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8 Questions around this concept.
If the radii of director circle of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b_1^2}=1$ are in the ratio $1: 2$ and $4 e_1^2-e_2^2=\lambda$ where $e_1$ and $e_2$ are eccentricities of ellipse and hyperbola respectively, then the value of $\lambda$ is
If e is the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a<b)$, then
Director Circle
The locus of the point through which perpendicular tangents are drawn to a given Hyperbola S = 0 is a circle called the director circle of the hyperbola.
The equation of the director circle of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2-b^2$
Proof:
Equation of tangent of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ in slope form is $y=m x+\sqrt{a^2 m^2-b^2}$
it passes through the point $(h, k)$
$
\begin{aligned}
& k=m h+\sqrt{a^2 m^2-b^2} \\
& (k-m h)^2=a^2 m^2-b^2 \\
& k^2+m^2 h^2-2 m h k=a^2 m^2-b^2 \\
& \left(h^2-a^2\right) m^2-2 h k m+k^2+b^2=0
\end{aligned}
$
This is quadratic equation in m , slope of two tangents are $m_1$ and $m_2$
$
\begin{aligned}
& \mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{k}^2+\mathrm{b}^2}{\mathrm{~h}^2-\mathrm{a}^2} \\
& -1=\frac{\mathrm{k}^2+\mathrm{b}^2}{\mathrm{~h}^2-\mathrm{a}^2} \\
& \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2-\mathrm{b}^2
\end{aligned}
$
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