Question

The distance between the foci of the hyperbola \( x^2-4y^2=16 \), is

• A. \( 2\sqrt{5} \)
• B. \( 4\sqrt{5} \)
• C. \( 4\sqrt{3} \)
• D. \( 5\sqrt{3} \)
• E. \( 5\sqrt{2} \)

Hint:

For a hyperbola \( \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 \), always remember \( c^2=a^2+b^2 \), and the distance between the foci is \( 2c \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the distance between the foci of a hyperbola. This requires converting the given equation to its standard form, identifying the parameters 'a' and 'b', calculating the eccentricity 'e', and then using the formula for the distance between the foci.
Step 2: Key Formula or Approach:
1. The standard form of a horizontal hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). 2. The eccentricity \(e\) is given by the formula \(e = \sqrt{1 + \frac{b^2}{a^2}}\). 3. The coordinates of the foci are \((\pm ae, 0)\). 4. The distance between the foci is \(2ae\).
Step 3: Detailed Explanation:
First, convert the given equation to the standard form.
\[ x^2 - 4y^2 = 16 \] Divide the entire equation by 16:
\[ \frac{x^2}{16} - \frac{4y^2}{16} = 1 \] \[ \frac{x^2}{16} - \frac{y^2}{4} = 1 \] Comparing this with the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we have:
\[ a^2 = 16 \implies a = 4 \] \[ b^2 = 4 \implies b = 2 \] Next, calculate the eccentricity \(e\):
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{16}} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \] Finally, calculate the distance between the foci, which is \(2ae\):
\[ \text{Distance} = 2 \times a \times e = 2 \times 4 \times \frac{\sqrt{5}}{2} = 4\sqrt{5} \] Step 4: Final Answer:
The distance between the foci of the hyperbola is \(4\sqrt{5}\). This corresponds to option (B).