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Let $P( \alpha, \beta)$ be a point on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ where foci are $F_1$ and $F_2$. Then $(PF_1 - PF_2)^2= ?$

I went for a brute for approach, the focii are located at $(\mp ae,0)$ so the distance of point from the general focus is given as:

$$ d = \sqrt{ (\alpha \pm ae )^2 + \beta^2} = \sqrt{\alpha^2 +(ae)^2 \pm 2 \alpha a e + \beta^2}$$

Now, I use the parameterization $ \alpha = a \cos \theta $ and $ \beta = b \sin \theta$, this leads to :

$$ \sqrt{a^2 + b^2 + (ae)^2 \pm 2 a^2 \cos \theta e} \ \ \color{red} *$$

Further I use the property that $(ae)^2 = a^2 -b^2$,

$$ \sqrt{2a^2 \pm 2a^2 \cos \theta e} =a \sqrt{2} \sqrt{1 \pm e \cos \theta}$$

Substituting $ \cos \theta = \frac{\alpha}{a}$

$$ a \sqrt{2} \sqrt{ 1 \pm e \cos \theta} = \sqrt{2a} \sqrt{a \pm e \alpha}$$

I'm not quite sure how to proceed anymore...

In the book it was given that the distance of point from focus is given $(a \pm e \alpha)$, did I make mistake? How do I correctly proceed?

tryst with freedom
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1 Answers1

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After the parameterization, the expression becomes $$\sqrt{a^2 \cos^2 \theta +(a^2-a^2 e^2) \sin^2 \theta +(ae)^2 \pm 2ea^2 \cos \theta} \\ = \sqrt{a^2 +a^2e^2 \cos^2 \theta \pm 2ea^2 \cos \theta } \\ = a\sqrt{1 +(e\cos \theta)^2 \pm 2e\cos\theta} \\ = a(1\pm e\cos\theta) \\ = a \pm e\alpha$$

Vishu
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