Question

A tangent is drawn to the curve $y=\frac{8}{x^2}$ at a point $A(x_1,y_1),$ where $x_1=2.$ The tangent cuts the $x-$ axis at point $B$, then the scalar product of the vectors $\overrightarrow{AB}$ and $\overrightarrow{OB}$ is

• A. $3$
• B. $-3$
• C. $6$
• D. $-6$

Answer 1

The correct option is A $3$

We have $x^{2}y=8$
Differentiating, we get

$\frac{dy}{dx}\times x^2+2xy=0$

$\Rightarrow \frac{dy}{dx}=\frac{-2y}{x}=\frac{-16}{{x_1}^3}=-2$

The equation of tangent thus becomes $y=-2x+c$

This passes through $(2,2),\Rightarrow c=6$

Now $A=(2,2);B=(3,0);O=(0,0)$

$\therefore \left(3\hat{i}\right).(\hat{i}-2\hat{j})=3$