Question

The curve y=(x-2)^2+1 has a minimum point at p.A point q on the curve is such that the slope of pq is 2 . Find the area bounded by the curve and the chord pq

Answer 1

Area is 4.67 sq. units.

Step-by-step explanation:

The vertex of the parabola y = (x - 2)² + 1 is at P(2,1) and let the point Q has coordinates (h,k).

Now, slope of PQ = $\frac{k - 1}{h - 2} = 2$ {Given}

⇒ k - 1 = 2h - 4

⇒ k = 2h - 3 .......... (1)

Now, Q(h,k) satisfies the equation of the parabola.

So, k = (h - 2)² + 1 ........... (2)

Now, solving equations (1) and (2) we get,

(h - 2)² + 1 = 2h - 3

⇒ h² - 4h + 4 + 1 = 2h - 3

⇒ h² - 6h + 8 = 0

⇒ (h - 2)(h - 4) = 0

So, h = 2 and h = 4

From equation (1) we get,

For h = 2, k = 1 and for h = 4, k = 5

So, the point Q is (4,5) as it can not be (2,1).

Now, area of the region on the coordinate plane bounded by the given parabola above the x-axis and between the ordinates x = 2 to x = 4 will be

$\int\limits^4_2 {[(x - 2)^{2} + 1]} \, dx$

= $\int\limits^4_2 {(x^{2} - 4x + 5)} \, dx$

= ${[\frac{x^{3}}{3} - 2x^{2} + 5x]}^{4} _{2}$

= 9.33 - 4.66

= 4.67 sq. units.