A) 8 points
B) Two circles
C) 4 lines
D) None of these
Correct Answer: A
Solution :
[a] \[{{({{x}^{2}}-{{a}^{2}})}^{2}}{{({{x}^{2}}-{{b}^{2}})}^{2}}+{{c}^{4}}{{({{y}^{2}}-{{a}^{2}})}^{2}}=0\] This being the sum of two perfect squares, each term must be zero. Hence, we get \[{{({{x}^{2}}-{{a}^{2}})}^{2}}{{({{x}^{2}}-{{b}^{2}})}^{2}}=0\] or \[({{x}^{2}}-{{a}^{2}})({{x}^{2}}-{{b}^{2}})=0\] or \[(x-a)(x+a)(x-b)(x+b)=0\] ? (1) and \[{{c}^{4}}{{({{y}^{2}}-{{a}^{2}})}^{2}}=0\] or \[{{c}^{2}}({{y}^{2}}-{{a}^{2}})=0\] or \[{{c}^{2}}(y+a)(y-a)=0\] ? (2) Equation no. (1) Holds good for \[x=\pm a\] or \[x=\pm b\] Equation no. (2) Is satisfied by \[y=\pm a\] As both of these should be simultaneously satisfied, the give equation represents 8 points which we get as a result of different combinations of (1) and (2), namely \[(\pm a,\,\,\pm a),(\pm b,\,\,\pm a).\]
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