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A circle is described upon $AA'$, the major axis of an ellipse as diameter.

P is a point lying on the circle.

Let $AP,A'P$ be joined cutting ellipse in $Q,Q'$

It is given that $\dfrac{AP}{AQ}+\dfrac{A'P}{A'Q}=3$

Find the eccentricity of the ellipse

I tried using parametric form of circle and solved equations of ellipse and PA',AP. It became very lengthy and I didn't get the answer too

Saby123
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Arjun Karthikeya K
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2 Answers2

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Quick and dirty: the question has a solution only if the equality $$ \dfrac{AP}{AQ}+\dfrac{A'P}{A'Q}=3 $$ holds for any $P$ on the circle. Choose then $P$ on the perpendicular bisector of $AA'$. If $O$ is the center of the circle and $H$ is the projection of $Q$ on $AA'$ we then get: $$ QH={2\over3}a,\quad OH={1\over3}a, $$ where I set $a=OA=OA'$. But we also have: $$ {OH^2\over a^2}+{QH^2\over b^2}=1, $$ where $b$ is the semi-minor axis. Plugging here the above results we get $$ {b^2\over a^2}={1\over2}, $$ whence $e^2=1/2$.

Intelligenti pauca
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Let $P = (a \cos t, a \sin t)$ be on the circle with radius $a$, and let the ellipse be:

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $ where $b \lt a$

The (parametric) equation of the line from $(a, 0)$ to $P$ is

$Q(s) = (a, 0) + s ( a (\cos t -1 ) , a \sin t) \\ = (a (1 + s (\cos t - 1)) , s a \sin t )$

$Q(s)$ is on the ellipse so

$(1 + s (\cos t - 1))^2 + (a/b)^2 s^2 \sin^2 t = 1 $

Expanding yields,

$2 s (\cos t - 1) + s^2 (\cos t - 1)^2 + a^2/b^2 s^2 \sin^2 t = 0$

divide by s^2, then

$2 (1/s) (\cos t - 1) + ( (\cos t - 1)^2 + (a/b)^2 \sin^2 t ) = 0 $

Hence,

$1/s_1 = 1/2 ( (\cos t - 1)^2 + (a/b)^2 \sin^2 t ) / (1 - \cos t ) $

repeating the same for the left vertex, gives

$1/s_2 = 1/2 ( (\cos t + 1)^2 + (a/b)^2 \sin^2 t ) / (1 + \cos t )$

Now, let $x = \cos t, y = \sin t$ then,

$3 = 1/s_1 + 1/s_2$

so that, by combining the two fractions, and noting that $1 - x^2 = y^2$, we get

$6 y^2 = (1 + x) ( (x - 1)^2 + (a/b)^2 y^2 ) + (1 - x) ( (x + 1)^2 + (a/b)^2 y^2 ) \\ = 2 x^2 + 2 + 2 (a/b)^2 y^2 + x ( - 4 x ) \\= -2 x^2 + 2 + 2 (a/b)^2 y^2 $

but $x^2 = 1- y^2$, so

$6 y^2 = -2 + 2 y^2 + 2 + 2(a/b)^2 y^2 = 2 ( 1 + (a/b)^2 ) y^2$

therefore,

$3 = 1 + (a/b)^2$

so that $(a/b)^2 = 2 \\(b/a)^2 = 1/2 \\1 - (b/a)^2 = 1/2 \\e = \dfrac{1}{\sqrt{2} } $

Hosam Hajeer
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