Question

The equation of the tangent at the point $\theta =\frac{\pi }{3}$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is

• A. $x+\sqrt{2}y=2\sqrt{2}$
• B. $2x+3\sqrt{3}y=12$
• C. $3x-4y+12\sqrt{2}=0$
• D. $x+y+12\sqrt{2}=0$

Answer 1

Answer: (B)

$2x+3\sqrt{3}y=12$

Elipse $:\frac{x^2}{9}+\frac{y^2}{4}=1$

Given$\theta =\frac{\pi }{3}$

For angle $\varphi$ and elipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, Tangent is: $\frac{x\cos \varphi }{a}+\frac{y\sin \varphi }{b}=1$

$\frac{x\cos \frac{\pi }{3}}{a}+\frac{y\sin \frac{\pi }{3}}{b}=1$

$\Rightarrow \frac{x\left(\frac{1}{2}\right)}{3}+\frac{y\left(\frac{\sqrt{3}}{2}\right)}{2}=1$

$\Rightarrow \frac{x}{6}+\frac{\sqrt{3}y}{4}=1$

$\Rightarrow 2x+3\sqrt{3}y=12$