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Question : 41 of 160
Marks:
+1,
-0
Solution:
Given that,
AP+BP=2a Using distance formula,
AP=√(x+ae)2+y2
and
BP=√(x−ae)2+y2 Now,
√(x+ae)2+y2+√(x−ae)2+y2=2a Squaring both sides,
(x+ae)2+y2+(x−ae)2+y2 +2√(x+ae)2+y2√(x−ae)2+y2=4a2 ⇒2x2+2y2+2a2e2−4a2 =−2√(x+ae)2+y2√(x−ae)2+y2 ⇒x2+y2+a2e2−2a2 =−√(x+ae)2+y2√(x−ae)2+y2 Squaring both sides,
a4e4−4a4e2+4a4+2a2x2e2−4a2x2 +2a2y2e2−4a2y2+x4+2x2y2+y4 =a4e4−2a2x2e2+2a2y2e2+x4+2x2y2+y4 ⇒4(a4−a4e2−a2x2−a2y2+a2x2e2)=0 ⇒a2(a2−a2e2−x2−y2+x2e2)=0 ⇒a2(1−e2)−x2−y2+x2e2=0 ⇒b2−x2−y2+x2(1−)=0 [∵b2=a2(1−e2)]
⇒x2[1−−1]−y2+b2=0 ⇒x2(−)−y2=−b2 ⇒+=1(divide by b2 )
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