CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

What is the point of contact between the hyperbola
x2a2y2b2=1 and
the tangent y=mx±a2m2b2.


  1. [±a2ma2m2b2,±b2a2m2b2]

  2. [±b2a2m2b2,±a2ma2m2b2]

  3. [±a2m2b2,±a2m2+b2]

  4. [±a2m,±b2]


Solution

The correct option is A

[±a2ma2m2b2,±b2a2m2b2]


Solving the equation of hyperbola and tangent will give the point of contact.

x2a2(mx±a2m2b2)2b2=1

b2x2a2(m2x2+a2m2b2±2mxa2m2b2=a2b2

 x=±2ma2a2m2b2±4m2a4(a2m2b2)4(b2a2m2)(a4m2)2(b2a2m2)

=±2ma2[a2m2b2±a2m2b2+b2a2m2]2(b2a2m2)

=±a2ma2m2b2(b2a2m2)

=a2ma2m2b2

y=b2a2m2b2

The required point of contact is
(a2ma2m2b2,b2a2m2b2)

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image