Question

Two tangents are drawn from a point $(- 2, -1)$ to the curve, $y^{2}=4x.$ If $\alpha$ is the angle between them, then $|\tan \alpha|$ is equal to :

• A. $\frac{1}{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $3$

Answer 1

Answer: (D)

Let equation of tangent from $(-2,-1)$ be $y +1= m ( x +2)$
$\Rightarrow y = m x +(2 m -1)$
Condition of tangency, $C =\frac{ a }{ m }$
i.e., $2 m -1=\frac{1}{ m }$
$\Rightarrow 2 m ^{2}- m -1=0 $
$(2 m +1)( m -1)=0 $
$m =-\frac{1}{2}, 1 $
Now, $|\tan \alpha|=\left|\frac{ m _{1}- m _{2}}{1+ m _{1} m _{2}}\right|$
$=\left|\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right|=3$