A) \[{{x}^{2}}-{{y}^{2}}={{p}^{2}}{{q}^{2}}\]
B) \[{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=pq\]
C) \[{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=\frac{1}{{{p}^{2}}{{q}^{2}}}\]
D) \[{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}\]
Correct Answer: D
Solution :
| [d] We know, \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\] |
| and \[\sec \theta =\frac{x}{p}\]and \[\tan \theta =\frac{y}{q}\] |
| \[\frac{{{x}^{2}}}{{{p}^{2}}}-\frac{{{y}^{2}}}{{{q}^{2}}}=1\] |
| \[{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}\] |
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