Question

The equation of the tangent to the ellipse $4x^2+3y^2=12$ at the point whose eccentric angle is $\frac{\pi }{4}$ is:

• A. $\sqrt{3}x+2y=2\sqrt{6}$
• B. $2x+\sqrt{3}y=2\sqrt{6}$
• C. $2x-\sqrt{3}y=2\sqrt{6}$
• D. none of these

Answer 1

Answer: (B)

$2x+\sqrt{3}y=2\sqrt{6}$

Given ellipse may be written as, $\frac{x^2}{3}+\frac{y^2}{4}=1$

$a=\sqrt{3},b=2$

Thus equation of tangent with given eccentric angle $\theta$ is given by,

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

$\frac{x\cos \pi /4}{\sqrt{3}}+\frac{y\sin \pi /4}{2}=1$

$\Rightarrow 2x+\sqrt{3}y=2\sqrt{6}$