The angle between the tangents drawn from the point (1, 4) to the parabola $y^2 = 4x$, is ______.
Show 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!
Step 1: Understanding the Concept:
The equation of a tangent to $y^2 = 4ax$ in terms of slope $m$ is $y = mx + \frac{a}{m}$. Step 2: Formula Application:
Here $4a = 4 \implies a = 1$. The tangent is $y = mx + \frac{1}{m}$.
Since it passes through $(1, 4)$: $4 = m(1) + \frac{1}{m}$. Step 3: Explanation:
$m^2 - 4m + 1 = 0$.
Let the slopes be $m_1$ and $m_2$.
Sum of slopes $m_1 + m_2 = 4$.
Product of slopes $m_1m_2 = 1$.
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| = \frac{\sqrt{(m_1+m_2)^2 - 4m_1m_2}}{1 + m_1m_2}$.
$\tan \theta = \frac{\sqrt{16 - 4}}{1 + 1} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
$\theta = \tan^{-1}(\sqrt{3}) = \pi/3$. Step 4: Final Answer:
The angle is $\pi/3$.