Question:
<p>If the line x - 2y = 12 is tangent to the ellipse <span class="math-tex">\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)</span> at the point <span class="math-tex">\(\left(3, \frac{-9}{2}\right)\)</span>, then the length of the latusrectum of the ellipse is</p>
<p style="display:inline"><span class="math-tex">\(8 \sqrt{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(12 \sqrt{2}\)</span></p>
<p style="display:inline">5</p>
<p style="display:inline">9</p>
Step-by-Step Solution
Key Concept: The equation of the tangent at a specific point on an ellipse is compared with the given line equation by matching ratios of their respective coefficients to find the semi-axes.
<p>Equation of given ellipse is <span class="math-tex">$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$</span> ...(i)<br />
Now, equation of tangent at the point <span class="math-tex">$\left(3,-\frac{9}{2}\right)$</span> on the ellipse (i) is<br />
<span class="math-tex">$\Rightarrow \quad \frac{3 x}{a^{2}}-\frac{9 y}{2 b^{2}}=1$</span> ...(ii) [<span class="math-tex">$\because$</span> the equation of the tangent to the ellipse <span class="math-tex">$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$</span> at the point (x<sub>1</sub>, y<sub>1</sub>) is <span class="math-tex">$\frac{xx_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=1$</span>]<br />
<span class="math-tex">$\because$</span> Tangent (ii) represent the line x - 2y = 12, so<br />
<span class="math-tex">$\frac{1}{\frac{3}{a^{2}}}=\frac{2}{\frac{9}{2 b^{2}}}=\frac{12}{1}$</span><br />
<span class="math-tex">$\Rightarrow$</span> a<sup>2</sup> = 36 and b<sup>2 </sup>= 27<br />
Now, Length of latusrectum = <span class="math-tex">$\frac{2 b_{2}}{a}=\frac{2 \times 27}{6}=9 \text { units }$</span></p>
Correct Answer: D