Discussion Forum

The triangle formed by the tangent to the curve $f(x)=x^{2}+bx-b$ at the point (1,1) and the coordinate axes lie in the first quadrant . if its area is 2 , then the value of b is 

Answer:

Explanation:

 Given curve is y=$f(x)=x^{2}+bx-b$ 

$\Rightarrow$  f'(x)=2x+b

 The equation of tangent at point (1,1) is 

$y-1=\left(\frac{dy}{dx}\right)_{(1,1)}(x-1)$

$\Rightarrow$    $y-1=(b+2)(x-1)$

$\Rightarrow$  $2+b)x-y=1+b$

$\Rightarrow$  $\frac{x}{\left(\frac{1+b}{2+b}\right)}-\frac{x}{(1+b)}=1$

So, $OA=\frac{1+b}{2+b}$

 and OB=-(1+b)

 Now, area  of $\triangle AOB= \frac{1}{2} \times \frac{(1+b)[-(1+b)]}{(2+b)}=2$

$\Rightarrow$     $4(2+b)+(1+b)^{2}=0$

$\Rightarrow$  $b^{2}+6b+9=0$

$\Rightarrow$    $(b+3)^{2} \Rightarrow b=-3$