Director Circle of Hyperbola: Equation, Formula, Examples

A hyperbola is a conic section with a set of points in a plane such that the distance from the fixed points are constant. The tangent of the hyperbola is a straight line touching the hyperbola at only one point without passing through it. This concept of tangent is used in director circles. we use the director circle to determine important properties of the hyperbola.

This article is about the director circle of the hyperbola wihich falls under the topic Two Dimensional Analytical Geometry.

What is a Director Circle of Hyperbola?

The director circle of a hyperbola is the locus of the point of intersection of the perpendicular tangents of the hyperbola.

Equation of the Director Circle of Hyperbola

The equation of the director circle of the hyperbola with centre as origin $(0,0)$ is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2-b^2$.

When the centre of the hyperbola is not at the origin but at $(h, k)$, then the equation becomes $(x-h)^2+(y-k)^2=a^2-b^2$
where $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively.

Case 1: For $\mathrm{a}>\mathrm{b}, \mathrm{e}<\sqrt{2}$, then the director circle of the hyperbola is real
Case 2: For $a<b, e>\sqrt{2}$, the radius of the circle is imaginary. In this case, there should not be any circle and no tangents at right angles can be drawn to the circles.
Case 3: For $a=b$, we have a director circle as $x^2+y^2=0$, which represents the point $(0,0)$. So, in this case, the centre is the only point from where we can draw a tangent at the right angle to the hyperbola.

Derivation of Equation

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 a quadratic equation in m, slope of two tangents are $m_1$ and $m_2$

$
\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} \\
& x^2+y^2=a^2-b^2
\end{aligned}
$

Solved Examples

Example 1: Find the equation of the diameter of hyperbola $16 x^2-9 y^2=144$, which is conjugate to the diameter whose equation is $x=2 y$.

Solution:
Let the equation of the diameter, which is conjugated to $x=2 y$ be $y=m_1 x$
As we know two diameters $y=m_1 x$ and $y=m_2 x$ are conjugates, if

$
\begin{aligned}
& m_1 m_2=\frac{b^2}{a^2} \\
& m_1 \times \frac{1}{2}=\frac{16}{9} \\
& m_1=\frac{32}{9}
\end{aligned}
$
$
\begin{aligned}
& \frac{r_1}{r_2}=\frac{1}{2} \\
& \Rightarrow \quad \frac{a \sqrt{2-e_1^2}}{a \sqrt{2-e_2^2}}=\frac{1}{2} \\
& \Rightarrow \quad \frac{2-e_1^2}{2-e_2^2}=\frac{1}{4} \\
& \Rightarrow \quad 8-4 e_1^2=2-e_2^2 \\
& \Rightarrow \quad 4 e_1^2-e_2^2=6=\lambda \\
& \lambda=6
\end{aligned}
$

Hence, the equation of the conjugate diameters is $y=\frac{32}{9} x$

Example 2: If radii of director circles of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2}-\frac{y^2}{(b)^2}=1$ are 2 r and r respectively and $\mathrm{e}_{\mathrm{e}}$ and $\mathrm{e}_{\mathrm{h}}$ be the eccentricities of the ellipse and the hyperbola respectively then:

Solution:
Eccentricity - $
e=\sqrt{1-\frac{b^2}{a^2}}
$
For the ellipse

$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$
Equation of director circles of ellipse and hyperbola are respectively.

