The condition that the straight line lx + my | vedclass

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Question

(a) Any normal to the hyperbola is

$\frac{{ax}}{{\sec 	heta }} + \frac{{by}}{{	an 	heta }} = {a^2} + {b^2}$ .....$(i)$

But it is given by $lx

Vedclass · Accepted Answer

$\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$

Vedclass · Answer

(a) Any normal to the hyperbola is

$\frac{{ax}}{{\sec 	heta }} + \frac{{by}}{{	an 	heta }} = {a^2} + {b^2}$ .....$(i)$

But it is given by $lx + my - n = 0$.....$(ii)$

Comparing $(i)$ and $(ii),$ we get

$\sec 	heta = \frac{a}{l}\left( {\frac{{ - n}}{{{a^2} + {b^2}}}} ight)$

and $	an 	heta = \frac{b}{m}\left( {\frac{{ - n}}{{{a^2} + {b^2}}}} ight)$

Hence eliminating $	heta $, we get

$\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$.

The condition that the straight line $lx + my = n$ may be a normal to the hyperbola ${b^2}{x^2} - {a^2}{y^2} = {a^2}{b^2}$ is given by

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