The area of the region enclosed by the curves $y=x^2-4x+4$ and $y^2=16-8x$ is:
- (1) $\frac{8}{3}$
- (2) $\frac{4}{3}$
- (3) 5
- (4) 8
Solution:
Curves are $y = (x-2)^2$ (Parabola 1, vertex at (2,0), opening up).
And $y^2 = -8(x-2)$ (Parabola 2, vertex at (2,0), opening left).
Intersection points:
Substitute $(x-2) = -y^2/8$ into first eq:
$y = (-y^2/8)^2 = y^4/64 \Rightarrow y(y^3 – 64) = 0$.
$y = 0$ or $y = 4$.
If $y=0, x=2$.
If $y=4, 16 = -8(x-2) \Rightarrow -2 = x-2 \Rightarrow x=0$.
Points are $(2,0)$ and $(0,4)$.
Standard area formula for parabolas $y^2=4ax$ and $x^2=4by$: Area $= \frac{16ab}{3}$.
Here, standard forms shifted to vertex (2,0).
$P_1: (x-2)^2 = y \Rightarrow 4b = 1 \Rightarrow b = 1/4$.
$P_2: y^2 = 8(2-x) \Rightarrow 4a = 8 \Rightarrow a = 2$.
Area $= \frac{16ab}{3} = \frac{16(2)(1/4)}{3} = \frac{8}{3}$.
Ans. (1)