Definite Integration Question 175
Question: The area of the region bounded by the parabola $ {{(y-2)}^{2}}=x-1 $ , the tangent of the parabola at the point (2, 3) and the x-axis is:
Options:
6
9
12
3
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The given parabola is $ {{(y-2)}^{2}}=x-1 $ Vertex (1, 2) and it intersects the x-axis at (5, 0) Also it gives $ y^{2}-4y-x+5=0 $
So, that equation of the tangent to the parabola at (2, 3) is $ y.3-2(y+3)-\frac{1}{2}(x+2)+5=0 $ or $ x-2y+4=0 $ which intersects the x-axis at (-4, 0).
In the figure shaded area is the required area. Let us draw PD perpendicular to y - axis. Then required area $ =Ar\Delta BOA+Ar\Delta OCPD $ $ -Ar(\Delta APD) $
$ =\frac{1}{2}\times 4\times 2+\int_0^{3}{xdy}-\frac{1}{2}\times 2\times 1$ $ 3+\int_0^{3}\left({(y-2)}^{2}+1\right)dy$ $ =3+[ \frac{{{(y-2)}^{3}}}{3}+y ]_0^{3}=3+[ \frac{1}{3}+3+\frac{8}{3} ] $
$ =3+6=9 $ Sq. units
BETA
