Question

The angle between the pair of tangents drawn from the point $(1,2)$ to the ellipse $3x^2+2y^2=5$ is

• A. ${\tan }^{-1}\left(\frac{12}{\sqrt{5}}\right)$
• B. ${\tan }^{-1}\left(\frac{\sqrt{5}}{12}\right)$
• C. ${\tan }^{-1}\left(\frac{12}{5}\right)$
• D. ${\tan }^{-1}\left(\frac{5}{12}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{12}{\sqrt{5}}\right)$

Equation of line through (1,2) is

$y-2=m(x-1)$

$\Rightarrow y=mx+(2-m)$

Equation for tangency is $c^2=a^2{m}^2+b^2$

$(2-m{)}^2=\frac{5}{3}{m}^2+\frac{5}{2}$

$4+m^2-4m=\frac{5}{3}{m}^2+\frac{5}{2}$

$\frac{2}{3}{m}^2+4m-\frac{3}{2}=0$

$m_1+m_2=\frac{-4}{\frac{2}{3}}=-6$

$m_1{m}_2=-\frac{-3/2}{2/3}=\frac{-9}{4}$

$(m_1-m_2{)}^2=(m_1+m_2{)}^2-4m_1{m}_2$

$\Rightarrow m_1-m_2-\sqrt{45}$

$\therefore \tan \alpha =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{4\sqrt{45}}{5}=\frac{3\times 4}{\sqrt{5}}=\frac{12}{\sqrt{5}}$

$\therefore \alpha ={\tan }^{-1}\left(\frac{12}{\sqrt{5}}\right)$