Test your skills 2

Calculus Level 5

Find the area of the region bounded by the parabola ( y − 2 ) 2 = x − 1 (y-2)^{2}=x-1 , the tangent to it at the point where the ordinate is 3 ,and the x x -axis.


The answer is 9.

Mohit Gupta by Mohit Gupta , 20, India

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1 solution

x = ( 3 − 2 ) 2 + 1 = 2 , ⟹ t h e p o i n t o f t a n g e n t c y i s T ( 2 , 3 ) . S l o p e o f t h e t a n g e n t m = d y d x = 1 2 ( 3 − 2 ) = 1 2 . ∴ t h e t a n g e n t i s y = 1 2 ( x − 2 ) , ⟹ x = 2 ( y − 3 ) + 2. δ A = ( x p − x t ) δ ( y ) = − { 2 ( y − 3 ) + 2 − { ( y − 2 ) 2 + 1 } } ∗ δ ( y ) . A = ∫ 0 3 − { 2 ( y − 3 ) + 2 − { ( y − 2 ) 2 + 1 } } ∗ d y = 9 x=(3-2)^2+1=2,\ \implies \ \ the\ point\ of\ tangentcy\ \ is\ T(2,3) .\\ Slope\ of\ the\ tangent\ m=\dfrac{dy}{dx}=\dfrac 1 {2(3-2)}=\frac 1 2.\\ \therefore\ the\ tangent\ is\ y=\frac 1 2(x-2),\ \ \implies\ x=2(y-3)+2.\\ \delta A=(x_p-x_t)\delta(y)=- \left \{2(y-3)+2- \{(y-2)^2+1\} \right\}*\delta(y).\\ \displaystyle A\ =\ \int_0^3\ - \left \{2(y-3)+2- \{(y-2)^2+1\} \right\}*dy= \Large\ \ \color{#D61F06}{9}
The minus signs in the last two expressions are to correct my previous mistakes .

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