Question:
<p>The equation 16x<sup>2</sup> - 3y<sup>2</sup> - 32x + 12 y - 44 = 0 represents a hyperbola:</p>
<p style="display:inline">the length of whose transverse axis is <span class="math-tex">\(4 \sqrt{3}\)</span></p>
<p style="display:inline">whose eccentricity is <span class="math-tex">\(\sqrt{\frac{19}{3}}\)</span></p>
<p style="display:inline">the length of whose conjugate axis is 4</p>
<p style="display:inline">whose centre is (-1, 2)</p>
Step-by-Step Solution
Key Concept: Transform the general equation into the standard form of a hyperbola by completing the squares for $x$ and $y$ terms to reveal its center, axes, and eccentricity.
<p>16(x<sup>2 </sup>- 2x)-3(y<sup>2 </sup>- 4y) = 44<br />
<span class="math-tex">$\Rightarrow$</span> 16(x - 1)<sup>2 </sup>−3(y - 2)<sup>2 </sup>= 44 + 16 − 12<br />
<span class="math-tex">$\Rightarrow$</span> 16(x - 1)<sup>2</sup> - 3(y - 2)<sup>2</sup> = 48<br />
<span class="math-tex">$\Rightarrow \frac{(\mathrm{x}-1)^{2}}{3}-\frac{(\mathrm{y}-2)^{2}}{16}$</span> = 1<br />
Here, centre (1, −2), length of transverse axis<br />
= 2a = 2 <span class="math-tex">$\times$</span> <span class="math-tex">$\sqrt{3}$</span> = <span class="math-tex">$2 \sqrt{3}$</span><br />
Length of conjugate axis = 2b = 2 <span class="math-tex">$\times$</span> 4 = 8<br />
and eccentricity, e = <span class="math-tex">$\sqrt{1+\frac{b^{2}}{a^{2}}}$</span><br />
= <span class="math-tex">$\sqrt{1+\left(\frac{4}{\sqrt{3}}\right)^{2}}$</span> = <span class="math-tex">$\sqrt{1+\frac{16}{3}}$</span><br />
= <span class="math-tex">$\sqrt{\frac{19}{3}}$</span></p>
Correct Answer: B