Question

The acute angle between the pair of tangents drawn to the ellipse 2x2 + 3y2 = 5 from the point (1, 3) is

• A. $\tan^{-1}(\frac{16}{7\sqrt5})$
• B. $\tan^{-1}(\frac{24}{7\sqrt{5}})$
• C. $\tan^{-1}(\frac{32}{7\sqrt5})$
• D. $\tan^{-1}(\frac{3+8\sqrt5}{35})$

Answer 1

Answer: (B)

$\tan^{-1}(\frac{24}{7\sqrt{5}})$

Answer 2

The correct answer is (B) : $\tan^{-1}(\frac{24}{7\sqrt{5}})$
2 x 2 + 3 y 2 = 5
Equation of tangent having slope m.
$y=mx±\sqrt{\frac{5}{2}m^2+\frac{5}{3}}$
which passes through (1, 3)
$3=m±\sqrt{\frac{5}{2}m^2+\frac{5}{3}}$
$\frac{5}{2}m^2+\frac{5}{3}=9+m^2−6m$
$\frac{3}{2}m^2+6m−\frac{22}{3}=0$
$9m^2+36m−44=0$
$m_1+m_2=−4,m_1m_2=−\frac{44}{9}$
$(m_1−m_2)^2=16+4×\frac{44}{9}=\frac{320}{9}$
Acute angle between the tangents is given by
$α=\tan^{−1}⁡|\frac{m_1−m_2}{1+m_1m_2}|$
$= \tan^{-1} |\frac{\frac{8\sqrt5}{3}}{1-\frac{44}{9}}|$
$= \tan^{-1} (\frac{24\sqrt5}{35})$
$α = \tan^{-1}(\frac{24}{7\sqrt5})$