A) \[{{y}^{2}}-{{b}^{2}}={{c}^{2}}({{x}^{2}}+{{a}^{2}})\]
B) \[{{y}^{2}}+{{a}^{2}}={{c}^{2}}({{x}^{2}}-{{b}^{2}})\]
C) \[{{y}^{2}}+{{b}^{2}}={{c}^{2}}({{x}^{2}}-{{a}^{2}})\]
D) \[{{y}^{2}}-{{a}^{2}}={{c}^{2}}({{x}^{2}}+{{b}^{2}})\]
Correct Answer: C
Solution :
| Let the slopes of the two tangents to the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] be cm and c/m. | ||
| Then the equation of the tangents are | ||
| \[y=cmx+\sqrt{{{a}^{2}}{{c}^{2}}{{m}^{2}}-{{b}^{2}}}\] | ? (i) | |
| and \[my-cx=\sqrt{{{a}^{2}}{{c}^{2}}-{{b}^{2}}{{m}^{2}}}\] | ? (ii) | |
| Squaring and subtracting (ii) from (i), we \[get\text{ }{{\left( y-cmx \right)}^{2}}-{{\left( my-cx \right)}^{2}}\] |
| \[={{a}^{2}}{{c}^{2}}{{m}^{2}}-{{b}^{2}}-{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{m}^{2}}\] |
| \[\Rightarrow \left( 1-{{m}^{2}} \right)\left( {{y}^{2}}-{{c}^{2}}{{x}^{2}} \right)\] |
| \[=-\left( 1-{{m}^{2}} \right)\left( {{a}^{2}}{{c}^{2}}+{{b}^{2}} \right)\]\[\Rightarrow {{y}^{2}}+{{b}^{2}}={{c}^{2}}\left( {{x}^{2}}-{{a}^{2}} \right)\] |
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