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If the line \mathrm{I x+m y+n=0} passes through the extremities of a pair of conjugate diameters of the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} then

Option: 1

\mathrm{a^2 l^2-b^2 m^2=0}


Option: 2

\mathrm{a^2 l^2+b^2 m^2=0}


Option: 3

\mathrm{a^2 l^2+b^2 m^2=n^2}


Option: 4

None of these


Answers (1)

The extremities of a pair of conjugate diameters of \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad \text { are }\ (a \sec \phi, b \tan \phi)} and \mathrm{(a \tan \phi, b \sec \phi)} respectively.
According to the question, since extremities of a pair of conjugate diameters lie on \mathrm{I x+m y+n=0}
\mathrm{\therefore \quad I(a \sec \phi)+m(b \tan \phi)+n=0 \Rightarrow I(a \tan \phi)+m(b \sec \phi)+n=0\ \ \ ........ (i)}
Then from \mathrm{(i)}, \mathrm{a l \sec \phi+b m \tan \phi=-n \text { or } a^2 l^2 \sec ^2 \phi+b^2 m^2 \tan ^2 \phi+2 a b l m \sec \phi \tan \phi=n^2 \ \ \ ...........(ii)}
And from \mathrm{(ii)}, \mathrm{a l \tan \phi+b m \sec \phi=-n \text { or } a^2 I^2 \tan ^2 \phi+b^2 m^2 \sec ^2 \phi+2 a b l m \sec \phi \tan \phi=n^2\ \ \ ........ (iii)}
Then subtracting \mathrm{(ii)} from \mathrm{(iii)}
\mathrm{\therefore \quad a^2 I^2\left(\sec ^2 \phi-\tan ^2 \phi\right)+b^2 m^2\left(\tan ^2 \phi-\sec ^2 \phi\right)=0 \text { or } \quad a^2 I^2-b^2 m^2=0}

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Ramraj Saini

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