# 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 $\dfrac {\sqrt {34}}{2}$.

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 $$x^2+y^2=10$$

WLOG the point$$(P)$$ be $$x=\sqrt8\cos(\pi/2- t),y=\sqrt{10}\sin(\pi/2- t)$$
Now we need $$34/4=(\sqrt8\sin t-0)^2+(\sqrt{10}\cos t-0)^2$$
• I believe that the eccentric angle is generally defined w/r the major axis, so don’t you need to take into account that for this ellipse, it’s the $y$-axis instead of the $x$-axis? – amd Jan 18 '20 at 22:30
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 $$\dfrac{y}{x} =\sqrt{\dfrac{c-1}{a=c}}$$.