Ellipse
12.0 Director circle
12.0 Director circle
The locus of the point of intersection of the tangents to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ which are perpendicular to each other is called director circle.
Proof: Let us assume any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ in terms of slope i.e., $$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $$
As shown in figure $17$, point $P(h,k)$ satisfies the equation of tangent.
$$\begin{equation} \begin{aligned} k = mh \pm \sqrt {{a^2}{m^2} + {b^2}} \\ k - mh = \pm \sqrt {{a^2}{m^2} + {b^2}} \\\end{aligned} \end{equation} $$
On squaring both sides, we get
$$\begin{equation} \begin{aligned} {(k - mh)^2} = {a^2}{m^2} + {b^2} \\ {k^2} + {m^2}{h^2} - 2mhk = {a^2}{m^2} + {b^2} \\ {m^2}({h^2} - {a^2}) - 2hkm + {k^2} - {b^2} = 0 \\\end{aligned} \end{equation} $$
which is a quadratic equation in $m$. Let slope of two tangents be ${m_1}$ and ${m_2}$.
$$\begin{equation} \begin{aligned} {m_1}{m_2} = \frac{{{k^2} - {b^2}}}{{{h^2} - {a^2}}} \\ - 1 = \frac{{{k^2} - {b^2}}}{{{h^2} - {a^2}}} \\ - {h^2} + {a^2} = {k^2} - {b^2} \\ {h^2} + {k^2} = {a^2} + {b^2} \\\end{aligned} \end{equation} $$
Hence, locus of point $P(h,k)$ is $${x^2} + {y^2} = {a^2} + {b^2}$$
which is the director circle of ellipse.
