A) A straight line
B) A circle
C) An ellipse
D) A hyperbola
Correct Answer: D
Solution :
Here \[\frac{x}{b}=\sec \varphi \] and \[\frac{y}{a}=\tan \varphi \] Therefore\[\frac{{{x}^{2}}}{{{b}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}}={{\sec }^{2}}\varphi -{{\tan }^{2}}\varphi \,\,\Rightarrow \,\,\frac{{{x}^{2}}}{{{b}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}}=1\], which is obviously a hyperbola.
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