Question

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

• A. ${\tan }^{−1}\left(\frac{12}{5}\right)$
• B. ${\tan }^{−1}(6\sqrt{5})$
• C. ${\tan }^{−1}\left(\frac{12}{\sqrt{5}}\right)$
• D. ${\tan }^{−1}(12\sqrt{5})$

Answer 1

Answer: (C)

Equation of tangent at $(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$ ..(i)
and Equation of pair of tangent to an ellipse from
$(x_1,y_1)$ is $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}−1\right)\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}−1\right)$
$={\left(\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}−1\right)}^2$ .. (ii)
and angle $θ$ between them $={\tan }^{−1}\frac{2\sqrt{h^2−ab}}{a+b}$ ... (iii)
Given equation of ellipse
$3x^2+2y^2=5$ and point $(1,2)$ .
It can be rewritten as, $\frac{x^2}{5/3}+\frac{y^2}{5/2}=1$
$∴$ Equation of tangent at $(1,2)$ is $\frac{x}{5/3}+\frac{2y}{5/2}=1$
$⇒$ $3x+4y=5$ [from (ii)]
$∴$ Joint equation of tangents $\left(\frac{x^2}{5/3}+\frac{y^2}{5/2}−1\right)\left(\frac{1}{5/3}+\frac{4}{5/2}−1\right)$
$={(3x+4y−5)}^2$ (from (ii))
$⇒$ $9x^2−4y^2−24xy+30x+40y−30=0$
$∴$ $a=9,b=−4,h=12,g=15$ (By comparing with $a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0)$
$∴$ $θ={\tan }^{−1}\left(\frac{2\sqrt{144+36}}{5}\right)={\tan }^{−1}\left(\frac{2â‹\dots 2â‹\dots 3\sqrt{5}}{5}\right)$
$θ={\tan }^{−1}(12/\sqrt{5})$