Question:
<p>The angle between the pair of tangents drawn to the ellipse 3x<sup>2</sup> + 2y<sup>2</sup> = 5 from the point (1, 2) is</p>
<p style="display:inline">tan<sup>-1</sup><span class="math-tex">\(\left(\frac{12}{5}\right)\)</span></p>
<p style="display:inline">tan<sup>-1</sup><span class="math-tex">\((6 \sqrt{5})\)</span></p>
<p style="display:inline">tan<sup>-1</sup><span class="math-tex">\(\left(\frac{12}{\sqrt{5}}\right)\)</span></p>
<p style="display:inline">tan<sup>-1</sup><span class="math-tex">\((12 \sqrt{5})\)</span></p>
Step-by-Step Solution
Key Concept: Use the joint equation of pair of tangents $SS_1 = T^2$ to obtain a general second-degree equation and then apply the angle formula $\tan \theta = \frac{2\sqrt{h^2 - ab}}{a+b}$.
<p>Here, SS<sub>1</sub> = T<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> (3x<sup>2</sup> + 2y<sup>2</sup> - 5) (3 + 8 - 5) = (3x + 4y - 5)<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> 9x<sup>2</sup> - 4y<sup>2</sup> - 24xy - 30x - 40y - 55 = 0<br />
Here, a = 9, b = -4 and h = -12<br />
tan <span class="math-tex">$\theta=2 \frac{\sqrt{h^{2}-a b}}{a+b}$</span>,<br />
<span class="math-tex">$\Rightarrow \theta=\tan ^{-1}\left(\frac{12}{\sqrt{5}}\right)$</span></p>
Correct Answer: C