Tag: auxiliary circle, director circle

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

For a standard hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the equation of director circle is given by:

$\begin{array}{l} equation\, \, of\, \, director\, \, circle\, \, { x^{ 2 } }+{ y^{ 2 } }={ a^{ 2 } }+4ab-\left( { 2ab-{ b^{ 2 } } } \right)  \ \Rightarrow { x^{ 2 } }+{ y^{ 2 } }={ a^{ 2 } }+{ b^{ 2 } }+2ab \ \Rightarrow { x^{ 2 } }+{ y^{ 2 } }={ \left( { a+b } \right) ^{ 2 } } \ Then,\, \, Radians\, \, of\, \, circle\, \, is\left( { a+b } \right)  \end{array}$

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

$P(a sec(t) , a tan(t))$ and $P'(a sec(t) , 0)$
$|PP'| = a^2 tan^2(t)$
The power of point P' relative to a circle $x^2 + y^2 = a^2$ is :

$(asec(t) - 0)^2 + (0-0)^2 - a^2$   (power of the point w.r.t. circle)

$= a^2sec^2(t) - a^2 = a^2(sec^2(t)-1) = a^2tan^2(t)$

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be the pole of the line

$lx+my+n=0$ with respect ot the hyperbola $\cfrac { { x }^{ 2 } }{ {

a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
Then the equation of the polar is
$\cfrac

{ { xx } _{ 1 } }{ { a }^{ 2 } } -\cfrac { { yy } _{ 1 } }{ { b }^{ 2 } }

=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Since $\left( { x } _{ 1 },{ y } _{ 1 } \right) $  is the pole of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Clearly $(i)$ and $(ii)$ represent the same line. Therefore,
$\therefore \quad \cfrac { { x } _{ 1 } }{ { a }^{ 2 }l } =\cfrac { { -y } _{ 1 } }{ { b }^{ 2 }m } =\cfrac { 1 }{ -n } $
${ x } _{ 1 }=\cfrac { { -a }^{ 2 }l }{ n } ,\quad { y } _{ 1 }=\cfrac { { b }^{ 2 }m }{ n } $
Hence the pole of the given line with respect of the given hyperbola is
$\left(- \cfrac { { a }^{ 2 }l }{ n } ,\cfrac { { b }^{ 2 }m }{ n }  \right) \quad $

The tangent equation to the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is 
$y=mx \pm \sqrt{a^2m^2-b^2}$
$\Rightarrow (y-mx)=\pm \sqrt{a^2m^2-b^2}$
On squaring both sides, we get
$\Rightarrow (y-mx)^2=(a^2m^2-b^2)$
$\Rightarrow (x^2-a^2)m^2-2xym+(y^2+b^2)=0$
Product of the slopes,$m _1m _2=\dfrac{(y^2+b^2)}{x^2-a^2}$
But given tangents are perpendicular to each other $\Rightarrow$ Their product of slopes equal to $-1.$
$\Rightarrow m _1m _2=-1$
$\Rightarrow \dfrac{(y^2+b^2)}{x^2-a^2}=-1$
$\Rightarrow x^2+y^2=(a^2-b^2)$
But given hyperbola equation as $\dfrac{x^2}{\cos\alpha^2}-\dfrac{y^2}{\sin\alpha^2}=1$
The required tangent equation is $x^2+y^2=\cos^2\alpha-\sin^2\alpha=\cos 2\alpha$
Since radius of circle is always greater than equal to zero.
$\Rightarrow \cos 2\alpha \geq 0$
But maximum value of $\cos$ is $1$.
$\Rightarrow 0 \leq \cos2\alpha \leq 1$
$\Rightarrow  \dfrac{\pi}{2} \leq 2\alpha \leq 0$
$\Rightarrow  \dfrac{\pi}{4} \leq \alpha \leq 0$
$\Rightarrow \alpha$has inifinite number of solutions.

Equation of any tangent in terms of slope $m$ is

$y = mx + (a^2m^2  b^2)$

It passes through $(h, k)$, so we have

$(k - mh)^2 = a^2m^2 - b^2$

So, $m^2(h^2 - a^2) - 2mhk + k^2 + b^2 = 0$

This is a quadratic in $m$

Let the slopes of tangents be $m _1$ and $m _2$.

then $m _1.m _2 = -1$.

So, $\dfrac{(k^2 + b^2)}{(h^2 - a^2)} = -1$

$(h^2 + k^2) = (a^2  b^2)$

Hence, the locus is $(x^2 + y^2) = (a^2  b^2)$ which is the director circle of $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1.$