Question

The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are 9 and $x = \pm \frac{4}{\sqrt{3}}$, respectively. Let the line $y - \sqrt{3}x + \sqrt{3} = 0$ touch this hyperbola at $(x_0, y_0)$. If $m$ is the product of the focal distances of the point $(x_0, y_0)$, then $4e^2 + m$ is equal to ________.

Answer 1

Correct Answer: 61

The problem asks for the value of \(4e^2 + m\), where \(e\) is the eccentricity of a hyperbola and \(m\) is the product of the focal distances of a specific point of tangency \((x_0, y_0)\). We are given the hyperbola's latus rectum, the equation of its directrices, and the equation of a tangent line.

Concept Used:

The solution uses the standard properties of a hyperbola with a horizontal transverse axis, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

1. Eccentricity Relation: \(b^2 = a^2(e^2 - 1)\).
2. Length of Latus Rectum: \( \frac{2b^2}{a} \).
3. Equations of Directrices: \( x = \pm \frac{a}{e} \).
4. Condition of Tangency: A line \(y = mx + c\) is tangent to the hyperbola if \(c^2 = a^2m^2 - b^2\).
5. Equation of Tangent: The equation of the tangent to the hyperbola at a point \((x_0, y_0)\) is \(\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1\).
6. Focal Distances: For any point \(P(x_0, y_0)\) on the hyperbola, the distances from the foci \(S(ae, 0)\) and \(S'(-ae, 0)\) are \(|ex_0 - a|\) and \(|ex_0 + a|\).
7. Product of Focal Distances: The product of the focal distances is \(m = |(ex_0 - a)(ex_0 + a)| = |e^2x_0^2 - a^2|\).

Note: There is an inconsistency in the problem statement. The latus rectum, directrix, and tangent line properties as given do not correspond to the same hyperbola. We will derive the hyperbola's parameters using the latus rectum and the tangent line condition, as they form a consistent system leading to a unique solution.

Step-by-Step Solution:

Step 1: Formulate equations from the given information.

The length of the latus rectum is 9:

\[ \frac{2b^2}{a} = 9 \implies b^2 = \frac{9a}{2} \quad \text{(Equation 1)} \]

The tangent line is \(y - \sqrt{3}x + \sqrt{3} = 0\), which can be written as \(y = \sqrt{3}x - \sqrt{3}\). For this line, the slope is \(m_{\text{tan}} = \sqrt{3}\) and the y-intercept is \(c = -\sqrt{3}\).

Using the condition for tangency, \(c^2 = a^2m_{\text{tan}}^2 - b^2\):

\[ (-\sqrt{3})^2 = a^2(\sqrt{3})^2 - b^2 \] \[ 3 = 3a^2 - b^2 \quad \text{(Equation 2)} \]

Step 2: Solve for the hyperbola's parameters \(a\) and \(b\).

Substitute Equation 1 into Equation 2:

\[ 3 = 3a^2 - \frac{9a}{2} \]

Multiply the entire equation by 2 to eliminate the fraction:

\[ 6 = 6a^2 - 9a \]

Rearrange into a quadratic equation:

\[ 6a^2 - 9a - 6 = 0 \]

Divide by 3:

\[ 2a^2 - 3a - 2 = 0 \]

Factor the quadratic equation:

\[ (2a + 1)(a - 2) = 0 \]

Since the semi-major axis length \(a\) must be positive, we have \(a = 2\).

Now, find \(b^2\) using Equation 1:

\[ b^2 = \frac{9(2)}{2} = 9 \]

So, the equation of the hyperbola is \(\frac{x^2}{4} - \frac{y^2}{9} = 1\).

Step 3: Calculate the square of the eccentricity, \(e^2\).

Using the relation \(b^2 = a^2(e^2 - 1)\):

\[ 9 = 4(e^2 - 1) \] \[ \frac{9}{4} = e^2 - 1 \] \[ e^2 = \frac{9}{4} + 1 = \frac{13}{4} \]

Step 4: Find the point of tangency \((x_0, y_0)\).

The equation of the tangent at \((x_0, y_0)\) is \(\frac{xx_0}{4} - \frac{yy_0}{9} = 1\). The given tangent line is \(\sqrt{3}x - y = \sqrt{3}\). Comparing the coefficients of the two forms of the tangent line equation:

\[ \frac{x_0/4}{\sqrt{3}} = \frac{-y_0/9}{-1} = \frac{1}{\sqrt{3}} \]

From the first and third parts:

\[ \frac{x_0}{4\sqrt{3}} = \frac{1}{\sqrt{3}} \implies x_0 = 4 \]

From the second and third parts:

\[ \frac{y_0}{9} = \frac{1}{\sqrt{3}} \implies y_0 = \frac{9}{\sqrt{3}} = 3\sqrt{3} \]

The point of tangency is \((x_0, y_0) = (4, 3\sqrt{3})\).

Step 5: Calculate \(m\), the product of the focal distances.

The formula for the product of focal distances is \(m = |e^2x_0^2 - a^2|\).

Substitute the values we found: \(e^2 = \frac{13}{4}\), \(x_0 = 4\), and \(a = 2\).

\[ m = \left| \left(\frac{13}{4}\right)(4)^2 - (2)^2 \right| \] \[ m = \left| \left(\frac{13}{4}\right)(16) - 4 \right| \] \[ m = |13 \times 4 - 4| = |52 - 4| = 48 \]

Step 6: Compute the final value of \(4e^2 + m\).

Substitute the values of \(e^2\) and \(m\):

\[ 4e^2 + m = 4\left(\frac{13}{4}\right) + 48 \] \[ = 13 + 48 = 61 \]

The value of \(4e^2 + m\) is 61.

Answer 2

Given:
\[\frac{2b^2}{a} = 9 \quad \text{and} \quad \frac{a}{e} = \frac{4}{\sqrt{3}}\]
The equation of the tangent \( y - \sqrt{3}x + \sqrt{3} = 0 \) can be rewritten for easier manipulation. The slope \( S \) of this line is:
\[S = \sqrt{3}\]
Using the condition of tangency, we find:
\[6 = 6a^2 - 9\]
\[\implies a = 2, \quad b^2 = 9\]
Thus, the equation of the hyperbola is:
\[\frac{x^2}{4} - \frac{y^2}{9} = 1\]
and for the tangent line:
\[y = \sqrt{3}x + \sqrt{3}\]
The point of contact \( (x_0, y_0) \) is:
\[(4, 3\sqrt{3}) = (x_0, y_0)\]
Now, calculating the eccentricity:
\[e = \sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}\]
The product of focal distances \( m \) is given by:
\[m = (x_0 + a)(x_0 - a)\]
\[m + 4e^2 = 20 \times \frac{13}{4} = 61\]
Note: There is a printing mistake in the equation of the directrix as \( x = \pm \frac{4}{\sqrt{3}} \).
Corrected equation: The correct equation for the directrix should be \( x = \pm \frac{4}{\sqrt{13}} \), as eccentricity must be greater than one, so this question might be a bonus.