If two tangents drawn from a point P to the parabola y² = 16(x - 3) are at right angles, then the lo

Q: Why is the locus of perpendicular tangents always the directrix? Is this a universal rule?

Yes, it is a beautiful and universal property of parabolas! If you take any point on the directrix and draw two tangents to the parabola, the angle between them will always be $90^\circ$. This is a standard result in coordinate geometry that saves you from doing lengthy algebraic calculations involving slopes.

Q: Can I solve this using the slope form of the tangent $y = mx + \frac{a}{m}$ instead of memorizing the directrix property?

Absolutely! You can use the tangent equation $y = mx + \frac{a}{m}$. If the tangents are perpendicular, their slopes $m_1$ and $m_2$ satisfy $m_1 m_2 = -1$. By expressing the intersection point $(h, k)$ in terms of $m$ and using the perpendicularity condition, you will arrive at the same result. However, remembering the directrix property is a powerful tool for competitive exams.

Q: I am confused about the shift. If the parabola is $y^2 = 16(x - 3)$, why is the directrix $x - 3 = -4$ and not $x = -4$?

Great catch! The directrix equation $X = -a$ applies to the standard form $Y^2 = 4aX$. In your case, the parabola is shifted. The variable $X$ in the standard formula corresponds to $(x - 3)$ in your equation. Therefore, you must set the entire shifted term $(x - 3)$ equal to $-a$. This is why the directrix becomes $x - 3 = -4$.

Question

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 :

Options:

(A) $x + 3 = 0$

(B) $x + 1 = 0$

(C) $x + 2 = 0$

(D) $x + 4 = 0$

Sigma Percentile · Accepted Answer

### The Geometry of Tangents and the Magic of the Directrix

Imagine you are standing before the elegant curve of a parabola. It is a shape defined by its symmetry and its relationship with a fixed point called the focus and a fixed line called the directrix.

Today, we are exploring a fascinating property: what happens when we draw two tangents to a parabola that meet at a perfect right angle? This is not just a random occurrence; it is a geometric dance that always leads us to the same destination.

### The Perpendicularity Condition

When we are given that two tangents drawn from a point $P$ to the parabola $y^2 = 16(x - 3)$ are at right angles, we are essentially being handed a key to a secret door.

In the world of coordinate geometry, the locus of the point of intersection of two perpendicular tangents to a parabola is always its directrix. This is a powerful, time-saving property that every JEE aspirant should keep in their toolkit.

Instead of grinding through complex algebra, we can simply find the equation of the directrix.

### Analyzing the Shifted Parabola

Our parabola is $y^2 = 16(x - 3)$. It is not centered at the origin, but that should not intimidate us.

We compare it to the standard form $Y^2 = 4aX$. By mapping the terms, we see that $Y = y$ and $X = x - 3$.

The coefficient of $X$ is $16$, so we set $4a = 16$, which gives us $a = 4$. This parameter $a$ is the distance from the vertex to the focus, and it is crucial for defining the directrix.

### The Final Calculation

For a standard parabola $Y^2 = 4aX$, the directrix is defined by the equation $X = -a$.

Since our parabola is shifted, we substitute our shifted coordinate $X = x - 3$ and our parameter $a = 4$ into this formula:

$$x - 3 = -4$$

Solving for $x$, we add $3$ to both sides, resulting in $x = -1$.

Rearranging this, we get **$x + 1 = 0$**. This is the locus of point $P$.

It is a vertical line, perfectly parallel to the y-axis, representing the directrix of our parabola. The elegance of this result lies in how the perpendicularity condition simplifies the entire problem into a single, clean equation.

Animated Solution for Mathematics - Conic Sections: 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 :

Select Answer:

Visualized Solution

Visualizing the Parabola $y^{2} = 16 (x - 3)$

Introducing Point $P$ and Tangents

The Perpendicularity Condition $9 0^{\circ}$

The Locus Property: Directrix

Standard Form Comparison $Y^{2} = 4 a X$

Finding the Parameter $a$

Directrix Equation $X = - a$

Substituting the Shifted Coordinates

Final Locus Equation $x + 1 = 0$

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Tangents and the Magic of the Directrix

The Perpendicularity Condition

Analyzing the Shifted Parabola

The Final Calculation

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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=16(x−3) are at right angles, then the locus of point P is :

Select Answer:

Visualized Solution

Visualizing the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=16(x−3)

Introducing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and Tangents

The Perpendicularity Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 90∘

The Locus Property: Directrix

Standard Form Comparison .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y2=4aX

Finding the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Directrix Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=−a