<p>The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum to the ellipse <span class="math-tex">\(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\)</span> is</p>
<p style="display:inline">18</p>
<p style="display:inline">27</p>
<p style="display:inline"><span class="math-tex">\(\frac{27}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{27}{2}\)</span></p>
Step-by-Step Solution
Key Concept: Exploit the symmetry of the ellipse to find the area of the resulting rhombus by multiplying the area of the triangle formed by one tangent and the coordinate axes by four.
<p>Given equation of ellipse is <span class="math-tex">$\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$</span><br />
<span class="math-tex">$\therefore \quad a^{2}=9, b^{2}=5 \Rightarrow a=3, b=\sqrt{5}$</span><br />
Now, <span class="math-tex">$e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$</span><br />
Foci = <span class="math-tex">$(\pm a e, 0)=(\pm 2,0) \text { and } \frac{b^{2}}{a}=\frac{5}{3}$</span><br />
<img alt="" data-imgur-src="Mnju2vA.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/Mnju2vA.png" style="width: 300px; height: 187px;" /><br />
<span class="math-tex">$\therefore$</span> Extermities of one of latusrectumk are <span class="math-tex">$\left(2, \frac{5}{3}\right) \text { and }\left(2, \frac{-5}{3}\right)$</span><br />
<span class="math-tex">$\therefore$</span> Equation of tangent at <span class="math-tex">$\left(2, \frac{5}{3}\right)$</span> is<br />
<span class="math-tex">$\frac{x(2)}{9}+\frac{y(5 / 3)}{5}=1$</span> or 2x + 3y = 9<br />
Since, Eq. (ii) intersects X and Y-axes at <span class="math-tex">$\left(\frac{9}{2},0\right)$</span> and (0, 3), respectively.<br />
<span class="math-tex">$\therefore$</span> Area of quadrilateral = <span class="math-tex">$4 \times \text { Area of } \Delta P O Q$</span><br />
= <span class="math-tex">$4\times\left(\frac{1}{2} \times \frac{9}{2} \times 3\right)=27 \mathrm{sq} \text { units }$</span></p>
Correct Answer: B