Question

The equation \(16x^{2}-3y^{2}-32x+12y-44=0\) represents a hyperbola whose...

• A. centre is \((-1,2)\)
• B. eccentricity is \(\sqrt{\frac{19}{3}}\)
• C. length of its transverse axis is \(4\sqrt{3}\)
• D. length of its conjugate axis is \(4\)

Hint:

Complete the squares to reach the standard form, then read off the centre, axes and eccentricity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Complete the squares to bring the equation to \(\dfrac{(x-h)^{2}}{a^{2}}-\dfrac{(y-k)^{2}}{b^{2}}=1\).

Step 2: Complete the squares
\(16(x^{2}-2x)-3(y^{2}-4y)=44\Rightarrow 16(x-1)^{2}-3(y-2)^{2}=44+16-12=48\)

Step 3: Standard form
Dividing by 48: \(\dfrac{(x-1)^{2}}{3}-\dfrac{(y-2)^{2}}{16}=1\), so \(a^{2}=3,\ b^{2}=16\), centre \((1,2)\).

Step 4: Eccentricity
\(e^{2}=1+\dfrac{b^{2}}{a^{2}}=1+\dfrac{16}{3}=\dfrac{19}{3}\), so \(e=\sqrt{\dfrac{19}{3}}\).

Step 5: Other options
Transverse axis \(=2a=2\sqrt{3}\) and conjugate axis \(=2b=8\), so only the eccentricity statement is correct.

Final Answer:
The eccentricity of the hyperbola is root of 19 by 3. \[ \boxed{\text{(B) }\text{eccentricity is }\sqrt{\dfrac{19}{3}}} \]