Key Concepts and Formulas for a Hyperbola

For a hyperbola centered at the origin $(0,0)$ with its transverse axis along the x-axis, its standard form is given by: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ Here:

• $a$ is the length of the semi-transverse axis (half the distance between vertices).
• $b$ is the length of the semi-conjugate axis.

From this standard form, we can define other important properties:

1. Eccentricity ($e$): This value quantifies the "stretch" of the hyperbola. It is always greater than 1 ($e > 1$) for a hyperbola. The relationship between $a$, $b$, and $e$ is given by: $e^2 = 1 + \frac{b^2}{a^2}$
2. Foci (plural of focus): These are two fixed points on the transverse axis. For this standard form, the foci are located at $(\pm ae, 0)$.
3. Directrices (plural of directrix): These are two fixed lines perpendicular to the transverse axis. For this standard form, the equations of the directrices are $x = \pm \frac{a}{e}$.

Crucial Correspondence between Foci and Directrices:

• The directrix $x = \frac{a}{e}$ corresponds to the focus $(ae, 0)$.
• The directrix $x = -\frac{a}{e}$ corresponds to the focus $(-ae, 0)$. It is vital to match the directrix with its corresponding focus, which means they lie on the same side of the hyperbola's center.

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Step-by-Step Solution

Step 1: Convert the Hyperbola Equation to Standard Form

The given equation of the hyperbola is $16x^2 - 9y^2 = 144$. Our first objective is to transform this equation into the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. This step is crucial because the standard form directly reveals the values of $a^2$ and $b^2$, which are fundamental for calculating all other properties of the hyperbola.

To achieve the standard form, we need the right-hand side of the equation to be 1. We do this by dividing every term in the equation by the constant term on the right-hand side, which is 144: $\frac{16x^2}{144} - \frac{9y^2}{144} = \frac{144}{144}$ Now, simplify the fractions: $\frac{x^2}{9} - \frac{y^2}{16} = 1$ By comparing this simplified equation with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we can identify the values of $a^2$ and $b^2$: $a^2 = 9 \implies a = 3$ $b^2 = 16 \implies b = 4$ (We take the positive roots for $a$ and $b$ because they represent lengths of axes.) Since the $x^2$ term is positive, the transverse axis of this hyperbola lies along the x-axis, confirming our choice of standard form.

Step 2: Calculate the Eccentricity ($e$)

With the values of $a^2$ and $b^2$ determined, our next step is to calculate the eccentricity $e$. The eccentricity is a key parameter that directly connects the definitions of the foci and directrices to the basic dimensions of the hyperbola. We need $e$ to find the coordinates of the focus.

Using the formula for the eccentricity of a hyperbola: $e^2 = 1 + \frac{b^2}{a^2}$ Substitute the values $a^2 = 9$ and $b^2 = 16$ into the formula: $e^2 = 1 + \frac{16}{9}$ To sum these, find a common denominator: $e^2 = \frac{9}{9} + \frac{16}{9}$ $e^2 = \frac{25}{9}$ Now, take the square root of both sides to find $e$. Since eccentricity is a positive value for a hyperbola ($e > 1$): $e = \sqrt{\frac{25}{9}}$ $e = \frac{5}{3}$ Notice that $e = \frac{5}{3} > 1$, which is consistent with the property of a hyperbola.

Step 3: Identify the Given Directrix and Its Corresponding Type

The problem states that $5x + 9 = 0$ is the directrix. To use this information, we need to rewrite this equation in the standard directrix form, $x = \text{constant}$. This will allow us to compare it with the general directrix formulas $x = \pm \frac{a}{e}$ and determine which focus it corresponds to.

From the given equation: $5x + 9 = 0$ $5x = -9$ $x = -\frac{9}{5}$ Now, we compare this with the standard forms of directrices, $x = \frac{a}{e}$ and $x = -\frac{a}{e}$. Since our given directrix is $x = -\frac{9}{5}$ (a negative value), it must correspond to the standard directrix $x = -\frac{a}{e}$. According to the crucial correspondence established earlier, the directrix $x = -\frac{a}{e}$ always corresponds to the focus $(-ae, 0)$. This matching of signs is vital for determining the correct focus.

Step 4: Calculate the Coordinates of the Corresponding Focus

We now have all the necessary components to find the coordinates of the corresponding focus. We know the focus is of the form $(-ae, 0)$, and we have calculated $a$ and $e$.

From Step 1, we found $a = 3$. From Step 2, we found $e = \frac{5}{3}$.

Now, let's calculate the product $ae$: $ae = (3) \times \left(\frac{5}{3}\right)$ $ae = 5$ Since the given directrix $x = -\frac{9}{5}$ corresponds to the focus $(-ae, 0)$, we substitute $ae = 5$ into this form: The corresponding focus is $(-5, 0)$.

Verification (Optional but Recommended): We can quickly verify that our calculated value for $-\frac{a}{e}$ matches the given directrix equation: $-\frac{a}{e} = -\frac{3}{5/3} = -\frac{3 \times 3}{5} = -\frac{9}{5}$ This matches the given directrix $x = -\frac{9}{5}$, which confirms the consistency of our calculations for $a$ and $e$ with the problem statement.

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Tips and Common Mistakes to Avoid

• Standard Form is Paramount: Always convert the hyperbola equation to its standard form first. Incorrectly identifying $a^2$ and $b^2$ will lead to errors in all subsequent calculations.
• Hyperbola vs. Ellipse Eccentricity: A common mistake is to confuse the eccentricity formula for a hyperbola with that of an ellipse.
  • For a hyperbola: $e^2 = 1 + \frac{b^2}{a^2}$ (note the '+' sign). Also, $e > 1$.
  • For an ellipse: $e^2 = 1 - \frac{b^2}{a^2}$ (note the '-' sign). Also, $0 < e < 1$.
• Matching Signs Correctly: Always ensure you match the directrix $x = \frac{a}{e}$ with focus $(ae, 0)$ and directrix $x = -\frac{a}{e}$ with focus $(-ae, 0)$. A directrix and its corresponding focus must lie on the same side of the center.
• Transverse Axis Orientation: In this problem, the $x^2$ term was positive, indicating the transverse axis lies along the x-axis. If the $y^2$ term were positive (e.g., $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$), the transverse axis would be along the y-axis. In such a case, the foci would be $(0, \pm ae)$ and the directrices would be $y = \pm \frac{a}{e}$. Always be mindful of the orientation.

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Summary and Key Takeaway

To determine the focus corresponding to a given directrix of a hyperbola:

1. Standardize the Equation: Convert the hyperbola's equation into its standard form ($\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2} - \frac{x^2