Question

Show that the curve (y-a)³ = a³ - 2a²x+ax², where a > 0, is always concave to the X-axis. How is it situated with respect to the Y-axis?​

Answer 1

Step-by-step explanation:

Let us find the Limits of integration.

(i)The curve is symmetrical about y−axis.

(ii)It passes through the origin and the tangents at the origin are x

2

=0 or x=0,x=0

∴ There is a cusp at the origin.

(iii)The curve has no asymptote.

(iv)The curve meets the x−axis at the origin only and meets the y−axis at (0,2a).From the equation of the curve, we have

x=

a

y

y(2a−y)

For y<0 or y>2a,x is imaginary.

Thus the curve entirely lies between y=0,x−axis and y=2a, which is shown in the figure.

∴ Area of the curve=2∫

0

2a

x.dy

Put y=2asin

2

θ∴dy=4asinθcosθdθ

⇒=

a

2

∫

0

2a

y

[y(2a−y)]

=

a

2

∫

0

2

π

2asin

2

θ

2asin

2

θ(2a−2asin

2

θ)

×4asinθcosθdθ

=32a

2

∫

0

2

π

sin

4

θcos

2

θdθ

=32a

2

6.4.2

3.1×1

.

2

π

=πa

2