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 '$a$'.
Asked
Active
Viewed 1,226 times
-1
-
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
1 Answers
0
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
- 8,115