Q. The distance of the centre of ellipse $x^{2} + 2y^{2} - 2 = 0$ to those tangents of the ellipse which are equally inclined from both the axes, is
A) $\frac{3}{\sqrt{2}}$
B) $\sqrt{\frac{3}{2}}$
C) $\frac{\sqrt{2}}{3}$
D) $\frac{\sqrt{3}}{2}$
Explanation
(d)
Equation of ellipse is $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1.$ General equation of tangent to the ellipse of slope $m$ is
$y = mx \pm \sqrt{2m^{2} + 1}$
Since, this is equally inclined to axes, so $m = \pm 1$. Thus, tangents are
$y = \pm x \pm \sqrt{2 + 1} = \pm x \pm \sqrt{3}$
Distance of any tangent from origin
$= \frac{|0 + 0 \pm \sqrt{3}|}{\sqrt{1^{2} + 1^{2}}} = \frac{\sqrt{3}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.