Question:
<p>The vertices of the hyperbola 9x<sup>2</sup> - 16y<sup>2</sup> - 36x + 96y - 252 = 0 are</p>
<p style="display:inline">(-6, 3) and (-6, -3)</p>
<p style="display:inline">(6, 3) and (-6, 3)</p>
<p style="display:inline">(3, 6) and (-3, 2)</p>
<p style="display:inline">(6, 3) and (-2, 3)</p>
Step-by-Step Solution
Key Concept: Transform the equation to standard form by completing the square and shift the vertices relative to the new center (h, k).
<p><img alt="" height="124" src="https://i.imgur.com/bFN2lxg.png" width="135" /><br />
We have, 9x<sup>2</sup> - 16y<sup>2</sup> - 36x + 96y - 252 = 0<br />
<span class="math-tex">$\Leftrightarrow \frac{(x-2)^{2}}{16}-\frac{(y-3)^{2}}{9}=1$</span><br />
<span class="math-tex">$\Rightarrow \frac{X^{2}}{16}-\frac{Y^{2}}{9}=1$</span><br />
The vertices are (X = <span class="math-tex">$\pm$</span>a, Y = 0)<br />
i.e.,(x - 2 = <span class="math-tex">$\pm$</span>4 , y - 3 = 0)<br />
The vertices of the hyperbola are (6, 3) and (-2 ,3).</p>
Correct Answer: D