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

The Correct Option is D

To determine the value of \(|\tan \alpha|\), where \(\alpha\) is the angle between the two tangents drawn from point \((-2, -1)\) to the parabola \(y^2 = 4x\), we follow these steps:

1. Identify the curve and point from which tangents are drawn. The given parabola is \(y^2 = 4x\), which is a standard parabola opening to the right and having its vertex at the origin.
2. The equation of the tangent to \(y^2 = 4x\) at any point \((x_1, y_1)\) on the parabola can be written using the parametric point form: \(yy_1 = 2(x + x_1)\).
3. For the tangents from the external point \((-2, -1)\), we substitute this into the equation of the tangent: \((*1*) yy_1 = 2(x + (-2))\). By simplifying, \(yy_1 = 2x - 4\).
4. Next, determine the equation of the tangents in the slope form because the focus is on the angle between tangents. Using a slope \(m\), the tangent has the form \(y = mx + \frac{a}{m}\), where \(a=1\) in this parabola:
   • For \(y^2 = 4ax\), \(a=1\); hence, the tangent equation is: \(y = mx + \frac{1}{m}\).
   • Find perpendicular distance formula between line and point: the point \((-2, -1)\) and line \(y = mx + \frac{1}{m}\) as: \(\frac{|(-1) - m(-2) - \frac{1}{m}|}{\sqrt{1 + m^2}}\).
   • Set equal with zero (since tangents meet one and only at the same circle/parabola/ is solution, symmetrical about their axis – angle calculation): leads us to a quadratic equation in \(m\).
5. Form the relationship solving \(m^2 + 2m + 1 = \pm 2\sqrt{1 + m^2}\) to get solutions for \(m\).
6. Let \(m_1, m_2\) be slopes; to find the angle between two slopes, use \(\tan \alpha = \left|\frac{m_1 - m_2}{1 + m_1 m_2} \right|\).
7. Solving yields \(m_1 = \frac{1}{2}, m_2 = 3\); \(\tan \alpha = 3\).

The correct option is therefore, \(3\).