This solution provides a comprehensive approach to finding the locus of a point from which two perpendicular tangents can be drawn to a parabola. We will explore two methods: first, by directly applying a key property of parabolas (the recommended approach for efficiency in JEE), and second, by deriving the result algebraically using the condition for perpendicular tangents.

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Key Concept: The Director Circle and the Directrix of a Parabola

For any conic section (parabola, ellipse, or hyperbola), the locus of the point of intersection of two tangents that are perpendicular to each other is known as its director circle. This is a fundamental property in coordinate geometry.

However, the parabola is a special case. For a parabola, this director circle degenerates into a straight line, which is precisely its directrix. This means that if you draw two tangents from a point P to a parabola such that these tangents are perpendicular to each other, then the point P must lie on the directrix of that parabola.

Therefore, to find the locus of point P from which two tangents drawn to the parabola $y^2 = 16(x-3)$ are at right angles, we simply need to find the equation of the directrix of this parabola.

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Understanding the Parabola Equation

The given equation of the parabola is $y^2 = 16(x-3)$. To work with this, we need to compare it with the standard form of a parabola whose axis is parallel to the x-axis: $(y-k)^2 = 4a(x-h)$ where $(h, k)$ represents the coordinates of the vertex of the parabola, and $a$ is the focal length (the distance from the vertex to the focus).

Let's extract these parameters from our given equation: $y^2 = 16(x-3)$

• Identifying $(h, k)$:
  • Comparing $(y-k)^2$ with $y^2$, we see that $k=0$.
  • Comparing $(x-h)$ with $(x-3)$, we see that $h=3$.
  • So, the vertex of the parabola is $(h, k) = (3, 0)$.
• Identifying $a$ (focal length):
  • Comparing $4a$ with $16$, we have $4a = 16$.
  • Dividing by 4, we get $a = 4$.

Since $a=4$ (which is positive), and the $y$-term is squared, this parabola opens to the right.

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Method 1: Using the Directrix Property (Recommended for Efficiency)

This method directly applies the key property identified above, making it very efficient for JEE exams.

Step 1: Recall the Directrix Equation for a Standard Parabola For a standard parabola of the form $Y^2 = 4AX$ (with its vertex at the origin and opening along the positive X-axis), the equation of the directrix is given by: $X = -A$ The directrix is a vertical line located at a distance '$A$' from the vertex, on the opposite side of the focus.

Step 2: Adapt for the Shifted Parabola Our parabola $y^2 = 16(x-3)$ is a shifted version of the standard parabola. To account for this shift, we replace $X$ with $(x-h)$ and $A$ with $a$ in the directrix equation. So, for a parabola of the form $(y-k)^2 = 4a(x-h)$, the equation of the directrix is: $(x-h) = -a$

Step 3: Substitute the Values and Determine the Locus Now, we substitute the values we found: $h=3$ and $a=4$ into the directrix equation $(x-h) = -a$: $x - 3 = -4$ To solve for $x$, add 3 to both sides: $x = -4 + 3$ $x = -1$ Thus, the locus of point P is the line $x = -1$. This can also be written as $x+1=0$.

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Method 2: Using the Condition for Perpendicular Tangents (Algebraic Derivation)

This method provides a deeper understanding by deriving the result algebraically, without directly relying on the director circle property.

Step 1: Write the General Equation of a Tangent to the Parabola The general equation of a tangent with slope $m$ to a standard parabola of the form $Y^2 = 4AX$ is given by: $Y = mX + \frac{A}{m}$ For our given parabola $y^2 = 16(x-3)$:

• We have $4A = 16$, so $A=4$.
• The variable $Y$ is replaced by $y$.
• The variable $X$ is replaced by $(x-3)$.

Therefore, the equation of a tangent to $y^2 = 16(x-3)$ with slope $m$ is: $y = m(x-3) + \frac{4}{m}$ This equation represents any tangent line to the parabola based on its slope $m$.

Step 2: Form a Quadratic Equation in 'm' for Tangents from P Let P be a point $(x_1, y_1)$ from which two tangents are drawn to the parabola. If the tangent passes through P$(x_1, y_1)$, then its coordinates must satisfy the tangent equation: $y_1 = m(x_1-3) + \frac{4}{m}$ Our goal is to find the slopes of the tangents drawn from P. To do this, we rearrange this equation into a quadratic in $m$. First, multiply the entire equation by $m$ (assuming $m \neq 0$, as a vertical tangent has an undefined slope, but if it were vertical, its perpendicular would be horizontal, which can be handled by this form): $my_1 = m^2(x_1-3) + 4$ Now, rearrange it into the standard quadratic form $Am^2 + Bm + C = 0$: $m^2(x_1-3) - my_1 + 4 = 0$ This quadratic equation has two roots, $m_1$ and $m_2$, which represent the slopes of the two tangents that can be drawn from the point P$(x_1, y_1)$ to the parabola.

Step 3: Apply the Condition for Perpendicular Tangents The problem states that the two tangents drawn from P are at right angles. This means the product of their slopes must be $-1$: $m_1 m_2 = -1$ From the quadratic equation $m^2(x_1-3) - my_1 + 4 = 0$, by Vieta's formulas, the product of the roots ($m_1 m_2$) is given by $\frac{\text{constant term}}{\text{coefficient of } m^2}$: $m_1 m_2 = \frac{4}{x_1-3}$ Now, we equate the two expressions for $m_1 m_2$: $\frac{4}{x_1-3} = -1$ This step is crucial as it connects the geometric condition (perpendicular tangents) to the algebraic properties of the quadratic equation.

Step 4: Solve for the Locus of P Solve the equation obtained in Step 3 for $x_1$: $4 = -(x_1-3)$ $4 = -x_1 + 3$ To isolate $x_1$, move $x_1$ to the left side and 4 to the right side: $x_1 = 3 - 4$ $x_1 = -1$ Since P is a general point $(x_1, y_1)$, its locus is given by replacing $x_1$ with $x$. So, the locus of P is $x = -1$, or $x+1=0$.

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Tips for Success & Common Pitfalls:

• Master the Standard Forms: Always start by identifying the correct standard form of the conic section. For a parabola, know the forms $(y-k)^2 = 4a(x-h)$ and $(x-h)^2 = 4a(y-k)$, and their respective directrix equations. Understanding the orientation (opens left/right or up/down) is key.
• **Correct