Q78 / 160 · AP EAMCET 2017 (26 Apr Shift 1) · Mathematics
The angle between the tangents drawn from the point $(1,2)$ to the ellipse $3 x^2+2 y^2=5$ is
- Correct answerA$\tan ^{-1}\left(\frac{12 \sqrt{5}}{5}\right)$
- B$\tan ^{-1}\left(\frac{12 \sqrt{5}}{13}\right)$
- C$\frac{\pi}{4}$
- D$-\frac{\pi}{4}$
Solution
Given Ellipse,
$$
3 x^2+2 y^2-5=0
$$
Then, equations of pairs of tangents drawn to the ellipse $S$ from the point $(1,2)$.
$$
\begin{aligned}
& S S_1=T^2 \\
& \left(3 x^2+2 y^2-5\right)\left(3(1)^2-2(2)^2-5\right) \\
& \quad=(3 x(1)+2 y(2)-5)^2 \\
& \left(3 x^2+2 y^2-5\right)(6)=(3 x+4 y-5)^2 \\
& 18 x^2+12 y^2-30=9 x^2+16 y^2+25 \\
& +24 x y-40 y-30
\end{aligned}
$$
$$
9 x^2-4 y^2-24 x y+30 x+40 y-55=0
$$
Now, the angle $\theta$ between these lines is given by
$$
\begin{aligned}
& \tan \theta=\frac{2 \sqrt{h^2-a b}}{a+b} \\
& \tan \theta=\frac{2 \sqrt{144+36}}{5}=\frac{2 \sqrt{180}}{5}=\frac{12 \sqrt{5}}{5} \\
& \therefore \quad \theta=\tan ^{-1}\left(\frac{12 \sqrt{5}}{5}\right)
\end{aligned}
$$
Hence, the correct answer is option: 'A'.