The locus of the foot of perpendicular drawn from the centre of the ellipse x^2+3y^2=6 on …

Question

The locus of the foot of perpendicular drawn from the centre of the ellipse $x^2+3y^2=6$ on any tangent to it is:

Accepted Answer

Write the ellipse as $\dfrac{x^2}{6}+\dfrac{y^2}{2}=1$, so $a^2=6,\ b^2=2$. The standard locus of the foot of the perpendicular from the centre onto a variable tangent $y=mx\pm\sqrt{a^2m^2+b^2}$ is
$(x^2+y^2)^2=a^2x^2+b^2y^2=6x^2+2y^2.$

Solutions · 1