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Prove that the equation of the circle ,having double contact with the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$(having eccentricity $e$) at the ends of a latus rectum,is $x^2+y^2-2ae^3x=a^2(1-e^2-e^4)$.

Since the ends of a latus rectum are $(ae,\frac{b^2}{a}),(ae,\frac{-b^2}{a})$.We have to find the equation of circle passing through $(ae,\frac{b^2}{a})$ and ,$(ae,\frac{-b^2}{a})$.I cannot find the equation of circle based on this equation.Any hints/suggestions will help me.Thanks.

diya
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HINT... you could find the equation of the normal to the ellipse at $(ae,\frac{b^2}{a})$ and find the point of intersection of this normal and the $x$ axis, which, due to symmetry, would be the centre of the circle

David Quinn
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  • I got the answer using this concept,but i could not understand why the intersection of the normal to the ellipse at $(ae,\frac{b^2}{a})$ would be the center of the circle? – diya Aug 27 '15 at 09:31
  • did you draw a picture? – David Quinn Aug 27 '15 at 09:43
  • I drew,sir,but i could not figure out. – diya Aug 27 '15 at 15:06
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    The normal to the ellipse is perpendicular to the tangent of the ellipse which is the same line as the tangent to the circle. The normal to the ellipse is therefore also the normal to the circle, in other words a radius of the circle. Due to symmetry, the $x$ axis is a diameter of the circle. – David Quinn Aug 27 '15 at 15:35
  • If any two curves have "double contact",then those two curves will have same tangent and same normal at the point of contact.Is it the basic definition of "double contact"? – diya Aug 28 '15 at 07:05
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    Two curves which touch share the same tangent at the point of contact, and as a consequence they share the same normal – David Quinn Aug 28 '15 at 09:25
  • I got it,thank you,sir. – diya Aug 28 '15 at 10:01