The angle between a pair of tangents drawn from a point P to the parabola y² = 4ax is 45°. Show that

Q: Wait, why do we use the slope form of the tangent $y = mx + \frac{a}{m}$ instead of the point-slope form?

Great question! We use the slope form because it directly links the slope $m$ of the tangent to the parameters of the parabola ($a$) and the coordinates of the point $P(h, k)$. By treating $m$ as the variable in a quadratic equation, we can use the properties of roots (sum and product) to relate the slopes of the two tangents to the coordinates of $P$ without needing to know the exact points of tangency.

Q: Why does the locus turn out to be a hyperbola? I thought it might be a circle or another parabola.

That is a natural intuition! The nature of the locus is determined by the final algebraic form. When we simplify the equation $(h+a)^2 = k^2 - 4ah$ and rearrange it, we get $(h+3a)^2 - k^2 = 8a^2$. This is of the form $\frac{(x-x_0)^2}{A^2} - \frac{(y-y_0)^2}{B^2} = 1$, which is the definition of a hyperbola. The negative sign between the squared terms is the key indicator that it is a hyperbola, not an ellipse or parabola.

Q: Is there a shortcut for this? Do I always have to derive the quadratic?

While deriving the quadratic is the most robust method, you can remember that the locus of the intersection of two tangents to a parabola $y^2 = 4ax$ that meet at an angle $\theta$ is given by the equation $(x+a)^2 \tan^2 \theta = (y^2 - 4ax)$. For $\theta = 45^\circ$, $\tan^2 \theta = 1$, which simplifies to $(x+a)^2 = y^2 - 4ax$. This leads to the same result, but deriving it ensures you understand the geometry behind the formula!

Question

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^2 = 4ax$ is $45^\circ$. Show that the locus of the point $P$ is a hyperbola.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe! Today, we are going to peel back the layers of a classic JEE Advanced problem.

Imagine you are standing in the coordinate plane, looking at the elegant curve of the parabola $y^2 = 4ax$. You pick an arbitrary point $P(h, k)$ outside this curve and draw two tangents to the parabola.

The problem asks us to prove that if the angle between these two tangents is fixed at $45^\circ$, then the point $P$ must trace out a hyperbola. Let us embark on this journey.

### The Tangent's Identity

To begin, we must capture the essence of a tangent line. In the world of parabolas, the most powerful tool for this is the slope form of the tangent:

$$y = mx + \frac{a}{m}$$

This equation is beautiful because it tells us that for any given slope $m$, there is a unique tangent line. Since our point $P(h, k)$ lies on these tangents, it must satisfy this equation.

By substituting $x = h$ and $y = k$, we get $k = mh + \frac{a}{m}$. This is our bridge between the geometry of the point and the algebra of the slopes.

### The Quadratic Bridge

Now, let us refine this relationship. Multiplying the entire equation by $m$, we obtain $mk = m^2h + a$, which rearranges into the quadratic equation:

$$m^2h - mk + a = 0$$

This is a pivotal moment! This quadratic equation in $m$ tells us that for any point $P(h, k)$, there are at most two tangents that can be drawn to the parabola.

The roots of this equation, $m_1$ and $m_2$, are precisely the slopes of these two tangents. Using Vieta's formulas, we immediately know:

$$m_1 + m_2 = \frac{k}{h} \quad 	ext{and} \quad m_1m_2 = \frac{a}{h}$$

### The Angle Constraint

We are given that the angle $	heta$ between these tangents is $45^\circ$. The formula for the angle between two lines is:

$$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$$

Since $	an 45^\circ = 1$, we have $1 = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$. To handle the absolute value and the difference of roots, we square both sides:

$$1 = \frac{(m_1 - m_2)^2}{(1 + m_1m_2)^2}$$

Using the algebraic identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2$, we can express everything in terms of our sum and product of roots.

### The Algebraic Transformation

Substituting our values, we get:

$$1 = \frac{(\frac{k}{h})^2 - 4(\frac{a}{h})}{(1 + \frac{a}{h})^2}$$

Simplifying the numerator and denominator, we find:

$$1 = \frac{\frac{k^2 - 4ah}{h^2}}{\frac{(h+a)^2}{h^2}}$$

The $h^2$ terms cancel out, leaving us with $(h+a)^2 = k^2 - 4ah$. Expanding this, we get $h^2 + 2ah + a^2 = k^2 - 4ah$, which rearranges to $h^2 + 6ah + a^2 - k^2 = 0$.

### Final Calculation

By completing the square for $h$, we arrive at:

$$(h+3a)^2 - k^2 = 8a^2$$

Replacing $h$ and $k$ with $x$ and $y$, we get the final locus:

$$(x+3a)^2 - y^2 = 8a^2$$

This is the standard equation of a **hyperbola**. We have successfully navigated the algebra to reveal the hidden geometric truth. Keep practicing, and remember: every equation tells a story!

Animated Solution for Mathematics - Conic Sections: The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

Visualized Solution

Visualizing the Parabola and Point $P$

Equation of Tangent in Slope Form

Forming the Quadratic in $m$

Sum and Product of Slopes

Applying the Angle Formula

Substituting $m_{1}$ and $m_{2}$ in Terms of $h, k$

Simplifying the Expression

Expanding and Rearranging

Completing the Square

The Final Locus: A Hyperbola

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Identity

The Quadratic Bridge

The Angle Constraint

The Algebraic Transformation

Final Calculation

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