Forming Triangle

Geometry Level 4

The triangle formed by the tangent to the curve f ( x ) = x 2 + b x − b f(x) = x^2 + bx - b at the point ( 1 , 1 ) (1, 1) and the coordinate axes, lie in the first quadrant. If its area is 2 unit 2 ^2 , what is the value of b b ?


The answer is -3.

Danish Ahmed by Danish Ahmed , 24, India

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1 solution

D i f f e r e n t i a t i n g f ( x ) , s l o p e o f t h e t a n g e n t , f ′ ( x ) = m = 2 x + b . ∴ m 1 , 1 = 2 + b . Differentiating f(x), slope\ of\ the\ tangent,\ f'(x)= m=2x+b.\ \ \therefore\ m_{1,1}=2+b.\\ S o t h e t a n g e n t i s y = ( 2 + b ) ( x − 1 ) + 1 = ( 2 + b ) x − ( b + 1 ) . So\ the\ tangent\ is\ \ y=(2+b)(x-1)+1=(2+b)x-(b+1).\\ ∴ t h e y − i n t e r c e p t , Y = − ( b + 1 ) , . . . . . . t h e x − i n t e r c e p t , X = b + 1 b + 2 . A r e a 1 2 ∗ Y ∗ X = 2. \therefore\ the \ y-intercept,\ Y =-(b+1),......\ the \ x-intercept,\ X =\dfrac{b+1}{b+2}.\\ Area\ \frac 1 2*Y*X=2.\\ ⟹ 4 = − ( b + 1 ) 2 b + 2 . S o l v i n g t h e q u a d r a t i c i n b , b 2 + 6 b + 9 = 0. b = − 3 . \implies\ 4=-\ \dfrac{(b+1)^2}{b+2}.\\ Solving\ the\ quadratic\ in \ b,\ \ b^2+6b+9=0.\ \ \ b=\Large\ \ \ \color{#D61F06}{-3}.\\

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