TS EAMCET · Maths · Hyperbola
Let \(\mathrm{S}\) be the focus of the hyperbola \(x^2-2 y^2=1\) lying on the positive \(\mathrm{X}\)-axis. Let \(\mathrm{P}(-1,1)\) be a given point. Then the area of the triangle formed by the line PS with the coordinate axes is (in sq. units)
- A \(\frac{\sqrt{2}}{2(\sqrt{2}+3)}\)
- B \(\frac{\sqrt{6}}{2(2+\sqrt{6})}\)
- C \(\frac{3}{2(2+\sqrt{6})}\)
- D \(\frac{\sqrt{3}}{2(\sqrt{2}+\sqrt{3})}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{2(2+\sqrt{6})}\)
Step-by-step Solution
Detailed explanation
Given hyperbola \(x^2-2 y^2=1\) can be written as \[ \Rightarrow \mathrm{a}=1, \mathrm{~b}=\frac{1}{\sqrt{2}} \text { hence, } \mathrm{e}=\sqrt{\frac{3}{2}} \] Now from fig. Area made by co-ordinate axes and the line \(\overline{P S}\) will be Area of…
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