A) \[{{a}^{2}}+{{b}^{2}}=2{{c}^{2}}\]
B) \[{{b}^{2}}-{{a}^{2}}=2{{c}^{2}}\]
C) \[{{a}^{2}}-{{b}^{2}}=2{{c}^{2}}\]
D) \[{{a}^{2}}-{{b}^{2}}={{c}^{2}}\]
Correct Answer: C
Solution :
| [c] Let \[({{x}_{1}},{{y}_{1}})\] be their point of intersection then |
| \[{{x}^{2}}_{1}-{{y}^{2}}_{1}={{c}^{2}}\] ? (1) \[\frac{{{x}^{2}}_{1}}{{{a}^{2}}}+\frac{{{y}^{2}}_{1}}{{{b}^{2}}}=1\] ? (2) |
| \[\Rightarrow {{x}^{2}}_{1}\left( \frac{1}{{{a}^{2}}}-\frac{1}{{{c}^{2}}} \right)+{{y}^{2}}_{1}\left( \frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}} \right)=0\] ? (3) |
| Now tangents to the curves are \[x{{x}_{1}}-y{{y}_{1}}={{c}^{2}}\] |
| And \[\frac{x{{x}_{1}}}{{{a}^{2}}}+\frac{y{{y}_{1}}}{{{b}^{2}}}=1\] |
| The tangents are perpendicular, so |
| \[\frac{{{x}_{1}}}{{{y}_{1}}}\times -\frac{{{b}^{2}}}{{{a}^{2}}}\frac{{{x}_{1}}}{{{y}_{1}}}=-1\Rightarrow {{b}^{2}}{{x}^{2}}_{1}-{{a}^{2}}{{y}^{2}}_{1}=0\] ? (4) |
| Eliminating \[{{x}^{2}}_{1}\] and \[{{y}^{2}}_{1}\] from (3) and (4) we get, |
| \[\frac{{{c}^{2}}-{{a}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}=\frac{-({{b}^{2}}+{{c}^{2}})}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}\Rightarrow {{a}^{2}}-{{b}^{2}}=2{{c}^{2}}\] |
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