Question

If $A$ is $(x_1,y_1)$ where $x_1=1$ on the curve $y=x^2+x+10$, the tangent at $A$ cuts the x-axis at $B$. The value of $\overline{OA}.\overline{AB}$ is

• A. $-148$
• B. $-352$
• C. $140$
• D. $148$

Answer 1

Answer: (A)

$-148$

$\to$ when $x_1=1,y_1=1+1+10=12$

$\frac{dy}{dx}=2x+1$ At $(x_1,y_1),\frac{dy}{dx}=$ slope of tangent =3

$\Rightarrow$ Tangent passes through (1,12) and has slope = 3

$\Rightarrow$ its equation : $y-12=3(x-1)$

$\Rightarrow y=3x+9$

At x-axis,$y=0\Rightarrow x=-3\Rightarrow BI(-3,0)$

$\overline{OA}=\hat{i}+12\hat{j}$ $\overline{AB}=-4\hat{i}-12\hat{j}$

$\Rightarrow \overrightarrow{OA}.\overrightarrow{AB}=-4-144=-148\Rightarrow \left(A\right)$