Question:
<p>Let the foci of a hyperbola be <span class="math-tex">\((1,14)\)</span> and <span class="math-tex">\((1,-12)\)</span>. If it passes through the point <span class="math-tex">\((1,6)\)</span>, then the length of its latus rectum is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{25}{6}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{144}{5}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{24}{5}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{288}{5}\)</span></p>
Step-by-Step Solution
Key Concept: The length of the transverse axis is found using the focal distance property |SP - S'P| = 2b, while the distance between foci determines the eccentricity and conjugate axis.
<p>Foci of hyperbola <span class="math-tex">$(1,14)(1,-12) $</span><br />
<span class="math-tex">$\therefore x=1$</span> be the transverse axis<br />
<span class="math-tex">$ \left|S P-S^{\prime} P\right|=2 b $</span><br />
<span class="math-tex">$ \Rightarrow 2 b=|8-18| $</span><br />
<span class="math-tex">$ \Rightarrow b=5 $</span><br />
<span class="math-tex">$ {SS}^{\prime}=2 b e $</span><br />
<span class="math-tex">$ 26=2 b e $</span><br />
<span class="math-tex">$ b e=13 $</span><br />
<span class="math-tex">$ e=\frac{13}{5} $</span><br />
<img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1756102668-7jdyzd.jpg" style="height:202px; width:200px" /><br />
<span class="math-tex">$\therefore$</span> Length of L R<br />
<span class="math-tex">$ =\frac{2 a^2}{b}=\frac{2 b^2\left(e^2-1\right)}{b} $</span><br />
<span class="math-tex">$ =2(5) \frac{(144)}{25}=\frac{288}{5} $</span></p>
Correct Answer: D