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The graphs of two identical ellipses with foci on the $x$-axis are such that each ellipse passes through the center of the other. The equation of the first ellipse is $4x^2+9y^2=36$. The second ellipse is directly to the right of the first. Find the equation of the second ellipse in standard form, and find their points of intersection.

My work:

The equation in standard form for the first ellipse is $\dfrac{x^2}9+\dfrac{y^2}4=1$. Since the ellipses are both on the x-axis, the second ellipse should be identical to the original ellipse and shifted $\sqrt 9=3$ units to the right. Then, the equation for the second ellipse would be $\dfrac{(x-3)^2}9+\dfrac{y^2}4=1$. Subtracting the second equation from the first, I get $x^2=x^2-6x+9$ and $x=\dfrac 23$. I feel like I have done something wrong here because I get different values for $y$ when substituting that value into the two equations. What did I do wrong?

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    Wrong way for the $x$-coordinates of the intersections: it's $3/2$ and not $2/3$! – 2017-02-03
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    And another trivial mistake by me. Thanks for spotting the mistake @ParclyTaxel – 2017-02-03
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    Now I get that $y=\sqrt 3$. Is this correct? – 2017-02-03
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    More properly $y=\pm\sqrt3$ (there are two intersections), but that is correct. – 2017-02-03

1 Answers 1

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Your process is right, but I think you have made a calculation mistake. Basically, we need to find $x $ from $$\frac {x^2}{9}+\frac {y^2}{4} = \frac {(x-3)^2}{9} + \frac {y^2}{4} $$ $$\Rightarrow \frac {x^2}{9} = \frac {(x-3)^2}{9} $$ $$\Rightarrow x = \pm (x-3)$$ giving us $x =\frac {3}{2} $ as an acceptable solution.

Now substituting this value to get $y $, we get $y = \pm \sqrt {3} $.

Thus the equation of the second ellipse is $\frac {(x-3)^2}{9} + \frac {y^2}{4} =1$ and the intersection points are $(\frac {3}{2}, \pm \sqrt {3}) $.

Hope it helps.