Find the eccentric angle of a point on the ellipse $\dfrac {x^2}{4}+\dfrac {y^2}{5}=2$ whose distance from the center is $\dfrac {\sqrt {34}}{2}$.

Question

Find the eccentric angle of a point on the ellipse $\dfrac {x^2}{4}+\dfrac {y^2}{5}=2$ whose distance from the center is $\dfrac {\sqrt {34}}{2}$.

My Attempt: The equation of ellipse is $$\dfrac {x^2}{4}+\dfrac {y^2}{5}=2$$ $$\dfrac {x^2}{8}+\dfrac {y^2}{10}=1$$ Length of major axis is $2b=2\sqrt {10}$ So, the semi major axis is of length $\sqrt {10}$ Now the equation of auxillary circle is $$x^2+y^2=10$$

Answer 1

WLOG the point$(P)$ be $x=\sqrt8\cos(\pi/2- t),y=\sqrt{10}\sin(\pi/2- t)$

Now we need $$34/4=(\sqrt8\sin t-0)^2+(\sqrt{10}\cos t-0)^2$$

Answer 2

If $x^2+ay^2 = c$ and $x^2+y^2 = 1$ then $(a-1)y^2 = c-1 $ so $y^2 =\dfrac{c-1}{a-1} $, $x^2 =1-y^2 =1-\dfrac{c-1}{a-1} =\dfrac{a-c}{a-1} $ so $\dfrac{y^2}{x^2} =\dfrac{c-1}{a=c} $ and $\dfrac{y}{x} =\sqrt{\dfrac{c-1}{a=c}} $.