Question

A line drawn from the orign touches the curve $y^2=x^2-x+1$ in the first quadrant. If the line is incline at an angle $\theta$ with the positive x-axis, then $\tan \theta$ is

• A. $\frac{\sqrt{3}}{2}$
• B. $\sqrt{\frac{3}{7}}$
• C. $\frac{2}{\sqrt{7}}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is A $\frac{\sqrt{3}}{2}$

Let $y-mx$ ……..(i)

$y^2=x^2-x+1$

$\Rightarrow m^2{x}^2=x^2-x+1$ ……(ii)

Let $(x_1,y_1)$ be the point of

$2y_1\frac{dy}{dx}=2x_1-1$

$\Rightarrow \frac{dy}{dx}=\frac{2x_1-1}{2y_1}$

$\Rightarrow m=\frac{2x_1-1}{2m{x}_1}$

$\Rightarrow 2m^2{x}_1-2x_1+1=0$

$\Rightarrow m^2=1-\frac{1}{2x_1}$

$(1-\frac{1}{2x_1})x_1^2=x_1^2-x_1+1$

$\Rightarrow x_1^2-\frac{x_1}{2}=x_1^2-x_1+1$

$\Rightarrow \frac{x_1}{2}=1$

$\therefore x_1=2$

$x_1^2=4-2+1$

$\Rightarrow x=\sqrt{3}$

$\therefore \sqrt{3}=2m$

$\Rightarrow \tan \theta =\frac{\sqrt{3}}{2}$.