the conjugate hyperbola.

Hope, it helps!