A) \[{{x}^{2}}+{{a}^{2}}={{b}^{2}}+{{y}^{2}}\]
B) \[{{x}^{2}}+{{a}^{2}}=2{{b}^{2}}-{{y}^{2}}\]
C) \[{{x}^{2}}-{{a}^{2}}={{b}^{2}}+{{y}^{2}}\]
D) \[{{x}^{2}}+{{a}^{2}}={{b}^{2}}-{{y}^{2}}\]
Correct Answer: D
Solution :
[d] Let P(x, y) be a point and \[A=(\alpha ,0),B=(-\alpha ,0).\] Now, \[P{{A}^{2}}={{(x-a)}^{2}}+{{y}^{2}}\] \[P{{B}^{2}}={{(x+a)}^{2}}+{{y}^{2}}\] Since the sum of the distances of the point \[P(x,y)\] from the points \[A(a,0)\] and \[B(-a,0)\] is \[2{{b}^{2}}.\] \[\therefore \,\,\,\,\,P{{A}^{2}}+P{{B}^{2}}=2{{b}^{2}}\] \[{{(x-a)}^{2}}+{{(y-0)}^{2}}+{{(x+a)}^{2}}+{{(y-0)}^{2}}=2{{b}^{2}}\] \[\Rightarrow {{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{x}^{2}}+{{a}^{2}}+2ax+{{y}^{2}}=2{{b}^{2}}\] \[\Rightarrow {{x}^{2}}+{{a}^{2}}+{{y}^{2}}={{b}^{2}}\] \[\Rightarrow {{x}^{2}}+{{a}^{2}}={{b}^{2}}-{{y}^{2}}\]You need to login to perform this action.
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