Key Concepts and Formulas

1. Standard Equation of a Hyperbola with Transverse Axis along the x-axis: When the vertices of a hyperbola are located at $(\pm a, 0)$, it signifies that the transverse axis lies along the x-axis, and the center is at the origin $(0,0)$. In this configuration, the standard equation of the hyperbola is given by: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ Here, $a$ represents the length of the semi-transverse axis (distance from the center to a vertex), and $b$ represents the length of the semi-conjugate axis. The value $a^2$ is always the denominator of the positive term.

2. Equation of the Normal to a Hyperbola: The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at a specific point $P(x_1, y_1)$ lying on the hyperbola is: $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$ This formula can be derived by first finding the slope of the tangent at $(x_1, y_1)$ using implicit differentiation. The slope of the normal is then the negative reciprocal of the tangent's slope. Finally, using the point-slope form of a line, $y - y_1 = m_{\text{normal}}(x - x_1)$, leads to this standard normal equation.

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Step-by-Step Solution

1. Determine the Orientation of the Hyperbola and the Value of $a^2$

• Why this step is taken: The given coordinates of the vertices are fundamental for identifying the orientation of the hyperbola and the value of its semi-transverse axis, '$a$'. This is the first step in constructing the hyperbola's equation.
• We are given that the hyperbola has vertices at $(\pm 6, 0)$.
• Since the vertices are of the form $(\pm a, 0)$, this immediately indicates that the transverse axis of the hyperbola lies along the x-axis, and its center is at the origin $(0,0)$.
• By comparing $(\pm a, 0)$ with $(\pm 6, 0)$, we can directly deduce the length of the semi-transverse axis: $a = 6$.
• Therefore, the square of the semi-transverse axis is $a^2 = 6^2 = 36$.
• At this stage, the standard equation of our hyperbola takes the form: $\frac{x^2}{36} - \frac{y^2}{b^2} = 1$ We still need to find $b^2$.

2. Use the Given Point P(10, 16) to Determine the Value of $b^2$

• Why this step is taken: A hyperbola is uniquely defined by its parameters $a^2$ and $b^2$. We have $a^2$, and we need $b^2$. The fact that the hyperbola passes through a specific point $P(10, 16)$ means that these coordinates must satisfy the hyperbola's equation. This allows us to substitute the point's coordinates into the equation and solve for the unknown parameter $b^2$.
• The hyperbola passes through the point $P(10, 16)$. We substitute $x=10$ and $y=16$ into the hyperbola's equation along with $a^2=36$: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $\frac{10^2}{36} - \frac{16^2}{b^2} = 1$
• Now, we perform the necessary algebraic calculations to find $b^2$: $\frac{100}{36} - \frac{256}{b^2} = 1$
• Simplify the fraction $\frac{100}{36}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 4: $\frac{25}{9} - \frac{256}{b^2} = 1$
• Isolate the term containing $b^2$ by moving the constant term to the right side of the equation: $\frac{256}{b^2} = \frac{25}{9} - 1$ $\frac{256}{b^2} = \frac{25 - 9}{9}$ $\frac{256}{b^2} = \frac{16}{9}$
• To solve for $b^2$, we can cross-multiply or rearrange the terms: $b^2 = \frac{256 \times 9}{16}$
• Simplify the expression. We know that $256 = 16 \times 16$: $b^2 = 16 \times 9$ $b^2 = 144$
• (Although $b=12$, only the value of $b^2$ is required for the normal equation formula.)

3. Write the Complete Equation of the Hyperbola (Optional but Good Practice)

• Why this step is taken: Explicitly writing the complete equation of the hyperbola confirms that we have successfully found both $a^2$ and $b^2$. It serves as a good intermediate check of our work and ensures we have all parameters correctly identified before moving to the final step.
• With $a^2 = 36$ and $b^2 = 144$, the equation of the hyperbola is: $\frac{x^2}{36} - \frac{y^2}{144} = 1$

4. Find the Equation of the Normal at Point P(10, 16)

• Why this step is taken: This is the final step where we apply the standard formula for the normal to the hyperbola using all the parameters ($a^2$, $b^2$) we've determined and the given point of interest $(x_1, y_1)$.
• We use the formula for the normal at $P(x_1, y_1) = (10, 16)$, with $a^2 = 36$ and $b^2 = 144$: $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$
• Substitute the values into the formula: $\frac{36x}{10} + \frac{144y}{16} = 36 + 144$
• Perform the addition on the right side and simplify the fractions on the left side: $\frac{18x}{5} + \frac{9y}{1} = 180$
• To eliminate the denominator and express the equation in a standard linear form (without fractions), multiply the entire equation by the least common multiple of the denominators (which is 5): $5 \left( \frac{18x}{5} \right) + 5 (9y) = 5 (180)$ $18x + 45y = 900$
• To simplify the equation further and match the format often seen in multiple-choice options, divide the entire equation by the greatest common divisor (GCD) of the coefficients (18, 45, and 900). The GCD of these numbers is 9: $\frac{18x}{9} + \frac{45y}{9} = \frac{900}{9}$ $2x + 5y = 100$

5. Compare with Options

• The derived equation of the normal is $2x + 5y = 100$.
• This equation perfectly matches option (A).

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Tips and Common Mistakes to Avoid:

• Correctly Identifying $a^2$ and $b^2$: For a hyperbola, $a^2$ is always the denominator under the positive term ($x^2$ or $y^2$), regardless of whether $a > b$ or $b > a$. The vertices $(\pm a, 0)$ or $(0, \pm a)$ directly give the value of '$a$'. Do not confuse this with ellipses where $a^2$ is always the larger denominator.
• Algebraic Precision: Be meticulous with calculations involving fractions, squaring, and multiplications. A single arithmetic error can lead to an incorrect $b^2$ or a wrong final equation. Double-check your work, especially when simplifying fractions.
• Normal vs. Tangent Formulas: Ensure you use the correct formula. The equation of the normal is distinct from the equation of the tangent. Remember:
  • Equation of Tangent: $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$
  • Equation of Normal: $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$
• Simplifying the Final Equation: Always simplify the equation of the line to its simplest integer form by dividing by the greatest common divisor of all coefficients. This makes it easier to compare with the given options.

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Summary and Key Takeaway:

This problem is a comprehensive test of your understanding of hyperbolas, specifically:

1. Geometric Interpretation: How to extract parameters (like $a$) from given geometric properties (like vertices).
2. Equation Formulation: How to use a point on the curve to determine the remaining parameter ($b^2$) and complete the hyperbola's equation.
3. Application of Formulas: How to correctly apply the standard formula for the normal to a hyperbola at a given point.
4. Algebraic Manipulation: Skill in simplifying equations to match standard forms.

Mastering the standard forms, properties, and associated formulas for all conic sections (parabola, ellipse, hyperbola) is fundamental for success in JEE Mathematics. This problem reinforces the importance of knowing both the defining characteristics and the standard equations related to tangents and normals for these curves.