The area of the region bounded by the parabola (y-2)^2=(x-1), the tangent to it at the poi…

Question

The area of the region bounded by the parabola $(y-2)^2=(x-1)$, the tangent to it at the point whose ordinate is $3$, and the $x$-axis is:

Accepted Answer

y = 3 $$\Rightarrow$$ x = 2

Point is (2, 3)

Diff. w.r.t x

2 (y - 2) y' = 1

$$\Rightarrow$$ $$y' = {1 \over {2(y - 2)}}$$

$$ \Rightarrow y{'_{(2,3)}} = {1 \over 2}$$

$$ \Rightarrow {{y - 3} \over {x - 2}} = {1 \over 2} \Rightarrow x - 2y + 4 = 0$$

Area $$ = \int\limits_0^3 {\left( {{{(y - 2)}^2} + 1 - (2y - 4)} \right)} \,dy$$

= 9 sq. units

Solutions · 1