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Question : 159 of 180
Marks:
+1,
-0
Solution:
We have,
y2=4xDifferentiating w.r.t.
x, we get
‌2y‌=4⇒‌=‌‌(‌)(1,2)=‌=1
Equation of normal to the curve at
(1,2) is
‌y−y1=‌(x−x1)⇒(y−2)=−‌(x−1)‌⇒y−2=−x+1⇒x+y=3‌‌ The line ‌x+y=3‌ meets the ‌X‌-axis at ‌x=3‌∴‌ Required area ‌=√4x‌dx+(3−x)‌dx‌=2[‌]01+[3x−‌]13‌=‌(1)+[9−‌−3+‌]=‌+(‌−‌)‌=‌+‌=‌+2=‌‌ sq units ‌‌
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