Question

The eccentricity of the hyperbola \( 16x^2 - 3y^2 - 32x - 12y - 44 = 0 \) is

• A. \( \sqrt{\dfrac{19}{3}} \)
• B. \( \sqrt{\dfrac{13}{19}} \)
• C. \( \dfrac{\sqrt{19}}{3} \)
• D. \( \dfrac{13}{\sqrt{19}} \)

Hint:

Always convert the given second-degree equation into standard form before identifying eccentricity.

Answer 1

The Correct Option is A

Step 1: Rearrange and group terms.
\[ 16x^2 - 32x - 3y^2 - 12y - 44 = 0 \] Step 2: Complete the squares.
\[ 16(x^2 - 2x) - 3(y^2 + 4y) - 44 = 0 \] \[ 16(x-1)^2 - 16 - 3(y+2)^2 + 12 - 44 = 0 \] Step 3: Simplify.
\[ 16(x-1)^2 - 3(y+2)^2 = 48 \] Step 4: Write in standard form.
\[ \frac{(x-1)^2}{3} - \frac{(y+2)^2}{16} = 1 \] Step 5: Identify parameters.
\[ a^2 = 3, \quad b^2 = 16 \] Step 6: Find eccentricity.
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{3}} = \sqrt{\frac{19}{3}} \] Step 7: Conclusion.
The eccentricity of the hyperbola is \( \sqrt{\dfrac{19}{3}} \).