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Question : 149 of 180
Marks:
+1,
-0
Solution:
Let
â–³ABC be an isosceles triangle inscribed in the ellipse
‌+‌=1Let coordinate of
A‌be(a,0).
Area of
△ABC,A=‌×AD×BC=‌×(a+x)×2y=y(a+x)=√b2(1−‌)(a+x)=‌√a2−x2(a+x)
‌⇒‌=‌[(a+x)‌×‌+√a2−x2]‌=‌[‌| −x(a+x)+(a2−x2) |
| √a2−x2 |
]‌‌ Put ‌‌=0‌⇒‌‌2x2+ax−a2=0‌⇒2x2+2ax−ax−a2=0‌⇒‌‌(2x−a)(x+a)=0‌⇒‌‌x=‌,−aSince,
‌<0, for
x=‌∴‌ Maximum area ‌‌=‌⋅(a+‌)√a2−‌‌=‌⋅‌(a)‌=‌‌ sq units ‌
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