The area of the region enclosed by the parabola ( y - 2 ) 2 = x - 1 , the line x - 2 y + 4 = 0 and the positive coordinate axes is [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The area of the region enclosed by the parabola ${(y - 2)}^{2} = x - 1,$ the line $x - 2 y + 4 = 0$ and the positive coordinate axes is ____________. [2024]

Ans.
(5)

Given, equation of parabola

${(y - 2)}^{2} = (x - 1)$ ...(i)

and line $x - 2 y + 4 = 0$ ...(ii)

From (ii), we have

$x - 2 (y - 2) = 0 \Rightarrow y - 2 = \frac{x}{2}$

Putting in (i), we get ${(\frac{x}{2})}^{2} = (x - 1)$

$\Rightarrow x^{2} - 4 x + 4 = 0$

$\Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x = 2$

From (ii), we have, $y = 3$

$∴$ Intersection point of (i) and (ii) is (2,3).

$∴$ $Required area = \int_{0}^{3} ({(y - 2)}^{2} + 1) d y - Area of triangle$

$= \int_{0}^{3} (y^{2} - 4 y + 5) d y - \frac{1}{2} \times 2 \times 1$

$= {[\frac{y^{3}}{3} - 2 y^{2} + 5 y]}_{0}^{3} - 1 = (9 - 18 + 15) - 1 = 6 - 1 = 5 sq. units$

Want Us to Email you About Special Offers & Updates?