A) \[\frac{1}{2}(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})\]
B) \[a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2}\]
C) \[\frac{1}{2}\,\,(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2})\]
D) \[\sqrt{a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}}\]
Correct Answer: A
Solution :
Let \[P(\alpha ,\beta )\] be the point which is equidistant to \[A({{a}_{1}},{{b}_{1}})\] and \[B({{a}_{2}},{{b}_{2}})\]. \[\therefore \] \[{{(PA)}^{2}}={{(PB)}^{2}}\] \[\Rightarrow {{(\alpha -{{a}_{1}})}^{2}}+{{(\beta -{{b}_{1}})}^{2}}={{(\alpha ={{a}_{2}})}^{2}}+{{(\beta -{{b}_{2}})}^{2}}\] \[\Rightarrow \] \[={{\alpha }^{2}}+a_{2}^{2}\,-2\alpha {{a}_{2}}\,+{{\beta }^{2}}+b_{2}^{2}-\,2\beta {{b}_{2}}\] Thus, the equation of locus (a,p) is But the given equation is Alternate Solution Since, the points and satisfy the equation so that ... (i) and ...(ii) On adding Eqs. (i) and (ii), we get \[({{a}_{1}}+{{a}_{2}})\,({{a}_{1}}-{{a}_{2}})\,+({{b}_{1}}+{{b}_{2}})\,({{b}_{1}}-{{b}_{2}})+2c=0\] \[\Rightarrow \,\,\,\,\,2c=-(a_{1}^{2}-a_{2}^{2}\,+b_{1}^{2}\,-b_{2}^{2})\] \[\Rightarrow \,\,\,\,c=\frac{1}{2}(a_{2}^{2}\,+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})\]You need to login to perform this action.
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