Question

A point on the ellipse $x^2+3y^2=9$, where the tangent is parallel to the line $y-x=0$, is

• A. $(\sqrt{3},\sqrt{2})$
• B. $(-\frac{3\sqrt{3}}{2},-\frac{\sqrt{3}}{2})$
• C. $(-\frac{3\sqrt{3}}{2},\frac{\sqrt{3}}{2})$
• D. $(-\sqrt{3},\sqrt{2})$

Answer 1

The correct option is C $(-\frac{3\sqrt{3}}{2},\frac{\sqrt{3}}{2})$

Given ellipse is, $\frac{x^2}{9}+\frac{y^2}{3}=1$
$\Rightarrow a^2=9,b^2=3$
Now equation of line parallel to $y-x=0$ is given by,
$y=x+c...\left(1\right)$
Since line (1) is tangent to the ellipse, $\therefore c^2=a^2{m}^2+b^2=12\Rightarrow c=\pm 2\sqrt{3}$
Thus line (1) is $x-y=\pm 2\sqrt{3}$
Comparing it with $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$
Point of intersection is $(-\frac{3\sqrt{3}}{2},\frac{\sqrt{3}}{2})$ or $(\frac{3\sqrt{3}}{2},-\frac{\sqrt{3}}{2})$
Since, there are two tangents possible parallel to $y-x=0$