A) \[y=\frac{32x}{9}\]
B) \[x=\frac{16}{9}y\]
C) \[y=\frac{16}{9}x\]
D) None of these
Correct Answer: A
Solution :
The diameters\[y={{m}_{1}}x\]and\[y={{m}_{2}}x\]are conjugate diameters of the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\], if\[{{m}_{1}}{{m}_{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\] Given equation is \[\frac{{{x}^{2}}}{9}\cdot \frac{{{y}^{2}}}{16}=1\] Here, \[{{a}^{2}}=9,\,\,{{b}^{2}}=16\] Also given,\[x=2y\] Then, \[{{m}_{1}}=\frac{1}{2}\] \[\because \] \[\frac{1}{2}\times {{m}_{2}}=\frac{16}{9}\] \[\Rightarrow \] \[{{m}_{2}}=\frac{32}{9}\] Hence, the required equation of diameter is \[y=\frac{32x}{9}\]
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