The area of the region bounded by the parabol | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given parabola: $(y-2)^{2} = x-1 \Rightarrow x = (y-2)^{2} + 1$.At ordinate $y=3$,$x = (3-2)^{2} + 1 = 2$. So,the point is $(2, 3)$.Differentiatin

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Given parabola: $(y-2)^{2} = x-1 \Rightarrow x = (y-2)^{2} + 1$.At ordinate $y=3$,$x = (3-2)^{2} + 1 = 2$. So,the point is $(2, 3)$.Differentiating $(y-2)^{2} = x-1$ with respect to $y$,we get $2(y-2) = \frac{dx}{dy}$.At $y=3$,$\frac{dx}{dy} = 2(3-2) = 2$.The equation of the tangent at $(2, 3)$ is $x - 2 = 2(y - 3) \Rightarrow x = 2y - 4$.The region is bounded by the parabola $x = (y-2)^{2} + 1$,the tangent $x = 2y - 4$,and the $x$-axis $(y=0)$.Integrating with respect to $y$ from $y=0$ to $y=3$:Area $= \int_{0}^{3} [((y-2)^{2} + 1) - (2y - 4)] dy$$= \int_{0}^{3} (y^{2} - 4y + 4 + 1 - 2y + 4) dy = \int_{0}^{3} (y^{2} - 6y + 9) dy$$= \int_{0}^{3} (y-3)^{2} dy = \left[ \frac{(y-3)^{3}}{3} ight]_{0}^{3} = 0 - (\frac{-27}{3}) = 9 	ext{ sq. units.}$

The area of the region bounded by the parabola $(y-2)^{2}=(x-1)$,the tangent to it at the point whose ordinate is $3$,and the $x$-axis is:

Similar Questions

The area of the figure bounded by the curves $y = |x - 1|$ and $y = 3 - |x|$ is ....... $sq. \text{ unit}$.

The area of the shaded region is ... sq. units.

The sine and cosine curves intersect infinitely many times,creating bounded regions of equal areas. The area of one such region is

The area bounded by the curves $y=\frac{1}{4}\left|4-x^2\right|$ and $y=7-|x|$ is

The volume of a solid obtained by revolving about the $y$-axis the region enclosed between the ellipse $x^2 + 9y^2 = 9$ and the straight line $x + 3y = 3$ in the first quadrant is: (in $\pi$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module