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

pi-π
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2 Answers2

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

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

lab bhattacharjee
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  • 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
  • @amd, please find the updated answer – lab bhattacharjee Jan 19 '20 at 04:09
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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}} $.

marty cohen
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