$
\begin{aligned}
& a^2+b^2=4 r^2 \ldots \ldots(i) \\
& a^2+b^2=r^2 \ldots \ldots(i i) \\
& a^2=\frac{5 r^2}{2}, b^2=\frac{3 r^2}{2} \\
& e_e^2=1-\frac{b^2}{a^2} \\
& \Rightarrow e_e^2=1-\frac{3 r^2}{2} \times \frac{2}{5 r^2}=1-\frac{3}{5}=\frac{2}{5}
\end{aligned}
$
$
\begin{aligned}
& e_h^2=1+\frac{b^2}{a^2} \\
& e_h^2=1+\frac{3}{5}=\frac{8}{5} \\
& \text { so }^{4 e_h^2-e_e^2}=4 \times \frac{8}{5}-\frac{2}{5}=\frac{30}{5}=6
\end{aligned}
$
Hence, the answer is $6$

Example 3: If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(\mathrm{a}>\mathrm{b})$ and $x^2-y^2=c^2$ cut at right angles, then

Solution:

$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad \Rightarrow \frac{d y}{d x}=-\frac{b^2 x}{a^2 y}
$

and, $x^2-y^2=c^2 \Rightarrow \frac{d y}{d x}=\frac{x}{y}$
The two curves will be cut at right angles if

$
\begin{aligned}
& \left(\frac{d y}{d x}\right)_{c_1} \times\left(\frac{d y}{d x}\right)_{c_2}=-1 \\
& \Rightarrow \quad-\frac{b^2 x}{a^2 y} \cdot \frac{x}{y}=-1 \quad \Rightarrow \quad \frac{x^2}{a^2}=\frac{y^2}{b^2} \\
& \Rightarrow \quad \frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{1}{2} \quad\left[u \operatorname{sing} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\right]
\end{aligned}
$

Substituting these values in $\mathrm{x}^2-\mathrm{y}^2=\mathrm{c}^2$, we get

$
\begin{aligned}
& \frac{a^2}{2}-\frac{b^2}{2}=c^2 \\
\Rightarrow & a^2-b^2=2 c^2
\end{aligned}
$
Hence, the answer is $\mathrm{a}^2-\mathrm{b}^2=2 \mathrm{c}^2$

Example 4: If the line $\mathrm{Ix}+\mathrm{my}+\mathrm{n}=0$ passes through the extremities of a pair of conjugate diameters of the hyperbola $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ then

Solution:
The extremities of a pair of conjugate diameters of $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad$ are $(\mathrm{a} \sec \phi, \mathrm{b} \tan \phi)$ and $(\mathrm{a} \tan \phi, \mathrm{b} \sec \phi)$ respectively.
According to the question, since the extremities of a pair of conjugate diameters lie on $\mathrm{Ix}+\mathrm{my}+\mathrm{n}=0$

$
\therefore \quad \mathrm{I}(\mathrm{a} \sec \phi)+\mathrm{m}(\mathrm{b} \tan \phi)+\mathrm{n}=0 \Rightarrow \mathrm{I}(\mathrm{a} \tan \phi)+\mathrm{m}(\mathrm{b} \sec \phi)+\mathrm{n}=0
$
Then from (i) al sec $\phi+\mathrm{bm} \tan \phi=-\mathrm{n}$ or $\mathrm{a}^2 \mathrm{l}^2 \sec ^2 \phi+\mathrm{b}^2 \mathrm{~m}^2 \tan ^2 \phi+2 \mathrm{ablmsec} \phi \tan \phi=\mathrm{n}^2$ $\qquad$
And from (ii), al $\tan \phi+\mathrm{bmsec} \phi=-\mathrm{n}$ or $\mathrm{a}^2 \mathrm{I}^2 \tan ^2 \phi+\mathrm{b}^2 \mathrm{~m}^2 \sec ^2 \phi+2 \mathrm{ablm} \sec \phi \tan \phi=\mathrm{n}^2$ $\qquad$

$\therefore \quad \mathrm{a}^2 \mathrm{I}^2\left(\sec ^2 \phi-\tan ^2 \phi\right)+\mathrm{b}^2 \mathrm{~m}^2\left(\tan ^2 \phi-\sec ^2 \phi\right)=0 \text { or } \quad \mathrm{a}^2 \mathrm{I}^2-\mathrm{b}^2 \mathrm{~m}^2=0
$
Hence, the answer is $\mathrm{a}^2 \mathrm{l}^2-\mathrm{b}^2 \mathrm{~m}^2=0$