Question

A comet is moving in a parabolic orbit around the sun which is at the focus of a parabola. When the comet is $80$ million kms from the sun, the line segment from the sun to the comet makes an angle of $\frac{\pi }{3}$ radians with the axis of the orbit. Find
(i) the equation of the comet's orbit
(ii) how close does the comet come nearer to the sun? (Take the orbit as open rightward).

Answer 1

Take the parabolics orbit as open rightward and the vertex at the origin.
Let $P$ be the position of the comet in which $FP=90$ million kms
In $\Delta FQF,\sin \frac{\pi }{3}=\frac{PQ}{PF}$
$\Rightarrow PQ=80\times \frac{\sqrt{3}}{2}=40\sqrt{3}$ million kms
$\Rightarrow \cos \frac{\pi }{3}=\frac{FQ}{PF}\Rightarrow FQ=80\times \frac{1}{2}=40$ million kms
$\therefore VQ=a+40$ if $VF=a$ also $P$ is $(VQ,PQ)=(a+40,40$ $\sqrt{3}$)
Since $P$ lie on the parabola $y^2=4ax$
$\Rightarrow (40\sqrt{3}{)}^2=4a(a+40)$
$\Rightarrow a^2+20a-1200=0$
$\Rightarrow (a+60)(a-20)=0$
$\Rightarrow a=20$ ....Since $a\ne -60$
$\therefore$ the equation of the orbit is $y^2=4ax$.
$\therefore y^2=80x$
The shortest distance between the sun and the cornet is $20$ million kms.