Question

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

• A. ${(x^2-y^2)}^2=6x^2+2y^2$
• B. ${(x^2-y^2)}^2=6x^2-2y^2$
• C. ${(x^2+y^2)}^2=6x^2+2y^2$
• D. ${(x^2+y^2)}^2=6x^2-2y^2$

Answer 1

The correct option is C ${(x^2+y^2)}^2=6x^2+2y^2$

$x^2+3y^2=6$

$\frac{x^2}{6}+\frac{y^2}{2}=1$ $(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1)$

equation of a tangent to the ellipse

$y=mx+\sqrt{a^2{m}^2+b^2}$

for given ellipse

$y=mx+\sqrt{6m^2+2}$ ---(1)

equation of the line through $(0,0)$ and perpendiculars to the above line

$y-0=\left(\frac{-1}{m}\right)x-0$

$m=-\frac{x}{y}$

Substituting slope in equation ----(1)

$y={\left(\frac{-x}{y}\right)}^x\sqrt{6\left(\frac{x^2}{y^2}\right)+2}$

$y^2=-x^2+\sqrt{6x^2+2y^2}$

$(x^2+y^2{)}^2=6x^2+2y^2$