Question

The angle between the tangents drawn from the point (1, 4) to the parabola $y^2 = 4x$, is ______.

• A. $\pi/6$
• B. $\pi/2$
• C. $\pi/3$
• D. $\pi/4$

Hint:

Using Vieta's formulas ($\alpha+\beta, \alpha\beta$) and the identity $(\alpha-\beta) = \sqrt{(\alpha+\beta)^2 - 4\alpha\beta}$ is vastly faster than manually using the quadratic formula to find the two individual slopes!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must find the angle between two intersecting tangents drawn from an external point to a standard horizontal parabola.

Step 2: Detailed Explanation:
The standard equation of a tangent with slope $m$ to a parabola $y^2 = 4ax$ is:
$y = mx + \frac{a}{m}$
For the given parabola $y^2 = 4x$, comparing with $y^2 = 4ax$ gives $a = 1$.
So the tangent equation becomes:
$y = mx + \frac{1}{m}$
We are given that this tangent physically passes through the external point $(1, 4)$. Substitute $x=1, y=4$:
$4 = m(1) + \frac{1}{m}$
Multiply the entire equation by $m$ to form a quadratic equation:
$4m = m^2 + 1$
$m^2 - 4m + 1 = 0$
This quadratic equation has two roots ($m_1$ and $m_2$), representing the slopes of the two individual tangents.
From Vieta's formulas relating to quadratic roots:
Sum of roots: $m_1 + m_2 = -\frac{b}{a} = 4$
Product of roots: $m_1 m_2 = \frac{c}{a} = 1$
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
We can express $(m_1 - m_2)$ algebraically using the sum and product:
$(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$
$(m_1 - m_2)^2 = (4)^2 - 4(1)$
$(m_1 - m_2)^2 = 16 - 4 = 12$
$|m_1 - m_2| = \sqrt{12} = 2\sqrt{3}$
Substitute this back into the angle formula:
$\tan \theta = \frac{2\sqrt{3}}{1 + 1} = \frac{2\sqrt{3}}{2} = \sqrt{3}$
The principal angle whose tangent is $\sqrt{3}$ is $60^\circ$ or $\pi/3$ radians.

Step 3: Final Answer:
The angle is $\pi/3$, matching option (c).