Question

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

Answer 1

Slope of line OP$=\frac{k}{h}$

$\therefore$ Slope of tangent$=\frac{-h}{k}$

Equation of tangent,

$(y-k)=\frac{-h}{k}(x-h)$

$y=\frac{-hx}{k}+\frac{h^2}{k}+k$ ………$\left(1\right)$

Now $\frac{x^2}{d^2}+\frac{y^2}{b^2}=1$

Equation of tangent is

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

$y+\frac{hx}{k}=\pm \sqrt{6m^2+2}$

$\frac{h^2+k^2}{k}=\pm \sqrt{6m^2+2}$ ………..[from $\left(1\right)$]

Squaring,

$\frac{(h^2+k^2{)}^2}{k^2}=6{\left(\frac{-h}{k}\right)}^2+2$

$\therefore (h^2+k^2{)}^2=6h^2+2k^2$

Thus equation of locus is

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

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

$\frac{x^2}{6}+\frac{y^2}{2}=1$

$\therefore a^2=6$

$b^2=2$

$m=-\frac{h}{k}$.