Question

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

Answer 1

Having double contact means both curves will have same slope at that point means they will have same tangent and normal lines.

Normal to circle at that point will also be normal to ellipse.

Normal to ellipse

$\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$

at $x_1=ae$ and $y_1=\frac{b^2}{a}$ that is ends of latus

rectum this will also be normal to circle

By symentary we can say centre will be an x axis\

$\frac{ax}{e}-ay=a^2-b^2$

X coordinate

$=a{e}^3$

$\therefore$ radius of circle would be distance between $(ae,\frac{b^2}{a})$

$(ae{3}^2,0)$ $1-e^2=\frac{b^2}{a^2}$

$r=\sqrt{(a{e}^3-ae{)}^2}+\frac{b^4}{a^2}$ $a^2(1-e^2)=\frac{b^4}{a^2}$

$r^2=a^2(e^6-e^4-e^2)$

Equation of circle is

$(x-a{e}^3{)}^2+y^2=a^2(e^6-e^4-e^2)$

$x^2+y^2-2a{e}^3.x=a^2(1-e^2-e^4)$

$m=2$