For what value of k, is the area bounded by y = x2
–3 and y = kx + 2 the least ? Calculate the area.
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
&⇒ a + b = k, ab =
Required area, A =
For least value of A, k = 0
&⇒ least value A = sq.
Find the area bounded by y = 2 - | 2 – x | and y= .
given curves are
the required area PQRSP = area PQRP + area PRSP
= sq. units
Find the area bounded by , y
= 7 - |x|.
y = 7 - |x| …..(2).
Rewriting (1) and (2)
Required area PQRSTUP = 2 area of PQRSP
= 2[ area of OVRSO – area of POQP – area of QRVQ]
= 32 sq. units.
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
Given parabola y= –x2–2x + 3
&⇒ Tangent at (2, -5) is
y + 5 = -6(x –2) or 6x + y – 7 = 0
&⇒ Required area MABM
= area DAML -area BLMB
= 12 -
= 12 -
= 12 - =
8/3 sq. units
parabola y2 = 2x divides the circle x2 + y2
= 8 in two parts. Find the ratio of the areas of both parts .
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 2and 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.
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 =
&⇒ y = ± , but we choose y = , as for arc OB, y
is positive. Same is the reasoning for circle)
&⇒ A1 =
= =sq. units.
Area of the circle =
p (2)2 = 8p sq. units.
Hence A2 =
8p - A1 = 7p - . Then the required
Note: Here B º (2, 2 ). The area A1 can
also be obtained by using the formula 2where x2=and x1