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A normal is drawn to the ellipse $\frac{x^2}{(a^2+2a+2)^2}+\frac{y^2}{(a^2+1)^2}=1$. If maximum radius of the circle centered at the origin and touching the normal is $5$, then find the possible values of '$a$'.

Ris97
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  • So you have $$\frac {x^2}{((a+1)^2+1)^2}+\frac {y^2}{(a^2+1)^2}=1$$ correct? What have you tried so far? – abiessu Dec 05 '13 at 15:43
  • I don't understand the question at all. How can a circle touch a normal of an ellipse? Should the normal be tangent to the circle? – Michael Hoppe Dec 05 '13 at 16:00
  • Try to use http://www.emathzone.com/tutorials/math-results-and-formulas/equations-of-tangent-and-normal-to-a-ellipse.html and the parametric form of http://planetmath.org/equationoftangentofcircle – lab bhattacharjee Dec 05 '13 at 18:41

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Hints:

This ellipse has valid values of $a$ in the positive and negative real numbers.

The equation of the circle centered at the origin and having maximum radius $5$ is where $r\in[0,5]$ and

$$x^2+y^2=r^2\implies \frac {x^2}{r^2}+\frac{y^2}{r^2}=1$$

Which direction does a "normal to an ellipse" travel in, relative to the center of the ellipse?

abiessu
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