Question

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

• A. a positive irrational number
• B. a positive integer
• C. a negative integer
• D. a negative irrational number

Answer 1

The correct option is C a negative integer

$\frac{dy}{dx}=2x+b$

The equation of the tangent to the curve at $(1,1)$ is $(b+2)x-y=b+1$

Area$=\frac{c^2}{2|ab|}=2$
$\Rightarrow \frac{(b+1{)}^2}{2|(b+2)(-1)|}=2$
$\Rightarrow (b+3{)}^2=0$
$\Rightarrow b=-3$

Another case

$(b+1{)}^2=4(b+2)$
$b^2-2b-7=0$
$b=1\pm 2\sqrt{2}$

Both $x$ and $y$ intercepts must be positive because given first quadrant
$y$ intercept $=-b-1$
So $b$ can't be $1+2\sqrt{2}$ because in this case $y$ intercept is negative

$x$ intercept $=\frac{b+1}{b+2}$
So $b$ can't be $1-2\sqrt{2}$ because in this case $x$ intercept is negative

So $b=-3$