A) \[y=\pm \,\,x\,\,\pm \sqrt{{{b}^{2}}-{{a}^{2}}}\]
B) \[y=\pm \,\,x\,\,\pm \sqrt{{{a}^{2}}-{{b}^{2}}}\]
C) \[y=\pm \,\,x\,\,\pm \sqrt{{{a}^{2}}+{{b}^{2}}}\]
D) \[y=\pm \,\,x\,\,\pm \left( {{a}^{2}}-{{b}^{2}} \right)\]
Correct Answer: B
Solution :
We have the hyperbolas \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] ?(i) and \[\frac{{{y}^{2}}}{{{a}^{2}}}-\frac{{{x}^{2}}}{{{b}^{2}}}=1\] ?(ii) Any tangent to the hyperbola Eq. (i), \[y=mx+c\] where \[c=\pm \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}\] ...(iii) But this tangent touches the parabola Eq. (ii), also \[\therefore \] \[\frac{{{(mx+c)}^{2}}}{{{a}^{2}}}-\frac{{{x}^{2}}}{{{b}^{2}}}=1\] \[\Rightarrow \] \[{{b}^{2}}({{m}^{2}}{{x}^{2}}+{{c}^{2}}+2mcx)-{{a}^{2}}{{x}^{2}}={{a}^{2}}{{b}^{2}}\] \[\Rightarrow \] \[({{b}^{2}}{{m}^{2}}-{{a}^{2}}){{x}^{2}}+2mcx{{b}^{2}}+{{b}^{2}}{{c}^{2}}-{{a}^{2}}{{b}^{2}}=0\] \[\Rightarrow \] \[({{b}^{2}}{{m}^{2}}-{{a}^{2}}){{x}^{2}}+2mc{{b}^{2}}x+{{b}^{2}}({{c}^{2}}-{{a}^{2}})=0\] For the tangency, it should have equal roots \[{{(2mc{{b}^{2}})}^{2}}=4({{b}^{2}}{{m}^{2}}-{{a}^{2}})\cdot {{b}^{2}}({{c}^{2}}-{{a}^{2}})\] \[\Rightarrow \] \[4{{m}^{2}}{{c}^{2}}{{b}^{4}}=4{{b}^{2}}({{b}^{2}}{{m}^{2}}{{c}^{2}}-{{b}^{2}}{{m}^{2}}{{a}^{2}}\] \[-{{a}^{2}}{{c}^{2}}+{{a}^{4}})\] \[\Rightarrow \] \[{{m}^{2}}{{c}^{2}}{{b}^{2}}={{b}^{2}}{{m}^{2}}{{c}^{2}}-{{b}^{2}}{{m}^{2}}{{a}^{2}}-{{a}^{2}}{{c}^{2}}+{{a}^{4}}\] \[\Rightarrow \] \[{{a}^{2}}{{c}^{2}}={{a}^{4}}-{{b}^{2}}{{m}^{2}}{{a}^{2}}\] \[\Rightarrow \] \[{{c}^{2}}={{a}^{2}}-{{b}^{2}}{{m}^{2}}\] \[\Rightarrow \] \[{{a}^{2}}{{m}^{2}}-{{b}^{2}}={{a}^{2}}-{{b}^{2}}{{m}^{2}}\][using Eq. (iii)] \[\Rightarrow \] \[({{a}^{2}}+{{b}^{2}}){{m}^{2}}={{a}^{2}}+{{b}^{2}}\] \[\Rightarrow \] \[{{m}^{2}}=1\] \[\Rightarrow \] \[m=\pm 1\] Hence, the equation of common tangent are \[y=\pm \,\,x\,\,\pm \sqrt{{{a}^{2}}-{{b}^{2}}}\]
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