Question

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

• A. $a^{2}l^{2} - b^{2}m^{2} = 0$
• B. $a^{2}l^{2} + b^{2}m^{2} = 0$
• C. $a^{2}l^{2} + b^{2}m^{2} = n^{2}$
• D. None of these

Answer 1

Answer: (A)

$a^{2}l^{2} - b^{2}m^{2} = 0$

Answer 2

The extremities of a pair of conjugate diameters of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ are $(a\sec\varphi,b\tan\varphi)$ and $(a\tan\varphi,b\sec\varphi)$ respectively.

According to the question, since extremities of a pair of conjugate diameters lie on $lx + my + n = 0$

$\therefore$ $l(a\sec\varphi) + m(b\tan\varphi) + n = 0$ ⇒ $l(a\tan\varphi) + m(b\sec\varphi) + n = 0$……(i)

Then from (i), $al\sec\varphi + bm\tan\varphi = - n$ or

$a^{2}l^{2}\sec^{2}\varphi + b^{2}m^{2}\tan^{2}\varphi + 2ablm\sec\varphi\tan\varphi = n^{2}$ ……(ii)

And from (ii), $al\tan\varphi + bm\sec\varphi = - n$ or

$a^{2}l^{2}\tan^{2}\varphi + b^{2}m^{2}\sec^{2}\varphi + 2ablm\sec\varphi\tan\varphi = n^{2}$……(iii)

Then subtracting (ii) from (iii)

$\therefore$ $a^{2}l^{2}(\sec^{2}\varphi - \tan^{2}\varphi) + b^{2}m^{2}(\tan^{2}\varphi - \sec^{2}\varphi) = 0$ or

$a^{2}l^{2} - b^{2}m^{2} = 0$