Definite Integration Question 425
Question: The area of the figure bounded by the parabola $ {{(y-2)}^{2}}=x-1 $ , the tangent to it at the point with the ordinate x = 3, and the x-axis is
Options:
A) 7 sq. units
B) 6 sq. units
C) 9 sq. units
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given parabola is $ {{(y-2)}^{2}}=x-1 $
$ \Rightarrow \frac{dy}{dx}=\frac{1}{2(y-2)} $ When $ y=3,x=2 $
$ \therefore \frac{dy}{dx}=\frac{1}{2(3-2)}=\frac{1}{2}. $ Tangent at (2, 3) is $ y-3=\frac{1}{2}(x-2) $ Or $ x-2y+4=0 $
$ \therefore $ Required area $ =\int_0^{3}{( {{(y-2)}^{2}}+1 )}dy-\int_0^{3}{(2y-4)dy} $
$ =| \frac{{{(y-2)}^{3}}}{3}+y |_0^{3}-|y^{2}-4y|_0^{3} $
$ =\frac{1}{3}+3+\frac{8}{3}-(9-12)=9 $ sq. units.
BETA
