Answer:
The true anomaly q will be such that the q=[tex]cos^{-1}(e)[/tex] where e is the eccentricity of the ellipse.
Explanation:
For ellipse velocity components and velocity is given as
[tex]v_r=\frac{\mu}{h} e \, sin\theta\\v_p=\frac{h}{r}=\frac{h}{\frac{h^2}{\mu}\frac{1}{1+ecos\theta}}\\v_{eclipse}=\sqrt{v_r^2 +v_p^2}\\v_{eclipse}=\sqrt{(\frac{\mu}{h} e \, sin\theta)^2 +(\frac{h}{\frac{h^2}{\mu}\frac{1}{1+ecos\theta}})^2}\\\\v_{eclipse}=\sqrt{(\frac{\mu}{h})^2(e^2+1+2ecos\theta)}\\\\v_{eclipse}^2=(\frac{\mu}{h})^2(e^2+1+2ecos\theta)[/tex]
For similar radius velocity of circle is given as
[tex]\\v_{circle}^2=\frac{\mu}{r}\\v_{circle}^2=\frac{\mu}{\frac{h^2}{\mu}\frac{1}{1+ecos\theta}}\\v_{circle}^2=\frac{\mu^2}{h^2}(1+ecos\theta)[/tex]
Now as per the condition
[tex]v_{eclipse}^2=v_{circle}^2\\(\frac{\mu}{h})^2(e^2+1+2ecos\theta)=\frac{\mu^2}{h^2}(1+ecos\theta)\\(e^2+1+2ecos\theta)=(1+ecos\theta)\\ecos\theta=-e^2\\cos\theta=-e\\theta=cos^{-1}(e)[/tex]
So the true anomaly q will be such that the q=[tex]cos^{-1}(e)[/tex] where e is the eccentricity of the ellipse.
