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I have an ellipse that is $$4x^2 + 9y^2 = 180 \Rightarrow \frac{x^2}{45} + \frac{y^2}{20} = 1$$ I need to find the points that are $2$ times closer to left focal point than right focal point. Let's say that $A$ is the unknown point, $A(x,y)$.

First, I wanted to find the two focal points. From the ellipse equation I could find that $$a^2 = 45, \quad b^2 = 20$$

And therefore $$a^2 - b^2 = 45-20 = 25 = c^2$$ The focal points are $$F_1(-5,0) \qquad F_2(5,0)$$

For the exercise I needed to do $$d_{AF_1} = \sqrt{\left(x-5\right)^2 + y^2}$$ $$d_{AF_2} = \sqrt{\left(x+5\right)^2 + y^2}$$ $$2\sqrt{\left(x+5\right)^2 + y^2} = \sqrt{\left(x-5\right)^2 + y^2}$$ And I get to $$3x^2+50x+75 + 3y^2 = 0$$ From the ellipse I get $$y^2 = \frac{180-4x^2}{9}$$

When I plug in the $y^2$ I get one more solution (according to WolframAlpha). Before I had $(-3,4)$ and $(-3,-4)$.

Now, when I plug it in

$$3x^2 + 50x+75+3\left(\frac{180-4x^2}{9}\right) = 0$$

I get according to WA, another solution: $x=-27$.

  1. Why do I get an additional solution when I plug in the $y^2$?

  2. How can I know if it's only $x=-3$, $x=-27$ or both solutions?

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    $x= -27$ does satisfy your equation, but it can't satisfy $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. – 2017-02-05

1 Answers 1

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Somewhere you seem to have introduced a new root: $x = -27$ does satisfy your last equation, which is equivalent to $$5x^2 + 150x + 405 = 0$$but it does not satisfy $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Where do you think this new root comes from?