Question:
<p>The equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13, is:</p>
<p style="display:inline">25x<sup>2</sup> - 144y<sup>2</sup> = 900</p>
<p style="display:inline">144x<sup>2</sup> + 25y<sup>2</sup> = 900</p>
<p style="display:inline">144x<sup>2</sup> - 25y<sup>2</sup> = 900</p>
<p style="display:inline">25x<sup>2</sup> + 144y<sup>2</sup> = 900</p>
Step-by-Step Solution
Key Concept: Use the fundamental identity $b^2 = (ae)^2 - a^2$ to solve for the semi-transverse axis $a$ using the given lengths of the conjugate axis $2b$ and focal distance $2ae$.
<p>Conjugate axis is 5 and distance between foci = 13<br />
<span class="math-tex">$\Rightarrow$</span> 2b = 5 and 2ae = 13<br />
Now, also we know for hyperbola<br />
b<sup>2</sup> = a<sup>2</sup>(e<sup>2</sup> - 1)<br />
<span class="math-tex">$\Rightarrow \frac{25}{4}=\frac{(13)^{2}}{4 e^{2}}$</span> (e<sup>2</sup> - 1)<br />
<span class="math-tex">$\Rightarrow \frac{25}{4}=\frac{169}{4}-\frac{169}{4 e^{2}}$</span> or e<sup>2</sup> = <span class="math-tex">$\frac{169}{144}$</span><br />
<span class="math-tex">$\Rightarrow$</span> e = <span class="math-tex">$\frac{13}{12}$</span> or a = 6, b = <span class="math-tex">$\frac 52$</span> or hyperbola is <span class="math-tex">$\frac{x^{2}}{36}-\frac{y^{2}}{\frac {25}4}$</span> = 1<br />
<span class="math-tex">$\Rightarrow$</span> 25x<sup>2</sup> - 144y<sup>2</sup> = 900</p>
Correct Answer: A