1. Problem Analysis and Essential Concepts

This problem requires us to find the equation of a tangent line to a specific hyperbola. We are given two crucial pieces of information: the hyperbola passes through a particular point $(4,6)$, and its eccentricity is $e=2$. To solve this, we will primarily rely on three fundamental concepts related to standard hyperbolas:

• Standard Equation of a Hyperbola: For a hyperbola centered at the origin with its transverse axis along the x-axis (the most common "standard" form unless otherwise specified), its equation is: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ Here, $a$ represents the length of the semi-transverse axis, and $b$ represents the length of the semi-conjugate axis. Note the minus sign, which distinguishes it from an ellipse.

• Eccentricity of a Hyperbola ($e$): The eccentricity quantifies the "openness" or "flatness" of the hyperbola. It is always greater than 1 for a hyperbola ($e > 1$). The relationship between $e$, $a$, and $b$ is given by: $e^2 = 1 + \frac{b^2}{a^2}$

• Equation of Tangent to a Hyperbola at a Point $(x_1, y_1)$: If a point $(x_1, y_1)$ lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the equation of the tangent line at this specific point is given by the 'T=0' form: $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$

Our strategy will be to use the given point and eccentricity to determine the unknown parameters $a^2$ and $b^2$ that define this specific hyperbola. Once these parameters are known, we can directly substitute them, along with the given point $(4,6)$, into the tangent equation formula.

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2. Step-by-Step Solution

Step 2.1: Forming the first equation using the given point (4,6)

• Concept Used: Any point that lies on a curve must satisfy the equation of that curve.
• Why this step? The point $(4,6)$ is stated to be on the hyperbola. This means if we substitute $x=4$ and $y=6$ into the general equation of the hyperbola, the equation must hold true. This allows us to establish the first algebraic relationship between our unknown parameters, $a^2$ and $b^2$.

Let the equation of the standard hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Substitute the coordinates of the given point $(x, y) = (4, 6)$ into this equation: $\frac{(4)^2}{a^2} - \frac{(6)^2}{b^2} = 1$ $\frac{16}{a^2} - \frac{36}{b^2} = 1 \quad \text{.........(1)}$

Step 2.2: Using the eccentricity to find a second relationship between $a^2$ and $b^2$

• Concept Used: The eccentricity formula for a hyperbola.
• Why this step? We have two unknown parameters, $a^2$ and $b^2$, in Equation (1). To solve for these two unknowns, we need a second independent equation. The given eccentricity provides this crucial second relationship, linking $a^2$ and $b^2$.

We are given that the eccentricity $e=2$. Recall the eccentricity formula for a hyperbola: $e^2 = 1 + \frac{b^2}{a^2}$. Substitute $e=2$ into the formula: $(2)^2 = 1 + \frac{b^2}{a^2}$ $4 = 1 + \frac{b^2}{a^2}$ Subtract 1 from both sides to isolate the ratio $\frac{b^2}{a^2}$: $3 = \frac{b^2}{a^2}$ Rearranging this equation gives us a direct relationship between $b^2$ and $a^2$: $b^2 = 3a^2 \quad \text{.........(2)}$

Step 2.3: Calculating the values of $a^2$ and $b^2$

• Concept Used: Solving a system of simultaneous equations.
• Why this step? Our immediate goal is to determine the specific equation of the hyperbola, which requires knowing the numerical values of $a^2$ and $b^2$. By substituting the expression for $b^2$ from Equation (2) into Equation (1), we can eliminate one variable and solve for the other.

