Question

Pn is the ordinate of any point p on the hyperbola x^2/a^2-y^2/b^2=1. If q divides ap in the ratio a^2:b^2, show that nq is perpendicular to ap, where aa' the transverse axis of the hyperbola.

Answer 1

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Answer 2

Answer:

Given:  

PN is the ordinate of any point P on the hyperbola

$\frac { { x }^{ 2 } }{ { a }^{ 2 } } \quad -\quad \frac { { y }^{ 2 } }{ { b }^{ 2 } } \quad =\quad 1$

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Solution:  

P and Q are the two points.

Assume that Co-ordinate of P be (x, y)

Similarly, for Q, the co-ordinates can be,

$\left\{ \left( \frac { a\quad +\quad x }{ 2 }  \right) ,\quad \left( \frac { y }{ 2 } \right)\right\}$

Q can also be represented as below,

$\left\{ \left( \frac { { a }^{ 2 }x\quad +\quad { b }^{ 2 }a }{ { a }^{ 2 }\quad +\quad { b }^{ 2 } } \right) ,\quad \left( \frac { { b }^{ 2 }y }{ { a }^{ 2 }\quad +\quad { b }^{ 2 } } \right) \right\}$

On equating the above two equations,

$\left\{ \left( \frac { a\quad +\quad x }{ 2 } \right) ,\quad \left( \frac { y }{ 2 } \right) \right\} \quad =\quad \left\{ \left( \frac { { a }^{ 2 }x\quad +\quad { b }^{ 2 }a }{ { a }^{ 2 }\quad +\quad { b }^{ 2 } } \right) ,\quad \left( \frac { { b }^{ 2 }y }{ { a }^{ 2 }\quad +\quad { b }^{ 2 } } \right) \right\}$

Equating for y coefficient,

$\Rightarrow \quad \frac { 1 }{ 2 } \quad =\quad \frac { { b }^{ 2 } }{ { a }^{ 2 }\quad +\quad { b }^{ 2 } }$

$\Rightarrow \quad { a }^{ 2 }\quad +\quad { b }^{ 2 }\quad =\quad 2{ b }^{ 2 }$

$\Rightarrow \quad { a }^{ 2 }\quad =\quad { b }^{ 2 }$

$\Rightarrow \quad a\quad =\quad b$

If a = b, then Q dividing AP into two equal halves.

Hence, it is proved that QN is perpendicular to AP.