A comet describe a parabola about the sun, show that the sum of the squares of the velocities at the extremities of a focal chord is constant.
I have no idea how to solve. Please help.
I only know that if the velocity of the comet at a point on the parabola be $v$ then $v^2=\frac{2\mu}{r}$, $\mu$ is any positive constant.
Let the mass of the comet be $m$ and its angular momentum in describing the parabola of latus rectum $4a$, be $L$. Since it is under the influence of a central force, its angular momentum is constant. Let the velocities at points $P$ and $Q$ be $v_1$ and $v_2$ ; the solar radii be $d_1$ and $d_2$; and the angles that the velocities make with the solar radii be $\theta_1$ and $\theta _2$ respectively.
Looking at the figure, we can make out 2 things :
$\theta_2=90^\circ-\theta_1$
$d_1\sin\theta_1\tan\theta_1=d_2\sin\theta_2\implies d_2=d_1\tan^2\theta_1$
Also, due to some property of parabola, $\dfrac{d_1d_2}{d_1+d_2}=a\implies a=d_1\sin^2\theta_1$
Now if we add up the velocities' squares: $$V=v_1^2+v_2^2=\left(\frac{L}{md_1\sin\theta_1}\right)^2+\left(\frac{L}{md_2\sin\theta_2}\right)^2=\frac{L^2}{m^2d_1^2\sin^2\theta_1}+\frac{L^2}{m^2d_2^2\sin^2\theta_2}$$ $$V=\frac{L^2}{m^2}\left( \frac{1}{d_1^2\sin^2\theta_1}+\frac{1}{d_2^2\cos^2\theta_1} \right)=\frac{L^2}{m^2}\left( \frac{1}{d_1^2\sin^2\theta_1}+\frac{1}{d_1^2\sin^2\theta_1\tan^2\theta_1} \right)$$ $$V=\frac{L^2}{m^2d_1^2\sin^2\theta_1}\left(1+\cot^2\theta_1\right)=\frac{L^2}{m^2d_1^2\sin^4\theta_1}=\frac{L^2}{a^2m^2}$$
Thus, we have shown that the sum of the squares of the velocities at the endpoints of a focal chord is equal to the square of the quotient of the angular momentum and the product of the mass and the semi-semi-latus rectum. Which is a constant.