Q96 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics
If the line $\mathrm{x} \cos \alpha+\mathrm{y} \sin x=2 \sqrt{3}$ is a tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{8}=1$ and $\alpha$ is an acute angle then $\alpha=$
- A$\frac{\pi}{6}$
- Correct answerB$\frac{\pi}{4}$
- C$\frac{\pi}{3}$
- D$\frac{\pi}{2}$
Solution
$x \cos \alpha+y \sin \alpha=2 \sqrt{3}$
$$
\frac{x^2}{16}+\frac{y^2}{8}=1
$$
We know that if $x \cos \alpha+y \sin \alpha=\mathrm{p}$ is tangent to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then
$$
p^2=a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha
$$
$(2 \sqrt{3})^2=16 \cos ^2 \alpha+8 \sin ^2 \alpha$
$$
12=16-16 \sin ^2 \alpha+8 \sin ^2 \alpha
$$
$$
8 \sin ^2 \alpha=4 \Rightarrow \sin \alpha= \pm \frac{1}{\sqrt{2}}
$$
but $\alpha$ is acute
$$
\therefore \sin \alpha=\frac{1}{\sqrt{2}} \Rightarrow \alpha=\frac{\pi}{4}
$$
Hence, the correct answer is option: 'B'.