Substitute $b^2 = 3a^2$ from Equation (2) into Equation (1): $\frac{16}{a^2} - \frac{36}{3a^2} = 1$ Simplify the second term on the left-hand side: $\frac{16}{a^2} - \frac{12}{a^2} = 1$ Combine the terms, as they share a common denominator: $\frac{16 - 12}{a^2} = 1$ $\frac{4}{a^2} = 1$ This directly gives us the value of $a^2$: $a^2 = 4$ Now, substitute the value of $a^2=4$ back into Equation (2) to find $b^2$: $b^2 = 3(4)$ $b^2 = 12$

Step 2.4: Writing the complete equation of the hyperbola (Optional but Recommended)

• Concept Used: Substituting specific parameters into the standard hyperbola equation.
• Why this step? While not strictly required to find the tangent, explicitly writing out the hyperbola's equation helps to consolidate our understanding of the specific curve we are dealing with and provides a good intermediate check of our calculations.

The equation of the hyperbola is: $\frac{x^2}{4} - \frac{y^2}{12} = 1$

Step 2.5: Determining the equation of the tangent at point (4,6)

• Concept Used: The formula for the tangent to a hyperbola at a point $(x_1, y_1)$ lying on the curve.
• Why this step? This is the final step to answer the problem's question. We now have all the necessary components: the point of tangency $(x_1, y_1) = (4, 6)$ and the hyperbola's parameters $a^2=4$ and $b^2=12$. We can directly apply the tangent formula.

Substitute $x_1=4$, $y_1=6$, $a^2=4$, and $b^2=12$ into the tangent formula $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$: $\frac{x(4)}{4} - \frac{y(6)}{12} = 1$ Simplify the terms in the equation: $\frac{4x}{4} - \frac{6y}{12} = 1$ $x - \frac{y}{2} = 1$ To eliminate the fraction and present the equation in a standard linear form ($Ax + By + C = 0$), multiply the entire equation by 2: $2 \left( x - \frac{y}{2} \right) = 2(1)$ $2x - y = 2$ Rearrange the terms to match the options format: $2x - y - 2 = 0$

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3. Crucial Tips and Common Pitfalls

• Master Standard Forms: Always begin by clearly recalling the standard forms for the hyperbola equation, its eccentricity, and the tangent equation. A common mistake is to confuse the hyperbola formula with that of an ellipse (which has a '+' sign).
• Sign Errors are Critical: The minus sign in the hyperbola's equation $\left(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\right)$ and its tangent formula $\left(\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1\right)$ is non-negotiable. Ensure you maintain it throughout calculations.
• Transverse Axis Orientation: While this problem implicitly assumes the transverse axis is along the x-axis, be aware that if it were along the y-axis, the standard equation would be $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, and the eccentricity and tangent formulas would adjust accordingly.
• Algebraic Precision: Be extremely careful with algebraic manipulations, especially when combining fractions and solving simultaneous equations. A small arithmetic error can lead to an incorrect final answer.
• Tangent at a Point vs. Tangent with a Given Slope: Remember that this problem asks for the tangent at a specific point on the curve. This uses the 'T=0' substitution method. If you were given the slope of the tangent, a different formula (or differentiation) would be required.
• Working with $a^2$ and $b^2$: It is often simpler and less prone to error to work directly with $a^2$ and $b^2$ as single variables throughout the calculation rather than finding $a$ and $b$ individually and then squaring them.

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4. Summary and Key Takeaways

This problem beautifully illustrates a systematic approach to solving conic section questions. The key takeaways are:

1. Translate Information into Equations: Each piece of given information (a point on the curve, eccentricity, etc.) provides a constraint that can be written as an algebraic equation involving the conic section's parameters.
2. Solve for Parameters: Use the system of equations formed in step 1 to determine the specific values of the parameters (like $a^2$ and $b^2$) that define the unique conic section.
3. Apply Specific Formulas: Once the conic section's parameters are known, apply the relevant formula (e.g., tangent equation, normal equation, focal properties) to directly answer the question.

By mastering these steps, you can confidently tackle a wide array of problems involving hyperbolas and other conic sections in JEE Mathematics.

Comparing our derived equation $2x - y - 2 = 0$ with the given options, it matches option (A). Therefore, the correct answer is (A).