Browse Questions

# The diameter of $16x^2-9y^2=144$ which is conjugate to $x=2y$ is

$\begin{array}{1 1}(A)\;y=\large\frac{16 x}{9} \\(B)\;y= \large\frac{32 x}{9} \\(C)\;x= \large\frac{16 y}{9} \\(D)\;x= \large\frac{32 y}{9} \end{array}$

Diameters $y=m_1x$
$y=m_2x$ are conjugate diameters of the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1. if m_1m_2= \large\frac{b^2}{a^2} Here a^2=9 b^2=16 m_1= \large\frac{1}{2} \therefore m_1m_2=\large\frac{b^2}{a^2} \qquad= \large\frac{1}{2}$$ (m_2)$
$\qquad= \large\frac{16}{9}$
$m_2=\large\frac{32}{9}$
Thus the required diameter is $y= \large\frac{32x }{9}$
Hence B is the correct answer.