Question:
<p>In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0, 5<span class="math-tex">\(\sqrt3\)</span>), then the length of its latus rectum is</p>
<p style="display:inline">10</p>
<p style="display:inline">6</p>
<p style="display:inline">5</p>
<p style="display:inline">8</p>
Step-by-Step Solution
Key Concept: Identify the vertical orientation of the ellipse from the Y-axis focus to correctly relate the axes using the property $b^2 - a^2 = (be)^2$.
<p>One of the focus of ellipse <span class="math-tex">$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$</span> is on Y-axis (0, 5<span class="math-tex">$\sqrt{3})$</span>)<br />
<span class="math-tex">$\therefore \quad b e=5 \sqrt{3}$</span> ...(i)<br />
[where e is eccentricity of ellipse]<br />
According to the question,<br />
2b - 2a = 10<br />
<span class="math-tex">$\Rightarrow$</span> b - a = 5 ...(ii)<br />
On squaring Eq. (i) both sides, we get<br />
<span class="math-tex">$b^{2e^2}$</span> = 75<br />
<span class="math-tex">$\Rightarrow b^{2}\left(1-\frac{a^{2}}{b^{2}}\right)=75 \quad\left[\because e^{2}=1-\frac{a^{2}}{b^{2}}\right]$</span><br />
<span class="math-tex">$\Rightarrow$</span> b<sup>2</sup> - a<sup>2</sup> = 75<br />
<span class="math-tex">$\Rightarrow$</span> (b + a)(b - a) = 75<br />
<span class="math-tex">$\Rightarrow$</span> b + a = 15 [from Eq. (ii)] ...(iii)<br />
On solving Eqs. (ii) and (iii), we get<br />
b = 10 and a = 5<br />
So, length of latusrectum is <span class="math-tex">$\frac{2 a_{2}}{b}=\frac{2 \times 25}{10}=5$</span> units</p>
Correct Answer: C