A) a circle
B) a parabola
C) an ellipse
D) a hyperbola
Correct Answer: A
Solution :
Let the two lines be taken as \[x\] and y axes, respectivley. Let \[C(\alpha ,\,\beta )\] be the centre of the ellipse of which \[{{S}_{1}}\,({{x}_{1}},\,\,{{y}_{1}})\] and \[{{S}_{2}}\,({{x}_{2}},\,{{y}_{2}})\] are foci. Then, \[{{x}_{1}}+{{x}_{2}}=2\alpha \] and \[{{y}_{1}}+{{y}_{2}}=2\beta \] Since, x-axis and y-axis are the tangents to the ellipse. \[\therefore \] \[{{x}_{1}}{{x}_{2}}={{b}^{2}}\] \[{{y}_{1}}{{y}_{2}}={{b}^{2}}\] Also, \[{{S}_{1}}S_{2}^{2}\,=4{{a}^{2}}{{e}^{2}}\] \[\Rightarrow \] \[{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}=4{{a}^{2}}{{e}^{2}}\] \[\Rightarrow \] \[{{({{x}_{1}}+{{x}_{2}})}^{2}}+{{({{y}_{1}}+{{y}_{2}})}^{2}}-4{{x}_{1}}{{x}_{2}}\] \[-4{{y}_{1}}{{y}_{2}}=4{{a}^{2}}{{e}^{2}}\] \[\Rightarrow \] \[4{{\alpha }^{2}}+4{{\beta }^{2}}-4{{b}^{2}}-4{{b}^{2}}=4{{a}^{2}}{{e}^{2}}\] \[\Rightarrow \] \[{{\alpha }^{2}}+{{\beta }^{2}}-{{a}^{2}}{{e}^{2}}=2{{b}^{2}}\] \[\Rightarrow \] \[{{\alpha }^{2}}+{{\beta }^{2}}=2{{b}^{2}}+{{a}^{2}}-{{b}^{2}}\] \[[\because \,{{b}^{2}}={{a}^{2}}\,\,(1-{{e}^{2}})]\] \[\Rightarrow \] \[{{\alpha }^{2}}+{{\beta }^{2}}={{a}^{2}}+{{b}^{2}}\] Locus \[(\alpha ,\,\beta )\] is \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\] Which represents a circle.You need to login to perform this action.
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