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

The correct option is B $2x+\sqrt{3}y=2\sqrt{6}$

Given, Equation of ellipse as $4x^2+3y^2=12$ eccentric angle $\theta =\frac{\pi }{4}$
$\Rightarrow \frac{x^2}{3}+\frac{y^2}{4}=1$
Let Point $P$ lies on ellipse as $P=(a\cos \theta ,b\sin \theta )\Rightarrow P=(\frac{\sqrt{3}}{\sqrt{2}},\frac{2}{\sqrt{2}})$
Tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(x_1,y_1)is\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$
$\therefore$Required Tangent equation is $\frac{x\times \left(\frac{\sqrt{3}}{\sqrt{2}}\right)}{3}+\frac{y\times \left(\frac{2}{\sqrt{2}}\right)}{4}=1$
$\Rightarrow \frac{x}{\sqrt{6}}+\frac{y}{2\sqrt{2}}=1$
$\Rightarrow 2x+\sqrt{3}y=2\sqrt{6}$