Key Concepts and Formulas Used

Before diving into the solution, let's recall the fundamental concepts required to solve this problem effectively.

1. Geometric Transformation: Reflection of a Point/Curve with Respect to the Line $y=x$:

   • When a point $P(x,y)$ is reflected across the line $y=x$, its mirror image, let's call it $P'(x',y')$, is found by simply interchanging its coordinates. That is, $x' = y$ and $y' = x$. So, $P'(y,x)$.
   • Consequently, to find the equation of the image of a curve defined by $F(x,y) = 0$ with respect to the line $y=x$, we replace every $x$ with $y$ and every $y$ with $x$ in the original equation. This yields the new equation $F(y,x) = 0$. This transformation essentially swaps the roles of the $x$ and $y$ axes.

2. Equation of Tangent to a Parabola at a Point on It:

   • The general equation of a parabola is $y^2 = 4ax$ (opens horizontally) or $x^2 = 4ay$ (opens vertically).
   • For a parabola of the form $y^2 = 4ax$, the equation of the tangent at a point $P(x_1, y_1)$ lying on the parabola is given by $yy_1 = 2a(x + x_1)$.
   • For a parabola of the form $x^2 = 4ay$, the equation of the tangent at a point $P(x_1, y_1)$ lying on the parabola is given by $xx_1 = 2a(y + y_1)$.
   • It is crucial to use the correct formula based on the orientation of the parabola. Both formulas are derived using differential calculus (finding the slope $\frac{dy}{dx}$ or $\frac{dx}{dy}$) or by the method of polar for a point on the curve (T=0).

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Step-by-Step Detailed Solution

The problem asks for the equation of the tangent to a curve $C$ at a specific point $P(2,1)$. The curve $C$ is defined as the mirror image of a given parabola with respect to the line $y=x$.

Step 1: Determine the Locus C (The Image Parabola)

• Understanding the Goal: Our first task is to find the algebraic equation that describes the locus $C$. This locus is formed by reflecting every point on the original parabola $y^2 = 4x$ across the line $y=x$.
• Applying the Reflection Rule: As established in our key concepts, to find the equation of the image of a curve $F(x,y)=0$ with respect to the line $y=x$, we simply interchange the variables $x$ and $y$ in the curve's equation.
  • The given original parabola is: $y^2 = 4x$.
  • To find the equation of its image, $C$, we perform the substitution: replace $y$ with $x$ and $x$ with $y$. $(x)^2 = 4(y)$ $x^2 = 4y$
• Equation of C: Therefore, the equation of the locus $C$ is $x^2 = 4y$. This is the equation of a standard parabola that opens upwards, with its vertex at the origin $(0,0)$.

Step 2: Verify if Point P(2,1) Lies on Curve C

• Understanding the Goal: We are asked to find the tangent to $C$ at point $P(2,1)$. The standard formula for the tangent to a conic section at a point $(x_1, y_1)$ is only valid if that point lies on the curve. If the point were external, a different approach (e.g., finding the chord of contact or using the slope form of the tangent) would be necessary. Hence, this verification step is crucial.
• Performing the Verification: We substitute the coordinates of $P(2,1)$ into the equation of $C$, which is $x^2 = 4y$.
  • Substitute $x=2$ and $y=1$: $(2)^2 = 4(1)$ $4 = 4$
  • Since the equation holds true ($4=4$), the point $P(2,1)$ indeed lies on the parabola $C$. This confirms that we can directly apply the tangent formula for a point on the curve.

Step 3: Apply the Tangent Equation Formula to C at P(2,1)

• Understanding the Goal: Now that we have the equation of $C$ and confirmed that $P(2,1)$ lies on it, we can use the appropriate tangent formula to find the equation of the line.
• Identifying Parabola Parameters: The equation of our parabola $C$ is $x^2 = 4y$. This is of the standard form $x^2 = 4ay$. We need to identify the value of the parameter $a$.
  • By comparing $x^2 = 4y$ with $x^2 = 4ay$, we can see that $4a = 4$.
  • Dividing both sides by 4, we get $a = 1$.
• Using the Tangent Formula: The equation of the tangent to a parabola of the form $x^2 = 4ay$ at a point $(x_1, y_1)$ on it is given by: $xx_1 = 2a(y + y_1)$
• Substituting Values: We have the parameter $a=1$, and the point of tangency is $P(x_1, y_1) = (2,1)$. Let's substitute these values into the tangent formula:
  • $x_1 = 2$
  • $y_1 = 1$
  • $a = 1$ $x(2) = 2(1)(y + 1)$
• Simplifying the Equation: Now, we simplify the equation to its standard linear form $Ax + By + C = 0$: $2x = 2(y + 1)$ Divide both sides by 2: $x = y + 1$ Rearrange the terms to get: $x - y - 1 = 0 \quad \text{or} \quad x - y = 1$

Thus, the equation of the tangent to the curve $C$ at the point $P(2,1)$ is $x - y = 1$.

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Tips for Success and Common Pitfalls to Avoid

• Mastering Geometric Transformations: Always clearly understand and internalize the rules for reflections and other geometric transformations. For reflection across $y=x$, it's a simple interchange of coordinates $(x,y) \to (y,x)$. However, rules differ for reflections across other lines (e.g., $x$-axis, $y$-axis, $y=-x$), and confusing them is a common mistake.
• Crucial Point Verification: It is an absolutely critical step to verify that the given point $P(x_1, y_1)$ actually lies on the curve before using the standard tangent formula for a point on the curve. If the point is external, you would need to use different methods (e.g., finding the equation of the chord of contact or using the slope form of the tangent $y=mx+a/m$ and conditions for tangency).
• Correct Tangent Formula for Parabola Orientation: Be extremely careful to use the correct tangent formula that matches the specific orientation of your parabola.
  • For $y^2 = 4ax$ (opens right/left), the tangent is $yy_1 = 2a(x+x_1)$.
  • For $x^2 = 4ay$ (opens up/down), the tangent is $xx_1 = 2a(y+y_1)$. Confusing these two formulas is a very frequent source of error in JEE problems.
• Algebraic Accuracy: Double-check all substitutions, arithmetic operations, and algebraic simplifications. Even a small calculation error can lead to an incorrect final answer, especially under exam pressure.

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Summary and Key Takeaway

This problem is a quintessential example testing two core concepts in coordinate geometry: (1) the ability to perform geometric transformations (specifically reflection across the line $y=x$) to find the equation of a transformed curve, and (2) the accurate application of the tangent formula for a parabola at a point lying on it. The solution process involved transforming the original parabola $y^2=4x$ into its image $x^2=4y$, verifying that the given point $P(2,1)$ lies on this new parabola, and then precisely using the tangent formula $xx_1 = 2a(y+y_1)$ with the correct parameters to arrive at the final equation $x-y=1$. Mastery of these fundamental ideas and meticulous execution are vital for success in JEE problems involving conic sections.