Example.1
For what value of k, is the area bounded by y = x2
–3 and y = kx + 2 the least ? Calculate the area.
Solution
The given curves are
y = x2 –3 ……(1)
y = kx + 2 ……(2)
Solving (1) and (2) we get
x2 – kx – 5 = 0
Let a, b be the roots of the
quadratic
&⇒ a + b = k, ab =
-5
Required area, A = ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image001.gif)
=
= ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image003.gif)
=![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image004.gif)
=
=![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image006.gif)
For least value of A, k = 0
&⇒ least value A =
sq.
units
Example.2
Find the area bounded by y = 2 - | 2 – x | and y=
.
Solution
The
given curves are
, ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image011.gif)
Hence
the required area PQRSP = area PQRP + area PRSP
=
![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image014.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image015.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image016.gif)
=
sq. units
Example.3
Find the area bounded by
, y
= 7 - |x|.
Solution
….(1)
y = 7 - |x| …..(2).
Rewriting (1) and (2)
![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image020.gif)
and
![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image022.gif)
Required area PQRSTUP = 2 area of PQRSP
= 2[ area of OVRSO – area of POQP – area of QRVQ]
=![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image023.gif)
=![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image024.gif)
=![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image025.gif)
=![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image026.gif)
= 32 sq. units.
Example.4
Find the area of the figure bounded by the parabola y =
-x2 – 2x +3, the line tangent to it at the point P(2, -5) and the
y-axis.
Solution
Given parabola y= –x2–2x + 3
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image027.gif)
&⇒
= -6
&⇒ Tangent at (2, -5) is
y + 5 = -6(x –2) or 6x + y – 7 = 0
&⇒ Required area MABM
= area DAML -area BLMB
= ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image029.gif)
= 12 - ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image030.gif)
= 12 - ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image031.gif)
= 12 -
=
8/3 sq. units ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image033.gif)
Example.5
The
parabola y2 = 2x divides the circle x2 + y2
= 8 in two parts. Find the ratio of the areas of both parts .
Solution
The graph of y2
= 2x ( a parabola having axis as positive x-axis and vertex at the origin)
and x2+ y2 = 8 ( a circle having radius 2
and centre at the
origin) are given in the adjacent figure:
Let the area of smaller part of circle be A1,
and that of bigger part be A2. We have to find A1/ A2. ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image035.gif)
The point B is a
point of intersection (lying in the first quadrant ) of the given parabola
and circle, whose coordinates can obtained by solving the two equation y2
= 2x and x2 + y2 = 8
&⇒ x2 + 2x = 8
&⇒ (x - 2) (x + 4) = 0
&⇒ x = 2, -4. x = – 4 is not possible .
as both points of
intersection have the same positive x-coordinate. .
Thus C º (2, 0 ).
Now A1 =
2(area(OBCO) + area (CBAC) )
&⇒
where y1and y2 are
respectively values of y from the equations of the parabola and the circle
or A1 =
(y2 =
2x
&⇒ y = ±
, but we choose y =
, as for arc OB, y
is positive. Same is the reasoning for circle)
&⇒ A1 = ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image040.gif)
=
=
sq. units.
Area of the circle =
p (2
)2 = 8p sq. units.
Hence A2 =
8p - A1 = 7p -
. Then the required
ratio is ![](http://www.quizsolver.com/radix/dth/notif/SUB_1_files/image044.gif)
=
.
Note: Here B º (2, 2 ). The area A1 can
also be obtained by using the formula 2
where x2=
and x1
=y2/2.