For what value of k, is the area bounded by y = x^{2}
–3 and y = kx + 2 the least ? Calculate the area.

The given curves are

y = x^{2} –3 ……(1)

y = kx + 2 ……(2)

Solving (1) and (2) we get

x^{2} – kx – 5 = 0

Let a, b be the roots of the
quadratic

&⇒ a + b = k, ab =
-5

Required area, A =

==

=

==

For least value of A, k = 0

&⇒ least value A = sq.
units

Find the area bounded by y = 2 - | 2 – x | and y= .

The given curves are

,

Hence the required area PQRSP = area PQRP + area PRSP

=

=

=

= sq. units

Find the area bounded by , y = 7 - |x|.

….(1)

y = 7 - |x| …..(2).

Rewriting (1) and (2)

and

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 =
-x^{2} – 2x +3, the line tangent to it at the point P(2, -5) and the
y-axis.

Given parabola y= –x^{2}–2x + 3

&⇒

&⇒= -6

&⇒ 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

The
parabola y^{2} = 2x divides the circle x^{2 }+ y^{2}
= 8 in two parts. Find the ratio of the areas of both parts .

The graph of y^{2}
= 2x ( a parabola having axis as positive x-axis and vertex at the origin)
and x^{2}+ y^{2} = 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 A_{1},
and that of bigger part be A_{2}. We have to find A_{1}/ A_{2}.

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 y^{2}
= 2x and x^{2} + y^{2} = 8

&⇒ x^{2 }+ 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 A_{1} =
2(area(OBCO) + area (CBAC) )

&⇒ where y_{1}and y_{2} are
respectively values of y from the equations of the parabola and the circle

or A_{1} = (y^{2} =
2x

&⇒ y = ± , but we choose y = , as for arc OB, y
is positive. Same is the reasoning for circle)

&⇒ A_{1} =

= =sq. units.

Area of the circle =
p (2)^{2} = 8p sq. units.

Hence A_{2 }=
8p - A_{1} = 7p - . Then the required
ratio is

= .

Note: Here B º (2, 2 ). The area A_{1} can
also be obtained by using the formula 2where x_{2}=and x_{1}
=y^{2}/2.