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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
$\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$
$\frac{{{l^2}}}{{{a^2}}} - \frac{{{m^2}}}{{{b^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$
$\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$
$\frac{{{l^2}}}{{{a^2}}} + \frac{{{m^2}}}{{{b^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$
Solution
(a) Any normal to the hyperbola is
$\frac{{ax}}{{\sec \theta }} + \frac{{by}}{{\tan \theta }} = {a^2} + {b^2}$ …..$(i)$
But it is given by $lx + my – n = 0$…..$(ii)$
Comparing $(i)$ and $(ii),$ we get
$\sec \theta = \frac{a}{l}\left( {\frac{{ – n}}{{{a^2} + {b^2}}}} \right)$
and $\tan \theta = \frac{b}{m}\left( {\frac{{ – n}}{{{a^2} + {b^2}}}} \right)$
Hence eliminating $\theta $, we get
$\frac{{{a^2}}}{{{l^2}}} – \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$.
Similar Questions
Let $H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$, where $a > b >0$, be $a$ hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4 \sqrt{3}$..
| List $I$ | List $II$ |
| $P$ The length of the conjugate axis of $H$ is | $1$ $8$ |
| $Q$ The eccentricity of $H$ is | $2$ ${\frac{4}{\sqrt{3}}}$ |
| $R$ The distance between the foci of $H$ is | $3$ ${\frac{2}{\sqrt{3}}}$ |
| $S$ The length of the latus rectum of $H$ is | $4$ $4$ |
The correct option is:
