Understanding the Problem and Key Concepts

This problem requires us to determine the equation of a specific hyperbola and then find the equation of its normal at a given point. We are provided with the eccentricity of the hyperbola and a point that lies on it. The final goal is to find the value of $\lambda - \beta$ by comparing our derived normal equation with a given general form.

To solve this, we will use the following fundamental properties and formulas for a hyperbola centered at the origin with its transverse axis along the x-axis:

1. Standard Equation: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
2. Eccentricity Relation: $b^2 = a^2(e^2 - 1)$
3. Equation of Normal at $(x_1, y_1)$: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$

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Step 1: Establish a Relationship Between the Semi-Axes ($a$ and $b$) Using the Given Eccentricity

Concept: The eccentricity $e$ of a hyperbola quantifies its shape and relates its semi-transverse axis ($a$) and semi-conjugate axis ($b$). For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the relationship is given by $b^2 = a^2(e^2 - 1)$.

Why this step?: We have two unknown parameters, $a^2$ and $b^2$, that define the hyperbola. By using the given eccentricity, we can establish a direct relationship between $a^2$ and $b^2$, effectively reducing the number of independent unknowns to one. This is the first piece of information we use to characterize our specific hyperbola.

Given: The eccentricity $e = \frac{5}{4}$.

Calculation: Substitute the value of $e$ into the eccentricity relation: $b^2 = a^2 \left( \left(\frac{5}{4}\right)^2 - 1 \right)$ First, square the eccentricity: $e^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16}$ Now, substitute this back into the equation: $b^2 = a^2 \left( \frac{25}{16} - 1 \right)$ To subtract 1, find a common denominator: $b^2 = a^2 \left( \frac{25}{16} - \frac{16}{16} \right)$ $b^2 = a^2 \left( \frac{25 - 16}{16} \right)$ $b^2 = a^2 \left( \frac{9}{16} \right)$ This gives us our first crucial relationship: $b^2 = \frac{9}{16}a^2 \quad \text{(Equation 1)}$

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Step 2: Use the Given Point on the Hyperbola to Form Another Equation

Concept: If a point $(x_1, y_1)$ lies on a curve, its coordinates must satisfy the equation of that curve.

Why this step?: We need a second independent equation involving $a^2$ and $b^2$ to form a system of equations. Substituting the coordinates of the given point into the standard hyperbola equation will provide this.

Given: The point $P\left( \frac{8}{\sqrt{5}}, \frac{12}{5} \right)$ lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Calculation: Substitute $x_1 = \frac{8}{\sqrt{5}}$ and $y_1 = \frac{12}{5}$ into the hyperbola's equation: $\frac{\left( \frac{8}{\sqrt{5}} \right)^2}{a^2} - \frac{\left( \frac{12}{5} \right)^2}{b^2} = 1$ Calculate the squares of the coordinates: $\left( \frac{8}{\sqrt{5}} \right)^2 = \frac{8^2}{(\sqrt{5})^2} = \frac{64}{5}$ $\left( \frac{12}{5} \right)^2 = \frac{12^2}{5^2} = \frac{144}{25}$ Substitute these squared values back into the equation: $\frac{\frac{64}{5}}{a^2} - \frac{\frac{144}{25}}{b^2} = 1$ This simplifies to: $\frac{64}{5a^2} - \frac{144}{25b^2} = 1 \quad \text{(Equation 2)}$

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Step 3: Determine the Specific Values of the Hyperbola Parameters ($a^2$ and $b^2$)

Concept: We now have a system of two equations (Equation 1 and Equation 2) with two unknowns ($a^2$ and $b^2$). We can solve this system to find the unique values that define our specific hyperbola.

Why this step?: Knowing the exact values of $a^2$ and $b^2$ is essential for writing the specific equation of the hyperbola and subsequently the equation of the normal.

Calculation: Substitute the expression for $b^2$ from Equation 1 ($b^2 = \frac{9}{16}a^2$) into Equation 2: $\frac{64}{5a^2} - \frac{144}{25 \left( \frac{9a^2}{16} \right)} = 1$ Simplify the denominator of the second term: $25 \left( \frac{9a^2}{16} \right) = \frac{225a^2}{16}$ So, the second term becomes: $\frac{144}{\frac{225a^2}{16}} = \frac{144 \times 16}{225a^2}$ Notice that $144$ and $225$ are both divisible by $9$: $144 \div 9 = 16$ and $225 \div 9 = 25$. So, the second term is: $\frac{16 \times 16}{25a^2} = \frac{256}{25a^2}$ Now, substitute this back into the combined equation: $\frac{64}{5a^2} - \frac{256}{25a^2} = 1$ To combine the terms on the left side, find a common denominator, which is $25a^2$: $\frac{64 \times 5}{25a^2} - \frac{256}{25a^2} = 1$ $\frac{320}{25a^2} - \frac{256}{25a^2} = 1$ $\frac{320 - 256}{25a^2} = 1$ $\frac{64}{25a^2} = 1$ Multiply both sides by $25a^2$: $64 = 25a^2$ Solve for $a^2$: $a^2 = \frac{64}{25}$ Now, use Equation 1 to find $b^2$: $b^2 = \frac{9}{16}a^2$ Substitute the value of $a^2$: $b^2 = \frac{9}{16} \times \frac{64}{25}$ $b^2 = \frac{9 \times 4}{25} \quad (\text{since } 64 \div 16 = 4)$ $b^2 = \frac{36}{25}$ So, the parameters of our specific hyperbola are $a^2 = \frac{64}{25}$ and $b^2 = \frac{36}{25}$.

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Step 4: Derive the Equation of the Normal at the Given Point

Concept: The equation of the normal line to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at a point $P(x_1, y_1)$ on the hyperbola is given by the formula $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.

Why this step?: This formula directly allows us to calculate the equation of the normal line using the specific parameters of the hyperbola ($a^2, b^2$) and the coordinates of the point of tangency $(x_1, y_1)$. This is the main objective before comparing it to the given form.

Given:

• Point $P(x_1, y_1) = \left( \frac{8}{\sqrt{5}}, \frac{12}{5} \right)$
• Hyperbola parameters: $a^2 = \frac{64}{25}$ and $b^2 = \frac{36}{25}$

Calculation: Substitute these values into the normal equation formula: $\frac{\left( \frac{64}{25} \right)x}{\frac{8}{\sqrt{5}}} + \frac{\left( \frac{36}{25} \right)y}{\frac{12}{5}} = \frac{64}{25} + \frac{36}{25}$ Let's simplify each term:

• First term (coefficient of $x$): $\frac{\frac{64x}{25}}{\frac{8}{\sqrt{5}}} = \frac{64x}{25} \times \frac{\sqrt{5}}{8} = \frac{8x\sqrt{5}}{25}$
• Second term (coefficient of $y$): $\frac{\frac{36y}{25}}{\frac{12}{5}} = \frac{36y}{25} \times \frac{5}{12} = \frac{3y}{5}$
• Right-hand side (constant term): $\frac{64}{25} + \frac{36}{25} = \frac{64+36}{25} = \frac{100}{25} = 4$ Now, combine the simplified terms to form the equation of the normal: $\frac{8\sqrt{5}x}{25} + \frac{3y}{5} = 4$ To eliminate the denominators and match the given form, multiply the entire equation by the least common multiple of $25$ and $5$, which is $25$: $25 \times \left( \frac{8\sqrt{5}x}{25} \right) + 25 \times \left( \frac{3y}{5} \right) = 25 \times 4$ $8\sqrt{5}x + 15y = 100$ This is the specific equation of the normal line at the given point.

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Step 5: Compare with the Given Normal Equation and Find $\lambda - \beta$

Concept: If two linear equations represent the same line, their corresponding coefficients and constant terms must be proportional. If we ensure the coefficient of a specific variable (e.g., $x$) is identical, then all other coefficients and the constant term must also be identical.

Why this step?: This is the final step to extract the values of $\beta$ and $\lambda$ required by the problem by directly comparing our derived equation to the given form.

Given Equation of the Normal: $8\sqrt{5}x + \beta y = \lambda$ Derived Equation of the Normal: $8\sqrt{5}x + 15y = 100$

Comparison:

• Comparing the coefficient of $x$: Both equations have $8\sqrt{5}$. This ensures a direct