Animated Solution for Mathematics - Conic Sections: The angle between the tangents drawn from the point $(1,4)$ to the parabola $y^2=4x$ is

Imagine you are standing at point $P(1,4)$ in the Cartesian plane. Before you lies the elegant curve of the parabola $y^2=4x$, a classic right-opening shape.

Your goal is to shine two laser beams from your position at $P$ such that they perfectly graze the parabola. These beams are our tangents, and we want to calculate the angle $\theta$ between them.

To capture these laser beams mathematically, we turn to the slope form of a tangent. For any parabola $y^2=4ax$, the equation of a tangent with slope $m$ is given by:

In our case, comparing $y^2=4x$ with $y^2=4ax$, we immediately identify $a=1$. Thus, our tangent equation becomes:

This equation represents every possible tangent to our parabola. However, we only care about the two that pass through our specific point $P(1,4)$.

By forcing the tangent to pass through $P(1,4)$, we substitute $x=1$ and $y=4$ into our equation:

Multiplying by $m$ to clear the fraction, we arrive at the following quadratic equation:

This quadratic equation is the heart of our problem. The two roots, $m_1$ and $m_2$, are the slopes of our two laser beams.

We do not need to solve for $m_1$ and $m_2$ individually using the quadratic formula. Instead, we use Vieta's formulas:

Now, we invoke the angle formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$:

We know $m_1{m}_2=1$, but we need the difference $|m_1-m_2|$. We use the algebraic identity:

Substituting our values, we get:

Therefore, $|m_1-m_2|=\sqrt{12}=2\sqrt{3}$. Plugging this into our angle formula:

Since $\tan \theta =\sqrt{3}$, we conclude that $\theta =\frac{\pi }{3}$ or $60^{\circ }$. We have successfully calculated the angle between our laser beams, proving that even the most complex geometric problems can be tamed with the right algebraic tools.