JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 10)
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 :
9
10
4
6
Explanation
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
_27th_August_Evening_Shift_en_10_2.png)
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
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