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: Write the hyperbola in standard form.
The given equation is \[ x^2-4y^2=16 \] To convert it into standard form, divide both sides by \( 16 \): \[ \frac{x^2}{16}-\frac{4y^2}{16}=1 \] \[ \frac{x^2}{16}-\frac{y^2}{4}=1 \] Now this is in the standard form \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \]

Step 2: Identify the values of \( a^2 \) and \( b^2 \).
Comparing \[ \frac{x^2}{16}-\frac{y^2}{4}=1 \] with \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] we get \[ a^2=16,\qquad b^2=4 \] Hence, \[ a=4,\qquad b=2 \]

Step 3: Recall the relation for the foci of a hyperbola.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] the foci are at \[ (\pm c,0) \] where \[ c^2=a^2+b^2 \] So we use \[ c^2=16+4=20 \]

Step 4: Find the value of \( c \).
\[ c=\sqrt{20}=2\sqrt{5} \] Thus the two foci are \[ (2\sqrt{5},0)\quad \text{and} \quad (-2\sqrt{5},0) \]

Step 5: Find the distance between the two foci.
The distance between \[ (2\sqrt{5},0)\quad \text{and} \quad (-2\sqrt{5},0) \] is \[ 2c \] Therefore, \[ 2c=2\cdot 2\sqrt{5}=4\sqrt{5} \]

Step 6: Verify using the distance formula.
Using the distance formula: \[ \sqrt{\left(2\sqrt{5}-(-2\sqrt{5})\right)^2+(0-0)^2} \] \[ =\sqrt{(4\sqrt{5})^2} \] \[ =4\sqrt{5} \] So the result is confirmed.

Step 7: Final conclusion.
Hence, the distance between the foci is \[ \boxed{4\sqrt{5}} \] Therefore, the correct option is \[ \boxed{(2)\ 4\sqrt{5}} \]