Question

A conic $C$ satisfies the differential equation $(1+y^2)dx-xydy=0$ and passes through the point $(1,0)$ An ellipse $E$ which is confocal with $C$ having its eccentricity equal to $\sqrt{2/3}$
(a) Find the length of the latus rectum of the conic $C$
(b) Find the equation of the ellipse $E$
(c) Find the locus of the point of intersection of the perpendicular tangents to the ellipse $E$.

Answer 1

Given a differential equation of a conic.

$(1+y^2)dx=xydx\Rightarrow \frac{dx}{x}=\frac{ydy}{1+y^2}$

Integrating both the sides we get

$\Rightarrow \ln x=\frac{1}{2}\ln (1+y^2)+\ln C$ where $\ln C$ is an integration constant.

$\Rightarrow 2\ln x=\ln (1+y^2)+2\ln C\Rightarrow \ln x^2=\ln (1+y^2)+\ln C^{{}^{\prime }}$ where $C^{{}^{\prime }}=C^2$

$\Rightarrow x^2=C^{{}^{\prime }}(1+y^2)$

Putting $x=1,y=0$ we get $C^{{}^{\prime }}=1$

$\therefore$ equation of $C$ is $x^2=1+y^2\Rightarrow x^2-y^2=1$ (hyperbola)

eccentricity of the above conic $r=e=\sqrt{1+\frac{1}{1}}=\sqrt{2}$

coordinates of the focus is $(+a,e,0)=(\pm \sqrt{2},0)$

$a)$ Length of the lactus rectum $=\frac{2b^2}{a}=2$units

$b)$ Given is an ellipse $E$ confocal with $C$ and eccentricity is $\sqrt{\frac{2}{3}}$

The distance between the two focus in an ellipse is $2ae=\left(2\sqrt{2}\right)$units.

$\therefore a=\sqrt{3}{e}^2=1-\frac{b^2}{a^2}=1-\frac{b^2}{3}\Rightarrow \frac{2}{3}=1-\frac{b^2}{3}\Rightarrow b^2=1\Rightarrow b=1$

$\therefore$ the equation of this ellipse $E$ is $\frac{x^2}{3}+\frac{4^2}{1}=1$

$c)$ The locus of the point of interaction of the perpendicular tangent to the ellipse is given by a circle

$x^2+y^2=a^2+b^2$ ($a,b$ are the major & minor axis of the ellipse)

$\therefore x^2+y^2=3+1=4$

Hence the required locus is $x^2+y^2=4$