If two tangents drawn from a point P to the parabola y² = 4x are at right angles, then the locus of

Q: Wait, why is it called a 'Director Circle' if the answer is a straight line?

That is a fantastic question! The term 'Director Circle' is a general name for the locus of the intersection of perpendicular tangents to any conic. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the locus is $x^2 + y^2 = a^2 + b^2$, which is indeed a circle. For a parabola, you can think of it as a circle with an infinite radius. As the radius tends to infinity, the curvature of the circle vanishes, and it becomes a straight line—the directrix. It is a beautiful example of how a general concept simplifies in a specific case!

Q: Can I derive this using the slope form of a tangent? Is that better?

Absolutely! You can definitely use the slope form. The equation of a tangent to $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$. If a point $(h, k)$ lies on this tangent, then $k = mh + \frac{a}{m}$, which rearranges to $m^2h - mk + a = 0$. If the two tangents are perpendicular, their slopes $m_1$ and $m_2$ satisfy $m_1m_2 = -1$. Using the product of roots formula for this quadratic in $m$, we get $\frac{a}{h} = -1$, which means $h = -a$. This confirms the locus is $x = -a$. Both methods are excellent, but knowing the 'Director Circle' property saves you precious time in the exam!

Q: Is this property true for all parabolas, or just $y^2 = 4x$?

It is true for all parabolas! The property that the locus of the intersection of perpendicular tangents is the directrix is a fundamental geometric characteristic of the parabola itself, regardless of its orientation. Whether the parabola is $y^2 = 4ax$, $x^2 = 4ay$, or even a rotated one, the locus of the intersection of perpendicular tangents will always be its directrix.

Question

If two tangents drawn from a point $P$ to the parabola $y^2 = 4x$ are at right angles, then the locus of $P$ is

Options:

(A) $2x + 1 = 0$

(B) $x = -1$

(C) $2x - 1 = 0$

(D) $x = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty plane. Before you, etched into the ground, is the elegant curve of a parabola, defined by the equation $y^2 = 4x$. It is a classic, right-opening curve, its vertex nestled comfortably at the origin $(0,0)$.

Now, imagine you pick an arbitrary point $P$ somewhere in this plane. From this point, you draw two lines that just barely kiss the curve—two tangents. The problem asks us to find the path, or the 'locus,' that point $P$ must trace if those two tangents are forced to meet at a perfect $90^{\circ}$ angle.

### The Concept of the Director Circle

In the world of conics, there is a powerful, almost magical concept known as the **Director Circle**. By definition, the locus of the point of intersection of perpendicular tangents to any conic section is called its Director Circle.

For an ellipse, this locus is a circle. For a hyperbola, it is also a circle. But the parabola is a rebel; it refuses to conform to the circular nature of its cousins.

When we apply the definition of the Director Circle to a parabola, the radius of that circle stretches to infinity. As the radius becomes infinite, the curve flattens out, and the circle degenerates into a straight line. That line is the **directrix** of the parabola.

### The Mathematical Execution

Let us translate this geometric intuition into the language of algebra. We are given the parabola $y^2 = 4x$. To find the directrix, we first need to identify the parameter $a$.

We compare our equation with the standard form $y^2 = 4ax$. By equating the coefficients, we see that:

$$4a = 4 \Rightarrow a = 1$$

This parameter $a$ is the distance from the vertex to the focus, and it is also the distance from the vertex to the directrix.

For a standard parabola $y^2 = 4ax$, the directrix is the vertical line defined by the equation $x = -a$. Since we have found that $a = 1$, the equation of our directrix becomes:

$$x = -1$$

This is the path! Any point $P$ that you choose on this line $x = -1$ will allow you to draw two tangents to the parabola $y^2 = 4x$ that are perfectly perpendicular to each other. If you move even a millimeter off this line, the angle between the tangents will shift, and the perpendicularity will be lost.

### Why This Matters

In the JEE Advanced, time is your most precious resource. While you could derive this using the slope form of the tangent—setting up a quadratic equation in $m$ and using the condition $m_1m_2 = -1$—knowing the property of the directrix allows you to solve this problem in seconds.

It transforms a complex analytical geometry problem into a simple identification of the directrix. Remember, mathematics is not just about calculating; it is about seeing the underlying structure of the universe.

The next time you see a parabola, don't just see a curve; see the directrix, the focus, and the hidden relationships that govern them. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic behind the math. The final locus is the line **$x = -1$**.

Animated Solution for Mathematics - Conic Sections: If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

Select Answer:

Visualized Solution

The Parabola

The External Point $P$ and Tangents

The Perpendicularity Condition

The Director Circle Concept

The Directrix as the Locus

Comparing with Standard Form

Finding the Parameter $a$

Equation of the Directrix

Substituting the Value of $a$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Concept of the Director Circle

The Mathematical Execution

Why This Matters

Similar Questions

If two tangents drawn from a point $P$ to the parabola $y^{2} = 16 (x - 3)$ are at right angles, then the locus of point $P$ is :

The point of intersection of the tangents at the ends of the latus rectum of the parabola $y^{2} = 4 x$ is.........

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

Let $C$ be the locus of the mirror image of a point on the parabola $y^{2} = 4 x$ with respect to the line $y = x$ . Then the equation of tangent to $C$ at $P (2, 1)$ is :

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.

Tangent is drawn to parabola $y^{2} - 2 y - 4 x + 5 = 0$ at a point $P$ which cuts the directrix at the point $Q$ . A point $R$ is such that it divides $QP$ externally in the ratio $1/2 : 1$ . Find the locus of point $R$ .

The tangent to the parabola $y^{2} = 4 x$ at the point where it intersects the circle $x^{2} + y^{2} = 5$ in the first quadrant, passes through the point :

The equation of a tangent to the parabola $y^{2} = 8 x$ is $y = x + 2$ . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two tangents drawn from a point P to the parabola y2=4x are at right angles, then the locus of P is