Question

In cartesian co-ordinates the point 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 the dot product $\overset{\rightarrow}{OA}.\overset{\rightarrow}{AB}$ is

• A. $- \frac{520}{3}$
• B. $- 148$
• C. 140
• D. 12

Answer 1

Answer: (B)

$- 148$

Answer 2

Given curve is $y = x^{2} + x + 10$ ......(i)

when $x = 1$, $y = 1^{2} + 1 + 10 = 12$

$\therefore$ $A \equiv (1,12)$; $\therefore$ $\overset{\rightarrow}{OA} = \mathbf{i} + 12\mathbf{j}$

From (i), $\frac{dy}{dx} = 2x + 1$

Equation of tangent at A is $y - 12 = \left( \frac{dy}{dx} \right)_{(1,12)}(x - 1)$

⇒ $y - 12 = (2 \times 1 + 1)(x - 1)$ ⇒ $y - 12 = 3x - 3$

$\therefore$ $y = 3(x + 3)$

This tangent cuts x-axis (i.e., $y = 0$) at $( - 3,0)$

$\therefore$ $B \equiv ( - 3,0)$

$\overset{\rightarrow}{OB} = - 3\mathbf{i} + 0.\mathbf{j} = - 3\mathbf{i}$; $\overset{\rightarrow}{OA}.\overset{\rightarrow}{AB} = \overset{\rightarrow}{OA}.(\overset{\rightarrow}{OB} - \overset{\rightarrow}{OA})$

=$(\mathbf{i} + 12\mathbf{j}).( - 3\mathbf{i} - \mathbf{i} - 12\mathbf{j})$

= $(\mathbf{i} + 12\mathbf{j}).( - 4\mathbf{i} - 12\mathbf{j})$

= $- 4 - 144 = - 148$