A) \[\frac{\sqrt{7}}{2}\]
B) \[\sqrt{\frac{5}{3}}\]
C) \[\sqrt{\frac{3}{2}}\]
D) \[\sqrt{2}\]
Correct Answer: C
Solution :
| Chord of contact of tangent from\[(-\,4,2)\]the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]is | ||
| \[\frac{-\,4x}{{{a}^{2}}}-\frac{2y}{{{b}^{2}}}=1\] | ? (i) | |
| And from (2, 1) is \[\frac{2x}{{{a}^{2}}}-\frac{y}{{{b}^{2}}}=1\] | ? (ii) | |
| Since, chord of contact are perpendicular. | ||
| \[\therefore \]\[\left( \frac{-\,4}{{{a}^{2}}} \right)\left( \frac{2}{{{a}^{2}}} \right)+\left( \frac{2}{{{b}^{2}}} \right)\left( \frac{1}{{{b}^{2}}} \right)=0\] | ||
| \[\frac{{{a}^{4}}}{{{b}^{4}}}=4\] | ||
| \[\Rightarrow \] \[{{a}^{2}}=2{{b}^{2}}\] |
| \[\Rightarrow \] \[\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{1}{2}\] |
| \[\Rightarrow \] \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\] |
| \[=\sqrt{1+\frac{1}{2}}=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}\] |
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